id string | solution string | answer string | metadata dict | problem string | candidates list | relevance_scores list | relevance_scores_full null |
|---|---|---|---|---|---|---|---|
ours_27905 | Let the distance between \( A \) and \( B \) be \( s \, \text{km} \) and the original speed of the train be \( v \, \text{km/hr} \). During the 6 hours before the halt, the first train covered \( 6v \, \text{km} \). The remaining distance of \( (s - 6v) \, \text{km} \) was covered at a speed of \( 1.2v \, \text{km/hr} ... | 600 \, \text{km} | {
"competition": "misc",
"dataset": "Ours",
"posts": null,
"source": "Problems in Elementary Mathematics - group_21.md"
} | A train left a station \( A \) for \( B \) at 13:00. At 19:00 the train was brought to a halt by a snow drift. Two hours later the railway line was cleared, and to make up for the lost time, the train proceeded at a speed exceeding the original speed by \( 20\% \) and arrived at \( B \) only one hour later. The next da... | [
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numina_10042778 | 3. The trains are moving in the same direction, so they can move in the direction of $A B$ or $B A$. Let's consider each of these cases.
The trains are moving in the direction of $A B$.
1) $40 \times 8=320$ km - the first train traveled;
2) $48 \times 8=384$ km - the second train traveled;
3) $384-320=64$ km - by thi... | 956 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 3. Two trains leave from two cities at the same time. The first one travels at 40 km/h, while the second one travels at 48 km/h. How far apart will these trains be from each other after 8 hours, if they are moving in the same direction and the distance between the cities is 892 km? | [
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aops_1184997 | [quote=arqady]
Hence, it remains to prove that $(a+b+c)^4\geq8\sum_{cyc}(a^3b+a^3c)$, which is obvious.[/quote]
Let $x=a^2+b^2+c^2, \ y=ab+bc+ca$.
$(a+b+c)^4=(x+2y)^2\ge 8xy=8\sum_{cyc}(a^3b+a^3c+a^2bc)\ge 8\sum_{cyc}(a^3b+a^3c)$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Given $a,b,c\\geq 0$, prove that $$\\sum_{cyc}^{}\\sqrt[3]{\\frac{a}{b+c}}\\geq 2$$",
"content_html": "Given <span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemsolving.com/3/0/f/30fa6... | Given \(a,b,c\ge 0\), prove that
\[
\sum_{\text{cyc}}\sqrt[3]{\frac{a}{b+c}}\ge 2.
\] | [
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numina_10734145 | Proof: Let \( A=\frac{1}{1+a^{4}}, B=\frac{1}{1+b^{4}}, C=\frac{1}{1+c^{4}}, D=\frac{1}{1+d^{4}} \), then \( a^{4}=\frac{1-A}{A}, b^{4}=\frac{1-B}{B}, c^{4}=\frac{1-C}{C}, d^{4}=\frac{1-D}{D} \).
Using the Arithmetic Mean-Geometric Mean Inequality, we get
\[
\begin{aligned}
& (B+C+D)(C+D+A)(D+A+B)(A+B+C) \\
\geqslant ... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "inequalities"
} | Example 7 Let $a, b, c, d>0$. When $\frac{1}{1+a^{4}}+\frac{1}{1+b^{4}}+\frac{1}{1+c^{4}}+\frac{1}{1+d^{4}}=1$, prove: $a b c d \geqslant 3$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | [
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numina_10117640 | Because $0<x_{k} \leqslant x_{n}<1 \quad(k=1,2, \cdots, n)$
So $0<\frac{1-x_{n}}{1-x_{k}} \leqslant 1 \quad \cdot(k-n$ when taking “=” sign $)$
So $\left(1-x_{n}\right)^{2} \frac{x_{k}^{k}}{\left(1-x_{k}^{k} 1\right)^{2}}=\frac{\left(1-x_{n}\right)^{2}}{\left(1-x_{k}\right)^{2}} \cdot \frac{x_{k}^{k}}{\left(1+x_{k}... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | II. (50 points)
Let $0<x_{1}<x_{2}<\cdots<x_{n}<1$, prove that:
$$
\left(1-x_{n}\right)^{2}\left[\frac{x_{1}}{\left(1-x_{1}^{2}\right)^{2}}+\frac{x_{2}^{2}}{\left(1-x_{2}^{3}\right)^{2}}+\cdots+\frac{x_{n}^{n}}{\left(1-x_{n}^{n+1}\right)^{2}}\right]<1
$$ | [
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numina_10049625 | ### 4.30 Method I. We have
$\sin 10 \alpha \sin 8 \alpha + \sin 8 \alpha \sin 6 \alpha - \sin 4 \alpha \sin 2 \alpha = \sin 8 \alpha \times$ $\times (\sin 10 \alpha + \sin 6 \alpha) - 2 \sin^2 2 \alpha \cos 2 \alpha = \sin 8 \alpha \cdot 2 \sin 8 \alpha \cdot$ $\cdot \cos 2 \alpha - 2 \sin^2 2 \alpha \cos 2 \alpha = 2... | 2\cos2\alpha\sin6\alpha\sin10\alpha | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | $4.30 \sin 10 \alpha \sin 8 \alpha + \sin 8 \alpha \sin 6 \alpha - \sin 4 \alpha \sin 2 \alpha$. | [
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ours_17896 | The given inequality can be transformed into:
\[
x^{3}(x+1) + y^{3}(y+1) + z^{3}(z+1) \geq \frac{3}{4}(x+1)(y+1)(z+1)
\]
By the AM-GM inequality, it suffices to prove a stronger inequality:
\[
x^{4} + x^{3} + y^{4} + y^{3} + z^{4} + z^{3} \geq \frac{1}{4} \left[(x+1)^{3} + (y+1)^{3} + (z+1)^{3}\right]
\]
Define \( ... | null | {
"competition": "imo",
"dataset": "Ours",
"posts": null,
"source": "IMO1998SL.md"
} | Let \( x, y, \) and \( z \) be positive real numbers such that \( x y z = 1 \). Prove that
\[
\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geq \frac{3}{4}
\] | [
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aops_576294 | [quote="oldbeginner"]If $a, b, c>0$ prove that
\[\frac{(a+b)(2b-c)}{a^2-ab+b^2}+\frac{(b+c)(2c-a)}{b^2-bc+c^2}+\frac{(c+a)(2a-b)}{c^2-ca+a^2}\le 6\][/quote]
After expanding we need to prove that
$\sum_{cyc}(a^5b+a^5c+3a^4b^2+a^4c^2-8a^4bc-4a^3b^3+2a^3b^2c+8a^3c^2b-4a^2b^2c^2)\geq0$,
which follows from two obvious ineq... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $a, b, c>0$ prove that\n\\[\\frac{(a+b)(2b-c)}{a^2-ab+b^2}+\\frac{(b+c)(2c-a)}{b^2-bc+c^2}+\\frac{(c+a)(2a-b)}{c^2-ca+a^2}\\le 6\\]",
"content_html": "If <img src=\"//latex.artofproblemsolving.com... | If \(a,b,c>0\), prove that
\[
\frac{(a+b)(2b-c)}{a^2-ab+b^2}+\frac{(b+c)(2c-a)}{b^2-bc+c^2}+\frac{(c+a)(2a-b)}{c^2-ca+a^2}\le 6.
\] | [
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numina_10731840 | Assume without loss of generality that $x \geqslant y \geqslant z$, it is easy to see that
$$\begin{array}{l}
\frac{x^{k+1}}{x^{k+1}+y^{k}+z^{k}} \geqslant \frac{y^{k+1}}{y^{k+1}+z^{k}+x^{k}} \geqslant \frac{z^{k+1}}{z^{k+1}+x^{k}+y^{k}} \\
z^{k+1}+x^{k}+y^{k} \geqslant y^{k+1}+z^{k}+x^{k} \geqslant x^{k+1}+y^{k}+z^{k}... | \frac{1}{7} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "inequalities"
} | Example 6.49 (2007 Serbia Mathematical Olympiad) $x, y, z>0, x+y+z=1$, prove that
$$\frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}}+\frac{y^{k+2}}{y^{k+1}+z^{k}+x^{k}}+\frac{z^{k+2}}{z^{k+1}+x^{k}+y^{k}} \geqslant \frac{1}{7}$$ | [
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aops_2813730 | Let me try.
$x_1+ x_2+ x_3=9 ... \Large{ \textcircled{1}} $
$ x_1 x_2 x_3=15 ... \Large{ \textcircled{2}} $
$ x_1- x_2= x_2- x_3 ... \Large{ \textcircled{3}} $
Insert $ \Large{ \textcircled{3}}$ into $ \Large{ ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "【National College Entrance Exam， old China】A$\\Sigma$-1\nSolve：\n$x^3-9x^2+23x-15=0$\nand, $ x_1- x_2=x_2-x_3$",
"content_html": "【National College Entrance Exam， old China】<span style=\"white-spac... | 【National College Entrance Exam， old China】A$\Sigma$-1
Solve：
$x^3-9x^2+23x-15=0$
and, $ x_1- x_2=x_2-x_3$ | [
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numina_10051913 | ## Solution.
$\sin ^{6} \alpha+\cos ^{6} \alpha=\sin ^{4} \alpha-\sin ^{2} \alpha \cdot \cos ^{2} \alpha+\cos ^{4} \alpha=1-3 \sin ^{2} \alpha \cdot \cos ^{2} \alpha=$ $=1-\frac{3}{4} \sin ^{2} 2 \alpha=\frac{1+3 \cos ^{2} 2 \alpha}{4} \Rightarrow A=\frac{4}{1+3 \cos ^{2} 2 \alpha}$.
From this, it is clear that $A$ t... | 4 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | ### 3.485 Find the maximum value of the expression
$$
A=\frac{1}{\sin ^{6} \alpha+\cos ^{6} \alpha} \text { for } 0 \leq \alpha \leq \frac{\pi}{2}
$$ | [
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numina_10050744 | Solution.
Rewrite this system in the form
$\left\{\begin{array}{l}(x-y)(x-y)(x+y)=45, \\ x+y=5\end{array} \Leftrightarrow\left\{\begin{array}{l}(x-y)^{2}(x+y)=45, \\ x+y=5 .\end{array} \Rightarrow\right.\right.$
$\Rightarrow(x-y)^{2}=9$, from which $x-y=-3$ or $x-y=3$. We obtain a combination of two systems:
1) $\l... | (4,1),(1,4) | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 6.086. $\left\{\begin{array}{l}(x-y)\left(x^{2}-y^{2}\right)=45, \\ x+y=5\end{array}\right.$
Solve the system of equations:
\[
\left\{\begin{array}{l}
(x-y)\left(x^{2}-y^{2}\right)=45, \\
x+y=5
\end{array}\right.
\] | [
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ours_4129 | Let \( a+b+c=ab+bc+ca=k \). Since \((a+b+c)^{2} \geq 3(ab+bc+ca)\), we have \( k^{2} \geq 3k \). Since \( k>0 \), it follows that \( k \geq 3 \).
We have \( bc \geq ca \geq ab \), so from the above relation, we deduce that \( bc \geq 1 \).
By AM-GM, \( b+c \geq 2\sqrt{bc} \) and consequently \( b+c \geq 2 \). The equa... | null | {
"competition": "bmo",
"dataset": "Ours",
"posts": null,
"source": "2019_bmo_shortlist-2.md"
} | Let \( a, b, c \) be real numbers such that \( 0 \leq a \leq b \leq c \). Prove that if
\[
a+b+c=ab+bc+ca>0,
\]
then \(\sqrt{bc}(a+1) \geq 2\). When does the equality hold? | [
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numina_10094681 | Analyzing the given two expressions, which are symmetric with respect to $x, y, z$, we can assume $x \leqslant y \leqslant z$.
Solution: Without loss of generality, let $x \leqslant y \leqslant z$, then
$$
\begin{aligned}
& x^{3}+y^{3}+z^{3}-x^{2}(y+z)-y^{2}(z+x)-z^{2}(x+y)+3 x y z \\
= & x^{3}-x^{2}(y+z)+x y z+y^{3}+z... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 7 Let $x, y, z \in \mathbf{R}^{\prime}$, compare $x^{3}+y^{3}+z^{3}+3 x y z$ with $x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)$. | [
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numina_10180045 | ## Solution:
Applying the inequality of means, we have:
$a^{2}+b c \geq 2 \sqrt{a^{2} b c} \quad \frac{1}{a^{2}+b c} \leq \frac{1}{2 \sqrt{a^{2} b c}} \quad \frac{a}{a^{2}+b c} \leq \frac{1}{2 \sqrt{b c}}$
$b^{2}+a c \geq 2 \sqrt{b^{2} a c} \quad \frac{1}{b^{2}+a c} \leq \frac{1}{2 \sqrt{b^{2} a c}} \quad \frac{b}{b... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | ## Problem 1
Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be strictly positive real numbers. Show that:
$\frac{\mathrm{a}}{\mathrm{a}^{2}+\mathrm{bc}}+\frac{\mathrm{b}}{\mathrm{b}^{2}+\mathrm{ac}}+\frac{\mathrm{c}}{\mathrm{c}^{2}+\mathrm{ab}} \leq \frac{1}{2}\left(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}+\frac{1}{\math... | [
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aops_436106 | [hide="Solution"]
If a quadratic has real roots, then its discriminant must be nonnegative. [hide="Reasoning"]
The quadratic formula is $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$. Its discriminant, $b^2-4ac$, is located inside the square root. What happens if the discriminant is negative?[/hide]
The discriminant of $x^2+bx+... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $x^2+bx+16$ has real roots, find all possible values of $b$. Express your answer in interval notation.",
"content_html": "If <img src=\"//latex.artofproblemsolving.com/9/7/5/9754a367606a271cab941d... | If \(x^2 + bx + 16\) has real roots, find all possible values of \(b\). Express your answer in interval notation. | [
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numina_10048390 | ## Solution.
Domain of definition: $x^{2}-4 \geq 0 \Leftrightarrow x \in(-\infty ;-2] \cup[2 ; \infty)$.
Let's write the equation in the form $2^{x+\sqrt{x^{2}-4}}-\frac{5}{2} \cdot 2^{\frac{x+\sqrt{x^{2}-4}}{2}}-6=0$. Solving it as a quadratic equation in terms of $2^{\frac{x+\sqrt{x^{2}-4}}{2}}$, we get $2^{\frac{x... | \frac{5}{2} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 7.215. $2^{x+\sqrt{x^{2}-4}}-5 \cdot(\sqrt{2})^{x-2+\sqrt{x^{2}-4}}-6=0$. | [
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21098,
5154,
61601,
3488,
5115,
7771,
19924,
8060,
... | null | null |
aops_80134 | [quote="Werwulff"][quote="nthd"]Prove that$\forall x,y,z \ge 0$
$\sum(\sqrt{x^2+yz}-\sqrt{y^2+zx})^2\le\sum(x-y)^2$
:) :)[/quote]
It is trivial and thus boring to show that this inequality is equivalent to
\[ \sum 2xy\leq\sum\sqrt{(x^2+yz)(y^2+zx)} \]
which can easily be proved using cauc... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Prove that$\\forall x,y,z \\ge 0$\r\n $\\sum(\\sqrt{x^2+yz}-\\sqrt{y^2+zx})^2\\le\\sum(x-y)^2$\r\n :) :)",
"content_html": "Prove tha<span style=\"white-space:nowrap;\">t<... | Prove that for all nonnegative real numbers \(x,y,z\),
\[
\sum_{\text{cyc}}\bigl(\sqrt{x^2+yz}-\sqrt{y^2+zx}\bigr)^2 \le \sum_{\text{cyc}}(x-y)^2.
\] | [
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6900,
45430,
696... | null | null |
numina_10234387 | 1. **Denote the sums for convenience:**
\[
\sum_{j=1}^{m} a_{ij} = x_i \quad \text{and} \quad \sum_{i=1}^{n} a_{ij} = y_j
\]
Then, the function \( f \) can be rewritten as:
\[
f = \frac{n \sum_{i=1}^{n} x_i^2 + m \sum_{j=1}^{m} y_j^2}{\left( \sum_{i=1}^{n} \sum_{j=1}^{m} a_{ij} \right)^2 + mn \sum_{i=... | \frac{m+n}{mn+n} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "aops_forum"
} | For two given positive integers $ m,n > 1$, let $ a_{ij} (i = 1,2,\cdots,n, \; j = 1,2,\cdots,m)$ be nonnegative real numbers, not all zero, find the maximum and the minimum values of $ f$, where
\[ f = \frac {n\sum_{i = 1}^{n}(\sum_{j = 1}^{m}a_{ij})^2 + m\sum_{j = 1}^{m}(\sum_{i= 1}^{n}a_{ij})^2}{(\sum_{i = 1}^{n}\su... | [
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24665,
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43837,
6749... | null | null |
aops_575416 | [quote="hungkg"]Let $a,b,c$ be nonnegative real numbers such as $ab+bc+ca=1$. Prove that
\[\frac{1}{{\sqrt {{a^2} + {b^2}} }} + \frac{1}{{\sqrt {{b^2} + {c^2}} }} + \frac{1}{{\sqrt {{c^2} + {a^2}} }} \ge 2 + \frac{1}{{\sqrt 2 }}.\][/quote]
[b]This is my proof for problem[/b]
Without of generality, we can assume that $... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $a,b,c$ be nonnegative real numbers such as $ab+bc+ca=1$. Prove that \n\\[\\frac{1}{{\\sqrt {{a^2} + {b^2}} }} + \\frac{1}{{\\sqrt {{b^2} + {c^2}} }} + \\frac{1}{{\\sqrt {{c^2} + {a^2}} }} \\ge 2 + \\f... | Let \(a,b,c\) be nonnegative real numbers such that \(ab+bc+ca=1\). Prove that
\[
\frac{1}{\sqrt{a^2+b^2}}+\frac{1}{\sqrt{b^2+c^2}}+\frac{1}{\sqrt{c^2+a^2}}\ge 2+\frac{1}{\sqrt2}.
\] | [
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... | null | null |
aops_277949 | [quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {a}{b^2 \plus{} b} \plus{} \frac {b}{c^2 \plus{} c} \plus{} \frac {c}{a^2 \plus{} a} \ge \frac {36(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca \plus{} 5}$[/quote]
hello,i can prove your p... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $ a,b,c$ be positive numbers such that: $ a\\plus{}b\\plus{}c\\equal{}1$.Prove that:\r\n$ \\frac{a}{b^2\\plus{}b}\\plus{}\\frac{b}{c^2\\plus{}c}\\plus{}\\frac{c}{a^2\\plus{}a} \\ge \\frac{36(a^2\\plus{... | Let \(a,b,c\) be positive numbers such that \(a+b+c=1\). Prove that
\[
\frac{a}{b^2+b}+\frac{b}{c^2+c}+\frac{c}{a^2+a}\ge \frac{36(a^2+b^2+c^2)}{ab+bc+ca+5}.
\] | [
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21158,
27650,... | null | null |
numina_10103728 | [Proof] Let $d=u_{k}-u_{k-1}$ be the common difference of the arithmetic sequence, then
$$
\begin{aligned}
t_{n} & =\frac{\sqrt{u_{2}}-\sqrt{u_{1}}}{u_{2}-u_{1}}+\frac{\sqrt{u_{3}}-\sqrt{u_{2}}}{u_{3}-u_{2}}+\cdots+\frac{\sqrt{u_{n}}-\sqrt{u_{n-1}}}{u_{n}-u_{n-1}} \\
& =\frac{\sqrt{u_{n}}-\sqrt{u_{1}}}{d} . \\
& =\frac... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | $3 \cdot 49$ Positive numbers $u_{1}, u_{2}, \cdots, u_{n}$ form an arithmetic sequence, prove:
$$
\begin{aligned}
t_{n} & =\frac{1}{\sqrt{u_{1}}+\sqrt{u_{2}}}+\frac{1}{\sqrt{u_{2}}+\sqrt{u_{3}}}+\cdots+\frac{1}{\sqrt{u_{n-1}}+\sqrt{u_{n}}} \\
& =\frac{n-1}{\sqrt{u_{1}}+\sqrt{u_{n}}} .
\end{aligned}
$$ | [
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50733,
1743... | null | null |
aops_3202306 | [quote=williamxiao][hide=sol?]Subtract equation 1 from equation 2 to get $(y+x)(y-x) + (y-x)z = 1, (x+y+z)(y-x) = 1$
Same with 3 and 2, $(y+z)(y-z) + (y-z)x = 1$, $(y+z+x)(y-z) = 1$
So $(x+y+z)(y-x) = (x+y+z)(y-z)$. That means $(y-x) = (y-z)$, so $x = z$.
We then have $(2x+y)(y-x) = 1$. Next, we have by substituting x... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find $x^{2}+y^{2}+z^{2}$ given\n$$\\begin{cases}\nx^{2}-yz=1 & \\\\\ny^{2}-xz=2 & \\\\\nz^{2}-xy=1\n\\end{cases}$$\n\n\n",
"content_html": "Find <img src=\"//latex.artofproblemsolving.com/e/e/f/eefec... | Find \(x^{2}+y^{2}+z^{2}\) given
\[
\begin{cases}
x^{2}-yz=1,\\[4pt]
y^{2}-xz=2,\\[4pt]
z^{2}-xy=1.
\end{cases}
\] | [
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... | null | null |
aops_1202710 | [quote=daisyxixi]
Let a, b and c be positive real such that $ab+bc+ca=2$.
Prove $abc(a+b+c+9abc)\le 4$.
[/quote]
Or
$abc(a+b+c)\le \frac{(ab+bc+ca)^2}{3} \ \ \Rightarrow \ \ abc(a+b+c)\le \frac{4}{3}$
$3\sqrt[3]{a^2b^2c^2}\le ab+bc+ca \ \ \Rightarrow \ \ 9a^2b^2c^2\le \frac{8}{3}$
$abc(a+b+c+9abc)= abc(a+b+c)+9a... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Hi all,\n\nHere is another inequality that I failed to prove...\n\nI want to ask if it is true that we have to try the problem by expressing the left side of the inequality as only the variable $abc$?\n\nP... | Let \(a,b,c>0\) satisfy \(ab+bc+ca=2\). Prove
\[
abc\bigl(a+b+c+9abc\bigr)\le 4.
\] | [
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ours_18647 | Solution: We use the identities \(\sin^2 x + \cos^2 x = 1\) and \(\cot^2 x + 1 = \csc^2 x\) to simplify the equation. Substituting these identities, the equation becomes:
\[
1 + \csc^2 x = \csc^2 x + \sec^2 x
\]
This simplifies to:
\[
1 = \sec^2 x
\]
Thus, \(\cos^2 x = \frac{1}{2}\), which implies \(\cos x = \pm \f... | \frac{\pi}{4}, \frac{3\pi}{4} | {
"competition": "jhmt",
"dataset": "Ours",
"posts": null,
"source": "AlgebraKey2006.md"
} | Find all \(x\) in \([0, \pi]\) inclusive so that
\[
\sin^2 x + \csc^2 x + \cos^2 x = \cot^2 x + \sec^2 x
\] | [
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534... | null | null |
numina_10089611 | ## Solution.
Note that $x$ must be different from zero for all expressions to make sense. 1 point
By moving all addends to the same side, we have
$$
\frac{x-m}{x^{2}}-2 \cdot \frac{x-m}{x}+x-m \geqslant 0
$$
1 point
Further simplification yields
$$
(x-m)\left(\frac{1}{x^{2}}-\frac{2}{x}+1\right) \geqslant 0
$$
o... | x\in\begin{cases}{1}\cup[,+\infty)& | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | ## Task A-1.1.
Depending on the real parameter $m$, determine for which real numbers $x$ the following inequality holds:
$$
\frac{x-m}{x^{2}}+x \geqslant 2\left(1-\frac{m}{x}\right)+m
$$ | [
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35559,... | null | null |
aops_281664 | [quote="ifai"]Suppose that $ \{a_n\}$ and $ \{b_n\}$ are two arithmetic sequences such that
$ \frac {S_n}{T_n} \equal{} \frac {3n \plus{} 2}{4n \minus{} 5},$
where $ S_n \equal{} a_1 \plus{} \cdots \plus{} a_n$ and $ T_n \equal{} b_1 \plus{} \cdots \plus{} b_n$. Find $ \frac {a_{2009}}{b_{2009}}$.[/quote]
$ S_{401... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Suppose that $ \\{a_n\\}$ and $ \\{b_n\\}$ are two arithmetic sequences such that\r\n\r\n$ \\frac{S_n}{T_n}\\equal{}\\frac{3n\\plus{}2}{4n\\minus{}5},$\r\n\r\nwhere $ S_n\\equal{}a_1\\plus{}\\cdots\\plus{}... | Suppose that \( \{a_n\}\) and \( \{b_n\}\) are two arithmetic sequences such that
\[
\frac{S_n}{T_n}=\frac{3n+2}{4n-5},
\]
where \(S_n=a_1+\cdots+a_n\) and \(T_n=b_1+\cdots+b_n\). Find \(\dfrac{a_{2009}}{b_{2009}}\). | [
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30696,
5746... | null | null |
aops_174069 | [quote="ElChapin"]For positive $ x_i$ such that $ \sum_{i \equal{} 1}^{n}{\sqrt {x_i}} \equal{} 1$
Prove that the following holds:
$ \sum_{i \equal{} 1}^{n}{x_i^2}\ge(\sum_{i \equal{} 1}^{n}{x_1})^3$
:)[/quote]
I think:(Using Holder)
$ (\sum_{i \equal{} 1}^{n}{x_i^2})(\sum_{i \equal{} 1}^{n}{\sqrt{x... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "For positive $ x_i$ such that $ \\sum_{i \\equal{} 1}^{n}{\\sqrt {x_i}} \\equal{} 1$\r\n\r\n\r\nProve that the following holds:\r\n\r\n $ \\sum_{i \\equal{} 1}^{n}{x_i^2}\\ge(\\sum_{i \\equal{} ... | For positive real numbers \(x_i\) such that \(\sum_{i=1}^n \sqrt{x_i}=1\), prove that
\[
\sum_{i=1}^n x_i^2 \ge \biggl(\sum_{i=1}^n x_i\biggr)^3.
\] | [
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aops_120113 | [quote="Beat"]$x,y,z>0$, $x+y+z=1$ : $\sum\frac{\sqrt{xy}}{z(\frac{1}{z}+1)}\le \frac{3}{4}$[/quote]
Let $x=a^{2},$ $y=b^{2}$ and $z=c^{2},$ where $a,$ $b$ and $c$ are positive numbers.
Hence, $\sum\frac{\sqrt{xy}}{z(\frac{1}{z}+1)}\le \frac{3}{4}\Leftrightarrow\sum_{cyc}\left(\frac{1}{4}-\frac{ab}{c^{2}+1}\right)\ge... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$x,y,z>0$, $x+y+z=1$ : $\\sum\\frac{\\sqrt{xy}}{z(\\frac{1}{z}+1)}\\le \\frac{3}{4}$",
"content_html": "<span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemsolving.com/5/e/5/5e5616447e... | Let \(x,y,z>0\) with \(x+y+z=1\). Prove that
\[
\sum_{\text{cyc}}\frac{\sqrt{xy}}{z\bigl(\frac{1}{z}+1\bigr)}\le \frac{3}{4}.
\] | [
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19828,
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numina_10085103 | ## Solution:
$$
\begin{aligned}
& a+b c=a(a+b+c)+b c=(a+b)(a+c) . \text { Similarly, } b+c a=(b+c)(b+a) \text { and } c+a b= \\
& (c+a)(c+b) . \text { Then: } \sqrt{a+b c}+\sqrt{b+c a}+\sqrt{c+a b}=\sqrt{(a+b)(a+c)}+\sqrt{(b+c)(b+a)} \\
& +\sqrt{(c+a)(c+b)} \leq \frac{a+b+a+c}{2}+\frac{b+c+b+a}{2}+\frac{c+a+c+b}{2}=\f... | =b==\frac{1}{3} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 8.8. Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Prove the inequality
$$
\sqrt{a+b c}+\sqrt{b+c a}+\sqrt{c+a b} \leq 2
$$
Find the numbers $a, b, c$ for which equality holds. | [
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40554,
37832,
16913,
... | null | null |
numina_10078672 | First, we use the arithmetic-geometric mean inequality on each term to simplify the overly dubious denominator of our fractions:
$$
\frac{a^{2}}{\frac{b+c}{2}+\sqrt{b c}}+\frac{b^{2}}{\frac{c+a}{2}+\sqrt{c a}}+\frac{c^{2}}{\frac{a+b}{2}+\sqrt{a b}} \geq \frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}
$$
It rema... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Let $a, b, c$ be three positive real numbers such that $a+b+c=1$.
Show that
$$
\frac{a^{2}}{\frac{b+c}{2}+\sqrt{b c}}+\frac{b^{2}}{\frac{c+a}{2}+\sqrt{c a}}+\frac{c^{2}}{\frac{a+b}{2}+\sqrt{a b}} \geq \frac{1}{2}
$$
## 2 Solution | [
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30122,
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32045,
70674,
2108... | null | null |
aops_1702426 | By AM-GM,
\begin{align*}
\frac{4-a}{b}+\frac{4-b}{c}+\frac{4-c}{d}+\frac{4-d}{a}&=\frac{b+c+d}b+\frac{c+d+a}c+\frac{d+a+b}d+\frac{a+b+c}a\\
&=4+\frac cb+\frac db+\frac dc+\frac ac+\frac ad+\frac bd+\frac ba+\frac ca\\
&\ge4+8\\
&=12
\end{align*} | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "[b]Easy version:[/b]\nLet $a,b,c,d>0$ such that $a+b+c+d = 4$. Prove that:\n\\[ \\frac{4-a}{b}+\\frac{4-b}{c}+\\frac{4-c}{d}+\\frac{4-d}{a} \\geq 12 \\]\n[b]Stronger version:[/b]\nLet $a,b,c,d>0$ such that... | Easy version:
Let \(a,b,c,d>0\) such that \(a+b+c+d = 4\). Prove that
\[
\frac{4-a}{b}+\frac{4-b}{c}+\frac{4-c}{d}+\frac{4-d}{a} \ge 12.
\]
Stronger version:
Let \(a,b,c,d>0\) such that \(a+b+c+d = 4\). Prove that
\[
\frac{4-a}{b}+\frac{4-b}{c}+\frac{4-c}{d}+\frac{4-d}{a} +4abcd \ge 16.
\] | [
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24964,
27307,
64384,
45929,
68596... | null | null |
ours_17179 | Using the AM-GM inequality, we have:
\[
\begin{aligned}
(n+k-1) x_{1}^{n} x_{2} \cdots x_{k} &\leq n x_{1}^{n+k-1} + x_{2}^{n+k-1} + \cdots + x_{k}^{n+k-1}, \\
(n+k-1) x_{1} x_{2}^{n} \cdots x_{k} &\leq x_{1}^{n+k-1} + n x_{2}^{n+k-1} + \cdots + x_{k}^{n+k-1}, \\
&\vdots \\
(n+k-1) x_{1} x_{2} \cdots x_{k}^{n} &\leq x... | null | {
"competition": "imo",
"dataset": "Ours",
"posts": null,
"source": "IMO1967SL.md"
} | Prove the inequality
\[ x_{1} x_{2} \cdots x_{k}\left(x_{1}^{n-1}+x_{2}^{n-1}+\cdots+x_{k}^{n-1}\right) \leq x_{1}^{n+k-1}+x_{2}^{n+k-1}+\cdots+x_{k}^{n+k-1}, \]
where \( x_{i} > 0 \) for \( i = 1, 2, \ldots, k \), \( k \in \mathbb{N} \), \( n \in \mathbb{N} \). | [
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... | null | null |
ours_3184 | Solution: Use the fact that the summand can be expressed as a telescoping series:
\[
\cot ^{-1}(1+1 / k) - \cot ^{-1}(1+1 /(k-1))
\]
This telescoping nature simplifies the sum to:
\[
\cot ^{-1}(1+1 / n)
\]
Thus, the answer is \(\cot ^{-1}(1+1 / n)\). | \cot ^{-1}(1+1 / n) | {
"competition": "alg_misc",
"dataset": "Ours",
"posts": null,
"source": "Telescoping Sums and Products - Po-Shen Loh - MOP 2003.md"
} | Evaluate:
$$
\sum_{k=1}^{n} \cot ^{-1}\left(2 k^{2}\right)=\cot ^{-1}(1+1 / n)
$$ | [
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aops_2156012 | We use the formula $d=rt.$ The distance that you travel is equal to your rate (basically your speed) times the time.
First, let's say that Sunny runs at a speed $r,$ so Moonbeam runs at a speed of $mr.$ Also, let's say that Sunny runs $d$ meters before Moonbeam catches up to him, which means that Moonbeam runs $d+h$ m... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Sunny runs at a steady rate, and Moonbeam runs $m$ times as fast, where $m$ is a number greater than $1$. If Moonbeam gives Sunny a headstart of $h$ meters, how many meters must Moonbeam run to overtake Su... | Sunny runs at a steady rate, and Moonbeam runs \(m\) times as fast, where \(m>1\). If Moonbeam gives Sunny a head start of \(h\) meters, how many meters must Moonbeam run to overtake Sunny? Express your answer in terms of \(h\) and \(m\). | [
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ours_6920 | Let \( f(0)=a \), where \( a \in \mathbb{R} \). Setting \( x=0 \) in the condition, we obtain \( f(a)=0 \). For \( y=a \), we have
\[
f(f(x))=x f(a+1).
\]
Assume first that \( f(a+1) \neq 0 \). Then \( f \) is injective. Indeed, if we assume that \( f(x_1)=f(x_2) \) for \( x_1 \neq x_2 \) and substitute these values ... | null | {
"competition": "bulgarian_tsts",
"dataset": "Ours",
"posts": null,
"source": "KBOM-All-2024-9-12 кл-sol (1).md"
} | Find all functions \( f: \mathbb{R} \rightarrow \mathbb{R} \) for which the following condition holds:
\[
f(f(x)+x f(y))=x f(y+1), \forall x, y \in \mathbb{R}.
\] | [
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18... | null | null |
aops_1915627 | [quote=MessingWithMath]How about the other way around, flipping the inequality sign?[/quote]
well by AMGM $1=\sum x_i \geq 5\sqrt[5]{\prod x_i}\implies \prod x_i \leq \left(\frac{1}{5}\right )^5\leq \left(\frac{3}{10}\right)^{10}$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $x_1,x_2,x_3,x_4,x_5\\in\\mathbb R^+$ such that $\\sum_{i=1}^5x_i=1$. Then prove or disprove that\n\n$$\\prod_{i=1}^{5}x_i\\geq(\\frac{3}{5})^{10}$$\n\n@below that's why I said Prove or [b]disprove[/b] ... | If \(x_1,x_2,x_3,x_4,x_5\in\mathbb{R}^+\) satisfy \(\sum_{i=1}^5 x_i=1\). Prove or disprove that
\[
\prod_{i=1}^{5} x_i \ge \left(\frac{3}{5}\right)^{10}.
\] | [
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aops_1773503 | $a^2+b^2+2 =\frac{(a^2+1)+(b^2+1)+(a^2+b^2)+2}{2}\geq a+b+ab+1=(a+1)(b+1)$
$\sqrt{a^2+b^2+2}+\sqrt{b^2+c^2+2}+\sqrt{c^2+a^2+2}\geq \sqrt{(a+1)(b+1)}+\sqrt{(b+1)(c+1)}+\sqrt{(c+1)(a+1)} \geq 3\sqrt[6]{8(a+1)(b+1)(c+1)}=6$
| null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "prove\n√⟨a²+b²+2⟩+√⟨b²+c²+2⟩+√⟨c²+a²+2⟩≥6\nif\na+b+c+ab+bc+ca+abc=7",
"content_html": "prove<br>\n√⟨a²+b²+2⟩+√⟨b²+c²+2⟩+√⟨c²+a²+2⟩≥6<br>\nif<br>\na+b+c+ab+bc+ca+abc=7",
"post_id": 11650399,
... | prove
√⟨a²+b²+2⟩+√⟨b²+c²+2⟩+√⟨c²+a²+2⟩≥6
if
a+b+c+ab+bc+ca+abc=7 | [
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... | null | null |
aops_1137661 | [quote=checkmatetang][quote]Julius has some spare change consisting of quarters, dimes and nickels. If the ratio of quarters to dimes is 3:4 and the ratio of quarters to nickels is 4:5, what is the ratio of dimes to nickels? Express your answer as a common fraction. I don't really know how to do this problem, so can I ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Julius has some spare change consisting of quarters, dimes and nickels. If the ratio of\n\nquarters to dimes is 3:4 and the ratio of quarters to nickels is 4:5, what is the ratio of dimes \n\nto nickels? E... | Julius has some spare change consisting of quarters, dimes, and nickels. If the ratio of quarters to dimes is \(3:4\) and the ratio of quarters to nickels is \(4:5\), what is the ratio of dimes to nickels? Express your answer as a common fraction.
When Mr. Tesla drives 60 mi/h, the commute from home to his office take... | [
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... | null | null |
numina_10100305 | 13
$$
\begin{aligned}
\frac{2 x y}{x+y-1} & =\frac{(x+y)^{2}-\left(x^{2}+y^{2}\right)}{x+y-1} \\
& =\frac{(x+y)^{2}-1}{x+y-1} \\
& =x+y+1 .
\end{aligned}
$$
Since
$$
\left(\frac{x+y}{2}\right)^{2} \leqslant \frac{x^{2}+y^{2}}{2}=\frac{1}{2},
$$
we have
$$
-\sqrt{2} \leqslant x+y \leqslant \sqrt{2}
$$
Thus,
$$
\frac{... | 1-\sqrt{2} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 13 If the real numbers $x, y$ satisfy $x^{2}+y^{2}=1$, then the minimum value of $\frac{2 x y}{x+y-1}$ is $\qquad$
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | [
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aops_3328316 | [hide=sol] Plugging in $x=y=0$ into the second property, we have $f(0)=\dfrac{2f(0)}{1+(f(0))^2}.$ Note $f(0)\neq 0,$ so $1=\dfrac{2}{1+(f(0))^2},$ and $(f(0))^2=1.$ It then follows that $f(0)=1$ or $f(0)=-1.$ If $f(0)=1,$ then plugging in $x=0$ yields $f(y)=\dfrac{1+f(y)}{1+f(y)}=1,$ and if $f(0)=-1,$ then plugging in... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all functions $f$ that map the set of real numbers into the set of real numbers, satisfying the following conditions: \n\n1) $|f(x)|\\ge 1$,\n\n2) $f(x+y)=\\frac{f(x)+f(y)}{1+f(x)f(y)}$ of all real va... | Find all functions \(f:\mathbb{R}\to\mathbb{R}\) satisfying:
1. \(|f(x)|\ge 1\) for all real \(x\).
2. \(f(x+y)=\dfrac{f(x)+f(y)}{1+f(x)f(y)}\) for all real \(x,y\). | [
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22013... | null | null |
aops_539329 | $\max(\sqrt{x}+\sqrt{y}+\sqrt{z})=1+\sqrt2+\sqrt3$.
Let $\sqrt{x}+\sqrt{y}+\sqrt{z}=A$ and $\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=B$.
Hence, $A^2=x+y+z+2B$ and $B^2=xy+xz+yz+2\sqrt{xyz}A$.
Id est, $A^2=x+y+z+2\sqrt{xy+xz+yz+2\sqrt{xyz}A}\leq6+2\sqrt{11+2\sqrt{6}A}$,
which gives $A\leq1+\sqrt2+\sqrt3$.
The equality occurs, when... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "let $x,y,z$ be non negative reals such that $x+y+z \\leq 6, xy+xz+yz \\leq 11, xyz \\leq 6$. Determine the maximum value of $\\sqrt{x}+\\sqrt{y}+\\sqrt{z}$.",
"content_html": "let <img src=\"//latex... | Let \(x,y,z\) be nonnegative real numbers such that
\[
x+y+z \le 6,\qquad xy+xz+yz \le 11,\qquad xyz \le 6.
\]
Determine the maximum value of \(\sqrt{x}+\sqrt{y}+\sqrt{z}\). | [
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numina_10034709 | 4. It will not catch up, as the dog can only catch the hare after 50 seconds. In this time, the hare can run 700 m, while the bushes are located 520 m away. | Itwillnotcatchup | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 4. A dog is chasing a hare at a speed of 17 m/s, the hare is running at a speed of $14 \boldsymbol{m} /$ s. The distance between them before the chase was 150 m. Will the dog catch the hare if there are bushes $520 \mathcal{M}$ from the hare where he can hide? | [
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22... | null | null |
aops_214281 | $ \sum_{cyc} {\frac {a}{b \plus{} cd}}\geq \sum_{cyc} {\frac {a}{b \plus{} 2c}} \equal{} \sum_{cyc} {\frac {a^2}{ab \plus{} 2ac}}\geq \frac {(a \plus{} b \plus{} c \plus{} d)^2}{ab \plus{} bc \plus{} cd \plus{} da \plus{} 2ac \plus{} 2bd}\geq\frac {4}{3}$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "For every $ 1\\leq a,b,c,d\\leq 2$, prove that\r\n\r\n$ \\frac{4}{3}\\leq \\frac{a}{b\\plus{}cd}\\plus{}\\frac{b}{c\\plus{}da}\\plus{}\\frac{c}{d\\plus{}ab}\\plus{}\\frac{d}{a\\plus{}bc}\\leq 2$.",
"... | For every real numbers \(a,b,c,d\) with \(1\le a,b,c,d\le 2\), prove that
\[
\frac{4}{3}\le \frac{a}{b+cd}+\frac{b}{c+da}+\frac{c}{d+ab}+\frac{d}{a+bc}\le 2.
\] | [
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numina_10190841 | Solution. The number $x$ is a solution to the above equation if and only if $x$ is a solution to the system
$$
\begin{aligned}
& \left(\frac{1}{\sin x}-\frac{1}{\cos x}\right)^{2}=8 \\
& \frac{1}{\sin x}-\frac{1}{\cos x} \geq 0
\end{aligned}
$$
From equation (1) we have
$$
\begin{aligned}
& \frac{1}{\sin ^{2} x}+\fr... | \frac{3\pi}{4},\frac{\pi}{12} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 2. In the interval $0 \leq x \leq \pi$ find the solutions to the equation
$$
\frac{1}{\sin x}-\frac{1}{\cos x}=2 \sqrt{2}
$$ | [
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6667,
36428,
6550... | null | null |
aops_606029 | Alternatively we can substitute $x=\frac{a}{b+c}$, $y=\frac{b}{c+a}$, $z=\frac{c}{a+b}$
Then clearly the condition gets satisfied
$(i)$Lower Bound
$\frac{3}{4} \leq \sum \frac{ab}{(b+c)(c+a)} \iff 6abc \le \sum ab^2 +\sum a^2b$
This is obvious
Upper Bound
$ \frac{ab}{(b+c)(c+a)} <1 \iff \sum ab(a+b) \le (a+b)(b+... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Suppose that $x,y ,z $ be positive real numbers such that $yz+zx+xy+2xyz=1$ .Prove that\n\\[(i) \\ \\frac{3}{4}\\le yz+zx+xy <1,\\]\n\\[(ii) \\ xyz\\le \\frac{1}{8}.\\]",
"content_html": "Suppose th... | Suppose that \(x,y,z\) are positive real numbers such that
\[
yz+zx+xy+2xyz=1.
\]
Prove that
\[
\text{(i)}\quad \frac{3}{4}\le yz+zx+xy<1,
\qquad
\text{(ii)}\quad xyz\le\frac{1}{8}.
\] | [
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aops_1297228 | [hide=Solution]We approach this by contradiction, suppose both are $0$. If $a+b+c=0$ then $a=-b-c$. WLOG suppose $|b|\ge |c|$. This means the expression $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is essentially $$\frac{1}{b}+\frac{1}{c}-\frac{1}{b+c}$$ $$=\frac{b^2+c^2+bc}{bc(b+c)}.$$ We are approaching this by contradictio... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $a$, $b$, and $c$ be nonzero real numbers. Prove that $a+b+c$ and $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}$ cannot both be $0$.",
"content_html": "Let <span style=\"white-space:nowrap;\"><img src=... | Let \(a\), \(b\), and \(c\) be nonzero real numbers. Prove that \(a+b+c\) and \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) cannot both be \(0\). | [
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6... | null | null |
aops_1944928 | [hide=Solution]Replace $2x=t$, and we will use \[sin3t=3sint-4si{{n}^{3}}t,\cos 3t=4{{\cos }^{3}}t-3\cos t\]. So we get: ${{\sin }^{3}}2x\cos 6x+{{\cos }^{3}}2x\sin 6x=3\sin t\cos t({{\cos }^{2}}t-{{\sin }^{2}})t=\frac{3}{2}\sin 2t\cos 2t=\frac{3}{4}\sin 4t=\frac{3}{4}\sin 8x$[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "The expression\n\\[\\sin^3 2x \\cos 6x + \\cos^3 2x \\sin 6x\\]\ncan be written in the equivalent form $a \\sin bx$ for some positive constants $a$ and $b.$ Find $a + b.$",
"content_html": "The expr... | The expression
\[
\sin^3(2x)\cos(6x)+\cos^3(2x)\sin(6x)
\]
can be written in the equivalent form \(a\sin(bx)\) for some positive constants \(a\) and \(b\). Find \(a+b\). | [
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... | null | null |
numina_10189851 | Solution. With simple trigonometric transformations, we obtain
$$
\begin{aligned}
& y_{1}=\sin ^{4} x+\cos ^{4} x=\frac{3+\cos 4 x}{4} \\
& y_{2}=\sin ^{6} x+\cos ^{6} x=\frac{5+3 \cos 4 x}{8}
\end{aligned}
$$
Therefore, these functions have a fundamental period equal to $\frac{2 \pi}{4}=\frac{\pi}{2}$, and it holds ... | \\frac{\pi}{12}+k\frac{\pi}{2},k\in\mathbb{Z} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 2. Given the functions $y_{1}=\sin ^{4} x+\cos ^{4} x$ and $y_{2}=\sin ^{6} x+\cos ^{6} x$. Prove that these functions have a fundamental period of $\frac{\pi}{2}$ and that $3 y_{1}-2 y_{2}=1$. For which values of $x$ is the first function $\frac{1}{16}$ greater than the second? | [
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numina_10156231 | SOLUTION
$\mathbf{E}$
From the information given in the question we see that $x$ satisfies the equation $x-\frac{1}{10}=\frac{x}{10}$. Multiplying both sides by 10 gives $10 x-1=x$, so $9 x=1$. So $x=\frac{1}{9}$. | \frac{1}{9} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 3. The number $x$ has the following property: subtracting $\frac{1}{10}$ from $x$ gives the same result as multiplying $x$ by $\frac{1}{10}$. What is the number $x$ ?
A $\frac{1}{100}$
B $\frac{1}{11}$
C $\frac{1}{10}$
D $\frac{11}{100}$
E $\frac{1}{9}$ | [
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53... | null | null |
numina_10105627 | (15) When $a=b=c=1$, we can get $k \geqslant 2$. Now we prove the inequality
$$
a b+b c+c a+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$
holds for all positive real numbers $a, b, c$.
By the AM-GM inequality, we have
$$
a b+\frac{1}{a}+\frac{1}{b} \geqslant 3 \sqrt[3]{a b \cdot \frac{1}{a} \cdot \f... | 2 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | (15) Find the smallest positive real number $k$, such that the inequality
$$
a b+b c+c a+k\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9,
$$
holds for all positive real numbers $a, b, c$. | [
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numina_10149319 | It is sufficient to show that (∑ a i ) 2 /(n - 1) - ∑ a i 2 ≤ 2a 1 a 2 . But this follows immediately from the Cauchy inequality for the two n-1 tuples: a 1 + a 2 , a 3 , a 4 , ... , a n ; and 1, 1, ... , 1: (∑ a i ) 2 <= (n - 1)( (a 1 + a 2 ) 2 + a 3 2 + ... + a n 2 ) = (n - 1) ∑ a i 2 + (n - 1) 2a 1 a 2 . 38th Putnam... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 38th Putnam 1977 Problem B5 a 1 , a 2 , ... , a n are real and b < (∑ a i ) 2 /(n - 1) - ∑ a i 2 . Show that b < 2a i a j for all distinct i, j. | [
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aops_126769 | It's not hard. By the [b]AM-GM Inequality[/b] we have:
$\sum\frac{x^{2}}{(2y+3z)(2z+3y)}\ge\sum\frac{x^{2}}{\left[\frac{5(y+z)}{2}\right]^{2}}=\frac{4}{25}\sum\left(\frac{x}{y+z}\right)^{2}\ge\frac{4}{25}\cdot\frac{\left(\sum\frac{x}{y+z}\right)^{2}}{3}\ge\frac{4}{25}\cdot\frac{\left(\frac{3}{2}\right)^{2}}{3}=\frac{3... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "For $a,b,c$ are positive numbers. Find minimum of expression:\r\n$A=\\frac{x^{2}}{(2y+3z)(2z+3y)}+\\frac{y^{2}}{(2x+3z)(2z+3x)}+\\frac{z^{2}}{(2x+3y)(2y+3x)}$",
"content_html": "For <img src=\"//late... | For positive numbers \(x,y,z\), find the minimum of
\[
A=\frac{x^{2}}{(2y+3z)(2z+3y)}+\frac{y^{2}}{(2x+3z)(2z+3x)}+\frac{z^{2}}{(2x+3y)(2y+3x)}.
\] | [
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numina_10192098 | $\mathrm{Al}$ and Bert must arrive at a town $22.5 \mathrm{~km}$ away. They have one bicycle between them and must arrive at the same time. Bert sets out riding at $8 \mathrm{~km} / \mathrm{h}$, leaves the bicycle and then walks at $5 \mathrm{~km} / \mathrm{h}$. Al walks at $4 \mathrm{~km} / \mathrm{h}$, reaches the bi... | 75 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | $\mathrm{Al}$ and Bert must arrive at a town $22.5 \mathrm{~km}$ away. They have one bicycle between them and must arrive at the same time. Bert sets out riding at $8 \mathrm{~km} / \mathrm{h}$, leaves the bicycle and then walks at $5 \mathrm{~km} / \mathrm{h}$. Al walks at $4 \mathrm{~km} / \mathrm{h}$, reaches the bi... | [
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numina_10180447 |
Solution: We use the trivial inequalities $a^{2}+1 \geq 2 a, b^{2}+1 \geq 2 b$ and $c^{2}+1 \geq 2 c$. Hence we obtain
$$
\frac{a^{2}+1}{b+c}+\frac{b^{2}+1}{c+a}+\frac{c^{2}+1}{a+b} \geq \frac{2 a}{b+c}+\frac{2 b}{c+a}+\frac{2 c}{a+b}
$$
$$
\frac{2 a}{b+c}+\frac{2 b}{c+a}+\frac{2 c}{a+b} \geq 3
$$
Adding 6 both sid... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} |
3. If $a, b, c$ are three positive real numbers, prove that
$$
\frac{a^{2}+1}{b+c}+\frac{b^{2}+1}{c+a}+\frac{c^{2}+1}{a+b} \geq 3
$$
| [
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ours_12154 | Let the speed of the sports plane be \(x \, \text{km/h}\). Then the speed of the fighter jet is \(3x \, \text{km/h}\). Since the sports plane flew \(x \, \text{km}\) in one hour, the fighter jet flew \((x + 200) \, \text{km}\) in half an hour, which means it flew \(2 \cdot (x + 200) \, \text{km}\) in one hour. Therefor... | null | {
"competition": "german_mo",
"dataset": "Ours",
"posts": null,
"source": "Loesungen_MaOlympiade_94.md"
} | A fighter jet flew \(200 \, \text{km}\) further in half an hour than a sports plane did in one hour. What was the speed of each of these two planes if the speed of the fighter jet was three times that of the sports plane? | [
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... | null | null |
aops_2157404 | [quote=NaPrai]By [b]AM-GM[/b], we have \begin{align*}\sum_{cyc}\frac{\sqrt{x_1x_2-1}}{x_2+x_3} &\le \frac{1}{2}\sum_{cyc}\sqrt{\frac{x_1x_2-1}{x_2x_3}} \\&= \frac{1}{2}\sum_{cyc}\sqrt{\frac{1}{x_3}\left(x_1-\frac{1}{x_2}\right)} \\&\le \frac{1}{4}\sum_{cyc}\left(x_1-\frac{1}{x_2}+\frac{1}{x_3}\right) \\&= \frac{1}{4}\s... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "For(n>2), $x_1,x_2,..,x_n$ positive real numbers not less than $k>0$ , prove that $$ \\frac{1}{4}\\sum_{cyc}x_1+\\frac{1-k^2}{4}\\sum_{cyc}\\frac{1}{x_1} \\geq \\sum_{cyc}\\frac{\\sqrt{x_1x_2-k^2}}{x_2+x_3... | For n > 2, let x_1, x_2, ..., x_n be positive real numbers with x_i ≥ k > 0 for all i. Prove that
\[
\frac{1}{4}\sum_{i=1}^n x_i+\frac{1-k^2}{4}\sum_{i=1}^n\frac{1}{x_i} \ge \sum_{i=1}^n\frac{\sqrt{x_i x_{i+1}-k^2}}{x_{i+1}+x_{i+2}},
\]
where indices are taken modulo n. | [
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aops_1202117 | Let $P(x,y)$ denote $f(xy)f(\frac{f(y)}{x})=1$ for all $x,y\in \mathbb{R}^+$
By $P(\frac{1}{x},f(y))$ give us $f(xf(f(y)))f(\frac{f(y)}{x})=1$ for all $x,y\in \mathbb{R}^+$
So $f(xy)=f(xf(f(y)))$ for all $x,y \in \mathbb{R}^+$
By $2)$ we get that if $f(a)=f(b),a<b$ then $f(x)=f(a)$ for all $x\in [a,b]$
If there exist $... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all functions $f:\\mathbb{R^{+}}\\rightarrow \\mathbb{R^{+}}$ such that\n$i/ f(xy)f(\\frac{f(y)}{x})=1$ $ \\forall x, y\\in \\mathbb{R^{+}}$\n$ii/ \\forall x\\geq y> 0$ then $f(x)\\leq f(y)$\n\n",
... | Find all functions \(f:\mathbb{R}^{+}\to\mathbb{R}^{+}\) such that
(i) \(f(xy)\,f\!\left(\dfrac{f(y)}{x}\right)=1\) for all \(x,y\in\mathbb{R}^{+}\).
(ii) For all \(x\ge y>0\) we have \(f(x)\le f(y)\). | [
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aops_3255223 | [quote=hangb6pbc]If $a \in R$, only $a=0$ work
$2=\left(\dfrac{sinx}{\sqrt{1-a+a^2}}+\dfrac{cosx}{\sqrt{1+a+a^2}} \right)^2 \leq (sin^2x+cos^2x)\left(\dfrac{1}{1-a+a^2}+\dfrac{1}{1+a+a^2} \right)=\dfrac{2(a^2+1)}{a^4+a^2+1}$
$\Rightarrow a^4 \leq 0 \Rightarrow a=0$[/quote]
Nice solution! ;-D | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Solve the equation and discuss:\n$ \\frac{ \\sin x} {\\sqrt{1 - a + a^2} } + \\frac{ \\cos x} {\\sqrt{1 + a + a^2} } = \\sqrt{2}$",
"content_html": "Solve the equation and discuss:<br>\n<img src=\"... | Solve the equation and discuss:
$ \frac{ \sin x} {\sqrt{1 - a + a^2} } + \frac{ \cos x} {\sqrt{1 + a + a^2} } = \sqrt{2}$ | [
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aops_1971996 | [quote=NTNT]:read: $ Let \enskip a,b,c>0 \enskip and \enskip (a+b)(b+c)(c+a)=1 \enskip. Prove \enskip that:$ \\
$ii. \enskip \left(\frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}+3\right)^2 \geq \frac{32}{3}(a+b+c)^3 \enskip $ :pilot: \\[/quote]
$RHS = \frac{32(a+b+c)^3}{3(a+b)(b+c)(c+a)} \leq 12 \frac{(a+b+c)^2}{... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "\n :read: $ Let \\enskip a,b,c>0 \\enskip and \\enskip (a+b)(b+c)(c+a)=1 \\enskip. Prove \\enskip that:$ \\\\ \n $i. \\enskip \\left(\\frac{2a}{b+c}+\\frac{2b}{c+a}+\\frac{2c}{a+b}+3\\right)^2 \\ge... | Let \(a,b,c>0\) and \((a+b)(b+c)(c+a)=1\). Prove that:
i. \[
\left(\frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}+3\right)^2 \ge 24\bigl(a^2+b^2+c^2+ab+bc+ca\bigr).
\]
ii. \[
\left(\frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}+3\right)^2 \ge \frac{32}{3}(a+b+c)^3.
\] | [
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numina_10101122 | 6. D Let $x=0, f(y)+f(-y)=0, f(-y)=-f(y)$, so $f(y)$ is an odd function. Also, by substituting $-y$ for $y$, we have $f(x-y)+f(x+y)=2 f(x) g(-y)$, thus $g(-y)=g(y)$. | D | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 6. Let $f(x), g(x)$ be two functions defined on $(-\infty,+\infty)$, for any real numbers $x, y$, satisfying $f(x+y) +$ $f(x-y)=2 f(x) \cdot g(y)$. If $f(0)=0$, but $f(x)$ is not identically zero, then
A. $f(x), g(x)$ are both odd functions
B. $f(x), g(x)$ are both even functions
C. $f(x)$ is an even function, $g(x)$ i... | [
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aops_2032931 | [quote=BestChoice123]For $x,y,z>0$ such that $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=2$. Show that $$(x+y)(y+z)(z+x)\ge 1$$[/quote]
There exists $a,b,c>0$ such that $ x=\frac{a}{b+c}, y=\frac{b}{c+a},z=\frac{c}{a+b}.$ So we need to [url=https://artofproblemsolving.com/community/c6h487722p2732787]prove that[/url] \[... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "For $x,y,z>0$ such that $\\frac{1}{x+1}+\\frac{1}{y+1}+\\frac{1}{z+1}=2$. Show that $$(x+y)(y+z)(z+x)\\ge 1$$",
"content_html": "For <img src=\"//latex.artofproblemsolving.com/5/e/5/5e5616447e444b385... | For \(x,y,z>0\) such that
\[
\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=2,
\]
show that
\[
(x+y)(y+z)(z+x)\ge 1.
\] | [
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aops_1887566 | [quote=White-Wolf]If $a,b,c\geq 0$ and $ab+bc+ac=3$ then prove that:
$4abc+9(a+b+c)\geq 31$[/quote]
The following inequality is also true :
If $a,b,c\geq 0$ and $ab+bc+ac=3$, for all $k \le 2 \sqrt{3}-3$ we have :
$$a+b+c+k\times abc\geq 3+k$$
(In your case $k=\frac49 \le 2 \sqrt{3}-3$) | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $a,b,c\\geq 0$ and $ab+bc+ac=3$ then prove that:\n\n$4abc+9(a+b+c)\\geq 31$",
"content_html": "If <img src=\"//latex.artofproblemsolving.com/3/0/f/30fa6a15477ea7489e3fe6af59cf06dc5d88be81.png\" cl... | If \(a,b,c\ge 0\) and \(ab+bc+ca=3\), prove that
\[
4abc+9(a+b+c)\ge 31.
\] | [
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... | null | null |
numina_10039181 | 34.19. The point is that if equality (3) holds for all real $x, y, x \neq y$, then
$$
\varphi\left(\frac{x+y}{2}\right)=\frac{\varphi(x)+\varphi(y)}{2}
$$
To prove this, replace $x$ with $x+y$ and $y$ with $x-y$ in (3). As a result, we get
$$
\frac{f(x+y)-g(x-y)}{2 y}=\varphi(x)
$$
for all real $x, y, y \neq 0$. Se... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 34.19. Prove that the functional equation
$$
\frac{f(x)-g(y)}{x-y}=\varphi\left(\frac{x+y}{2}\right)
$$
can be reduced to the functional equation (2) | [
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... | null | null |
aops_1149761 | [quote=loopback]Givn $a+b+c=0$ and $ a^2+b^2+c^2=1$ find maximum value of ${(9abc)}^2$[/quote]
Because If $a,\,b,\,c$ are real number such that $a+b+c=0$ then \[(a^2+b^2+c^2)^3 \geqslant 54a^2b^2c^2.\] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Givn $a+b+c=0$ and $ a^2+b^2+c^2=1$ find maximum value of ${(9abc)}^2$",
"content_html": "Givn <img src=\"//latex.artofproblemsolving.com/3/7/e/37e4c3e85710cff4f69bfec2adaff5ea38111a3a.png\" class=\"... | Given \(a+b+c=0\) and \(a^2+b^2+c^2=1\). Find the maximum value of \((9abc)^2\). | [
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numina_10160476 | 4. Answer: (D)
If $x \leq 0$, then $|x|=-x$, and we obtain from $|x|+x+5 y=2$ that $y=\frac{2}{5}$. Thus $y$ is positive, so $|y|-y+x=7$ gives $x=7$, which is a contradiction since $x \leq 0$. Therefore we must have $x>0$. Consequently, $|x|+x+5 y=2$ gives the equation
$$
2 x+5 y=2 \text {. }
$$
If $y \geq 0$, then $|... | 3 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 4. If $x$ and $y$ are real numbers for which $|x|+x+5 y=2$ and $|y|-y+x=7$, find the value of $x+y$.
(A) -3
(B) -1
(C) 1
(D) 3
(E) 5 | [
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numina_10126023 | [Solution] Rewrite the original system of equations as
$$
\left\{\begin{array}{l}
6=x^{2}-(y-z)^{2}=(x-y+z)(x+y-z), \\
2=y^{2}-(z-x)^{2}=(y-z+x)(y+z-x), \\
3=z^{2}-(x-y)^{2}=(z-x+y)(z+x-y) .
\end{array}\right.
$$
(1) $\times$ (2) $\times$ (3) gives
$$
36=(x-y+z)^{2}(y-z+x)^{2}(z-x+y)^{2} \text {, }
$$
Thus, $\quad(x-y... | \begin{pmatrix}x_{1}=\frac{5}{2},y_{1}=\frac{3}{2},z_{1}=2\\x_{2}=\frac{-5}{2},y_{2}=\frac{-3}{2},z_{2}=-2\end{pmatrix} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 4・129 Solve the system of equations $\left\{\begin{array}{l}x^{2}=6+(y-z)^{2}, \\ y^{2}=2+(z-x)^{2}, \\ z^{2}=3+(x-y)^{2} .\end{array}\right.$ | [
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aops_535042 | Set $a=\dfrac{2x}{y+z},b=\dfrac{2y}{x+z},c=\dfrac{2z}{x+y}=>ab+bc+ac+abc=4$ (you can check it).
We can prove that $a+b+c+abc\geq 4<=>\sum x^{3}+3xyz\geq \sum xy(x+y)$
Which is true by Schur inequality. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "find the max & min of $a+b+c+abc$ where $a,b,c\\in\\mathbb{R}^{+}$ and $ab+bc+ca+abc =4$.",
"content_html": "find the max & min of <img src=\"//latex.artofproblemsolving.com/1/a/9/1a9a299c16f3f11... | Find the maximum and minimum of
\[
a+b+c+abc
\]
subject to \(a,b,c\in\mathbb{R}^+\) and
\[
ab+bc+ca+abc=4.
\] | [
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numina_10734119 | Prove $\left(a^{2}+b c\right)\left(b^{2}+c a\right)\left(c^{2}+a b\right) \leqslant 1$
$$\begin{array}{l}
\Leftrightarrow(a+b)^{2}(a+c)^{2}(b+c)^{2} \geqslant 4\left(a^{2}+b c\right)\left(b^{2}+c a\right)\left(c^{2}+a b\right) \\
\Leftrightarrow(a-b)^{2}(a-c)^{2}(b-c)^{2}+\sum_{\text {sym }}\left(4 a^{3} b^{2} c+\frac{... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "inequalities"
} | Example 49 Non-negative real numbers $a, b, c$ satisfy $(a+b)(b+c)(c+a)=2$. Prove:
$$\left(a^{2}+b c\right)\left(b^{2}+c a\right)\left(c^{2}+a b\right) \leqslant 1$$ | [
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numina_10103851 | 7. $[-2,+\infty)$.
Let $f(x)=\sin ^{2} x+a \sin x+a+3$, then
$$
f(x)=\left(\sin x+\frac{a}{2}\right)^2+a+3-\frac{a^{2}}{4} \geqslant 0 .
$$
Since $x \in\left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$, thus, $\sin x \in\left[-\frac{1}{2}, 1\right]$.
Let $t=\sin x$, then $-\frac{1}{2} \leqslant t \leqslant 1$. Let
$$
g(t)=... | [-2,+\infty) | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 7. For any $x \in\left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$, the inequality $\sin ^{2} x+a \sin x+a+3 \geqslant 0$ always holds. Then the range of the real number $a$ is $\qquad$ . | [
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aops_336316 | first u write the given expressions as 1/x+1/y=5/6
1/y+1/z=7/10
this means 1/x-1/z=5/6-7/10
also from last equation , 1/x+1/z=8/15
from these get 1/x. then u will get 1... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "I am having a nightmare trying to solve this, yet it seems so simple\r\n\r\n[quote]Find x,y and z when....\n\n5xy/(x+y) =6\n\n7yz/(y+z) =10\n\n8zx/(z+x) =15[/quote]\r\n\r\nThank you",
"content_html... | Find \(x,y,z\) satisfying
\[
\frac{5xy}{x+y}=6,\qquad \frac{7yz}{y+z}=10,\qquad \frac{8zx}{z+x}=15.
\] | [
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aops_215205 | [quote="stergiu"]If $ a , b , c$ are positive numbers with sum $ 1$ , prove that :
$ \frac {ab}{1 \plus{} c} \plus{} \frac {bc}{1 \plus{} a} \plus{} \frac {ca}{1 \plus{} b} \leq \frac {1}{4}$
[/quote]
[quote="arqady"]
$ \sum_{cyc}\frac {ab}{1 \plus{} c}\leq\sum_{cyc}\frac {\sqrt {ab}}{1 \plus{} c}\leq\frac {1}{4}.$ ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $ a , b , c$ are positive numbers with sum $ 1$ , prove that :\r\n\r\n $ \\frac {ab}{1 \\plus{} c} \\plus{} \\frac {bc}{1 \\plus{} a} \\plus{} \\frac {ca}{1 \\plus{} b} \\leq \\frac {1}{4}$\r\n\r\n Babi... | If \(a,b,c\) are positive numbers with \(a+b+c=1\), prove that
\[
\frac{ab}{1+c}+\frac{bc}{1+a}+\frac{ca}{1+b}\le\frac{1}{4}.
\] | [
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aops_1160690 | We can first get that Rowena can paint $\frac{1}{14}$ every hour and Ruby can paint $\frac{1}{6}$ of the room every hour. We can set up the following equations: $\frac{1}{14}x+\frac{1}{6}y=\frac{1}{2}$ and $\frac{1}{14}y+\frac{1}{6}x=1$. Solving for $x$ and $y$ gives us $\boxed{(x,y)=(\frac{231}{40},\frac{21}{40})}... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Rowena can paint a room in $14$ hours, while Ruby can paint it in $6$ hours. If Rowena paints for $x$ hours and Ruby paints for $y$ hours, they will finish half of the painting, while if Rowena paints for ... | Rowena can paint a room in $14$ hours, while Ruby can paint it in $6$ hours. If Rowena paints for $x$ hours and Ruby paints for $y$ hours, they will finish half of the painting, while if Rowena paints for $y$ hours and Ruby paints for $x$ hours they will paint the whole room. Find the ordered pair $(x,y)$. | [
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aops_625060 | [quote="achilles04"]If $3x+4y+z=5$ where $x,y,z $ real numbers, then find the minimum value of $26(x^2+y^2+z^2)$ .[/quote]
Use cauchy-swarz inequality to find the minimum of x^2+y^2+z^2. Then the problem is done. :) | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $3x+4y+z=5$ where $x,y,z $ real numbers, then find the minimum value of $26(x^2+y^2+z^2)$ .",
"content_html": "If <img src=\"//latex.artofproblemsolving.com/d/6/6/d660975885da0d1c850f87c63dae48808... | If \(3x+4y+z=5\) where \(x,y,z\) are real numbers, find the minimum value of \(26(x^2+y^2+z^2)\). | [
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numina_10100184 | $$
\begin{array}{l}
\sum\left(\frac{a^{2}}{c}+\frac{b^{2}}{c}\right)+7 \sum a \\
\geqslant \frac{\left(\sum a\right)^{3}}{\sum a b}+\frac{2\left(\sum a b\right)^{2}}{a b c} \\
\Leftrightarrow \sum \frac{(a-b)^{2}}{c} \\
\geqslant \frac{\left(\sum a\right)^{3}-3\left(\sum a\right)\left(\sum a b\right)}{\sum a b}+ \\
\fr... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | High $\mathbf{5 2 2}$ Let $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\begin{aligned}
\sum & \left(\frac{a^{2}}{c}+\frac{c^{2}}{a}\right)+7 \sum a \\
& \geqslant \frac{\left(\sum a\right)^{3}}{\sum a b}+\frac{2\left(\sum a b\right)^{2}}{a b c},
\end{aligned}
$$
where, “ $\sum$ ” denotes the cyclic sum. | [
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aops_1589788 | It is simple, just start from $sin^{2}x+cos^{2}x=$1, thus sin$^{4}x+cos^{4}x=1-\frac{sin^{2}2x}{2}$, and $sin^{8}x-cos^{8}x=(sin^{2}x-cos^{2}x)(sin^{4}x+cos^{4}x),$ so we got the equation $cos2xsin^{2}2x=0$ (we are used $cos2x=cos^{2}x-sin^{2}x=1-2sin^{2}x)$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Solve the equation: $sin^8x-cos^8x-2sin^2x+1=0$",
"content_html": "Solve the equation: <img src=\"//latex.artofproblemsolving.com/8/a/9/8a9c2c7afe60ce29e5f042c0bad0b781465f07cc.png\" class=\"latex\" ... | Solve the equation:
\[
\sin^8 x - \cos^8 x - 2\sin^2 x + 1 = 0.
\] | [
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numina_10128063 | 【Solution】Solution: Assume all are soccer balls,
$$
96 \div 6=16 \text { (units), } 4 \times 6=24 \text { (people), }
$$
Basketballs: $24 \div(6-3)$,
$$
\begin{array}{l}
=24 \div 3, \\
=8 \text { (units); }
\end{array}
$$
Soccer balls: $20-8=12$ (units);
Answer: There are 12 soccer balls.
Therefore, the answer is: 12... | 12 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 10. (3 points) The school has a total of 20 soccer balls and basketballs, which can accommodate 96 students playing at the same time. Each soccer ball is shared by 6 students, and each basketball is shared by 3 students. Among them, there are $\qquad$ soccer balls. | [
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numina_10162807 | IV/2. If we substitute $y=-x$ into the equation, we get $f(f(x)-x)=f(0)+f(2012)$. Since the right side is a constant and $f$ is an injective function, $f(x)-x$ must also be a constant. Therefore, $f(x)=x+c$ for some real number $c$. If we substitute this into the original equation, we get $x+y+2c=x+y+2c+2012$, which gi... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 2. Prove that there does not exist an injective function $f: \mathbb{R} \rightarrow \mathbb{R}$ for which
$$
f(f(x)+y)=f(x+y)+f(2012) \quad \text { for all } x, y \in \mathbb{R}
$$ | [
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numina_10067757 | Substitute the values $x=0$ and $x=-1$ into the given equation. We get: $\left\{\begin{array}{c}-f(-1)=f(0), \\ -2 f(0)=-1+f(-1)\end{array}\right.$. Therefore, $2 f(-1)=-1+f(-1)$, which means $f(-1)=-1$.
## Answer
-1. | -1 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Folklore
The function $f(x)$ is defined for all $x$, except 1, and satisfies the equation: $(x-1) f\left(\frac{x+1}{x-1}\right)=x+f(x)$. Find $f(-1)$. | [
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... | null | null |
aops_1492517 | $$\tan \left(\frac{\pi}{4}+\frac y2\right)=\tan^3 \left(\frac{\pi}{4}+\frac x2\right)$$
Applying the tangent addition formula, we have:
$$\frac{1+\tan\frac y2}{1-\tan\frac y2}=\left(\frac{1+\tan\frac x2}{1-\tan\frac x2}\right) ^3$$
$$\frac{1+\frac {\sin \frac y2}{\cos \frac y2}}{1-\frac {\sin \frac y2}{\cos \frac y2}}=... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If tan(π/4+y/2) =tan[sup]3[/sup]( π/4+x/2)\nProve that:\nSiny/sinx = 3+sin[sup]2[/sup]x/1+3sin[sup]2[/sup]x\n",
"content_html": "If tan(π/4+y/2) =tan<sup>3</sup>( π/4+x/2)<br>\nProve that:<br>\nSiny/... | If \(\tan\!\left(\frac{\pi}{4}+\frac{y}{2}\right)=\tan^3\!\left(\frac{\pi}{4}+\frac{x}{2}\right)\), prove that
\[
\frac{\sin y}{\sin x}=\frac{3+\sin^2 x}{1+3\sin^2 x}.
\] | [
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aops_254900 | [quote="Brut3Forc3"]If $ \frac {xy}{x \plus{} y} \equal{} a, \frac {xz}{x \plus{} z} \equal{} b, \frac {yz}{y \plus{} z} \equal{} c$, where $ a,b,c$ are other than zero, then $ x$ equals:
$ \text{(A)}\, \frac {abc}{ab \plus{} ac \plus{} bc} \qquad\text{(B)}\, \frac {2abc}{ab \plus{} bc \plus{} ac} \qquad\text{(C)}\, \f... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $ \\frac {xy}{x \\plus{} y} \\equal{} a, \\frac {xz}{x \\plus{} z} \\equal{} b, \\frac {yz}{y \\plus{} z} \\equal{} c$, where $ a,b,c$ are other than zero, then $ x$ equals:\r\n\r\n$ \\textbf{(A)}\\ \\f... | If \(\dfrac{xy}{x+y}=a,\ \dfrac{xz}{x+z}=b,\ \dfrac{yz}{y+z}=c\), where \(a,b,c\) are nonzero, then \(x\) equals:
(A) \(\dfrac{abc}{ab+ac+bc}\)
(B) \(\dfrac{2abc}{ab+bc+ac}\)
(C) \(\dfrac{2abc}{ab+ac-bc}\)
(D) \(\dfrac{2abc}{ab+bc-ac}\)
(E) \(\dfrac{2abc}{ac+bc-ab}\) | [
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... | null | null |
aops_115303 | [quote="Phelpedo"]If $a+b+c=1$, find (with proof) the minimum and maximum possible value of $ab+bc+ca$.[/quote]
I think a minimum requires positive reals. :huh:
[hide="Because"] We can have the numbers be $-n$, $0$, $n+1$, $n\to\infty$. Thus, $ab+bc+ca=-n^{2}-n$
$\lim_{n\to\infty}-n^{2}-n=-\infty$ [/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $a+b+c=1$, find (with proof) the minimum and maximum possible value of $ab+bc+ca$.",
"content_html": "If <span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemsolving.com/0/4/6/0463a9... | If \(a+b+c=1\), find (with proof) the minimum and maximum possible values of \(ab+bc+ca\). | [
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numina_10152868 | $$
\begin{array}{l}
x+y=(\sqrt{x}+\sqrt{y})^{2}-2 \sqrt{x y} \\
\Rightarrow(\sqrt{x}+\sqrt{y})^{2}=18+2 \sqrt{x y} \leq 18+2\left(\frac{x+y}{2}\right)=36 \quad(\mathrm{GM} \leq \mathrm{AM})
\end{array}
$$
$\sqrt{x}+\sqrt{y} \leq 6=d$ (It is easy to get the answer by letting $x=y$ in $x+y=18$ )
Remark The original quest... | 6 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | G3.4 Let $x \geq 0$ and $y \geq 0$. Given that $x+y=18$. If the maximum value of $\sqrt{x}+\sqrt{y}$ is $d$, find the value of $d$. (Reference: 1999 FGS.2) | [
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numina_10048013 | ## Solution.
$$
\begin{aligned}
& 4 \cos \alpha \cos \varphi \cos (\alpha-\varphi)-2 \cos ^{2}(\alpha-\varphi)-\cos 2 \varphi= \\
& =2 \cos (\alpha-\varphi)(2 \cos \alpha \cos \varphi-\cos (\alpha-\varphi))-\cos 2 \varphi=
\end{aligned}
$$
$$
\begin{aligned}
& =2 \cos (\alpha-\varphi)(\cos \alpha \cos \varphi-\sin \a... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 3.223. $4 \cos \alpha \cos \varphi \cos (\alpha-\varphi)-2 \cos ^{2}(\alpha-\varphi)-\cos 2 \varphi=\cos 2 \alpha$. | [
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aops_1486733 | [quote=AOPS12142015][hide=Expereince with Problem 11]
Wait I actually think I got this problem except I couldn't get my answer into the cotangent form. I just used a lot of double angle and some other trig properties.
[/hide][/quote]
[hide=Hint]
You can write $\tan(\alpha)$ as $\cot(\alpha)-2\cot(2\alpha)$. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "The sum $\\sum\\limits_{n=1}^{102} (2^{n-1} \\tan 2^{2n-1}A)$ can be written in the form $x \\cot A + y \\cot z A$ where $x$ and $y$ and $z$ are integers. \n\nCompute the last three digits of $x+y+z$.\n\n[... | The sum
\[
\sum_{n=1}^{102} \bigl(2^{n-1}\tan(2^{2n-1}A)\bigr)
\]
can be written in the form \(x\cot A + y\cot(zA)\) where \(x\), \(y\), and \(z\) are integers. Compute the last three digits of \(x+y+z\). | [
29524,
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... | null | null |
numina_10050537 | Solution.
$$
\begin{aligned}
& \frac{\left(\sin ^{2} \alpha+\tan ^{2} \alpha+1\right)\left(\cos ^{2} \alpha-\cot ^{2} \alpha+1\right)}{\left(\cos ^{2} \alpha+\cot ^{2} \alpha+1\right)\left(\sin ^{2} \alpha+\tan ^{2} \alpha-1\right)}= \\
& =\frac{\left(\sin ^{2} \alpha+\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}+1\right)... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 3.048. $\frac{\left(\sin ^{2} \alpha+\tan^{2} \alpha+1\right)\left(\cos ^{2} \alpha-\cot^{2} \alpha+1\right)}{\left(\cos ^{2} \alpha+\cot^{2} \alpha+1\right)\left(\sin ^{2} \alpha+\tan^{2} \alpha-1\right)}=1$. | [
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15156,
46625... | null | null |
aops_410907 | Since $\cos 3x=4\cos^3 x-3\cos x$, we have
$A=(4\cos^29^\circ-3)(4\cos^227^\circ-3)={\cos(3\cdot 9^\circ)\over\cos 9^\circ}\cdot{\cos(3\cdot 27^\circ)\over\cos 27^\circ}$
$A={\cos 81^\circ\over\cos 9^\circ}={\sin 9^\circ\over\cos 9^\circ}=\tan 9^\circ$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Prove that\n\\[(4\\cos^2 9^\\circ - 3)(4\\cos^2 27^\\circ - 3) = \\tan 9^\\circ\\]",
"content_html": "Prove that<br>\n<img src=\"//latex.artofproblemsolving.com/6/a/a/6aa604b2a086744ce39d38fc3aad5b8d... | Prove that
\[
(4\cos^2 9^\circ - 3)(4\cos^2 27^\circ - 3) = \tan 9^\circ.
\] | [
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aops_2456171 | [quote=Quantum_fluctuations]Let $a_1, a_2, a_3, \cdots , a_n , b_1, b_2, b_3, \cdots b_n, c_1, c_2, c_3, \cdots c_n$ be $3n$ nonnegative real numbers.
Find the largest real constant $k$ for which the inequality
$$\left(a_1^2+ a_2^2+ a_3^2+ \cdots + a_n^2 \right) \left( b_1^2+ b_2^2+ b_3^2+ \cdots + b_n^2\right)\lef... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $a_1, a_2, a_3, \\cdots , a_n , b_1, b_2, b_3, \\cdots b_n, c_1, c_2, c_3, \\cdots c_n$ be $3n$ nonnegative real numbers. \n\nFind the largest real constant $k$ for which the inequality\n\n $$\\left(a_... | Let \(a_1,\dots,a_n,b_1,\dots,b_n,c_1,\dots,c_n\) be \(3n\) nonnegative real numbers. Find the largest real constant \(k\) for which the inequality
\[
\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)\left(\sum_{i=1}^n c_i^2\right)
\ge k\left(\sum_{i=1}^n a_i b_i c_i\right)^2
\]
always holds. | [
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397... | null | null |
aops_3320402 | [hide=Very very nice. Thank ytChen.][quote=ytChen][quote=sqing]Let $ a,b, c>0 $ and $ 8a^2+8b^2=c^2.$ Prove that$$ \frac{ab}{ c^2}+\frac{ c}{ a}+\frac{c}{b} \geq \frac{129}{16}$$[/quote]
[quote=sqing]Let $ a,b, c>0 $ and $ 3a^2+3b^2=c^2.$ Prove that
$$ \frac{ab}{ c^2}+\frac{ c}{ a}+\frac{c}{b} \geq \frac{1}{6}+2\sqr... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $ a,b, c>0 $ and $ 8a^2+8b^2=c^2.$ Prove that$$ \\frac{ab}{ c^2}+\\frac{ c}{ a}+\\frac{c}{b} \\geq \\frac{129}{16}$$",
"content_html": "Let <img src=\"//latex.artofproblemsolving.com/e/1/d/e1d... | Let \(a,b,c>0\) and \(8a^2+8b^2=c^2\). Prove that
\[
\frac{ab}{c^2}+\frac{c}{a}+\frac{c}{b}\ge\frac{129}{16}.
\] | [
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numina_10050558 | ## Solution.
$$
\begin{aligned}
& \sin ^{2}\left(\frac{\alpha}{2}+2 \beta\right)-\sin ^{2}\left(\frac{\alpha}{2}-2 \beta\right)=\frac{1-\cos (\alpha+4 \beta)}{2}-\frac{1-\cos (\alpha-4 \beta)}{2}= \\
& =\frac{1}{2}-\frac{\cos (\alpha+4 \beta)}{2}-\frac{1}{2}+\frac{\cos (\alpha-4 \beta)}{2}=\frac{1}{2}(\cos (\alpha-4 \... | \sin\alpha\sin4\beta | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 3.071. $\sin ^{2}\left(\frac{\alpha}{2}+2 \beta\right)-\sin ^{2}\left(\frac{\alpha}{2}-2 \beta\right)$. | [
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aops_1139038 | Let $P(x,y,z)$ be the inequality above.
$P(0,0,0)\Rightarrow\left(f(0)-\frac{1}{2}\right)^2\leq0$. Hence $f(0)=\frac{1}{2}$.
$P(1,1,1)\Rightarrow\left(f(1)-\frac{1}{2}\right)^2\leq0$. Hence $f(1)=\frac{1}{2}$.
$P(x,0,0)\Rightarrow f(x)\leq\frac{1}{2}\forall x \in \mathbb{R}$
$P(1,y,0)\Rightarrow f(y)\geq\frac{1}{2}\for... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all $f :\\mathbb{R} \\Rightarrow \\mathbb{R}$ such that\n$\\frac{f(xy)}{2}+\\frac{f(xz)}{2}-f(x)f(yz) \\geq \\frac{1}{4}$ $\\forall x,y,z \\in \\mathbb{R}$",
"content_html": "Find all <img src=\... | Find all functions \(f:\mathbb{R}\to\mathbb{R}\) such that
\[
\frac{f(xy)}{2}+\frac{f(xz)}{2}-f(x)f(yz)\geq\frac{1}{4}\qquad\forall x,y,z\in\mathbb{R}.
\] | [
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144... | null | null |
numina_10049636 | 4.44 Applying formulas (4.2), (4.3), (4.1), and (4.13) sequentially to the left side of the equation, we find
$$
A=\operatorname{tg} 2 \alpha+\operatorname{ctg} 2 \alpha+\operatorname{tg} 6 \alpha+\operatorname{ctg} 6 \alpha=\frac{\sin 2 \alpha}{\cos 2 \alpha}+\frac{\cos 2 \alpha}{\sin 2 \alpha}+
$$
$$
\begin{aligned... | \frac{8\cos^{2}4\alpha}{\sin12\alpha} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | $4.44 \operatorname{tg} 2 \alpha+\operatorname{ctg} 2 \alpha+\operatorname{tg} 6 \alpha+\operatorname{ctg} 6 \alpha=\frac{8 \cos ^{2} 4 \alpha}{\sin 12 \alpha}$.
$4.44 \tan 2 \alpha+\cot 2 \alpha+\tan 6 \alpha+\cot 6 \alpha=\frac{8 \cos ^{2} 4 \alpha}{\sin 12 \alpha}$. | [
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37... | null | null |
numina_10734432 | 122. Proof: Since
$$a^{2}+2 b^{2}+3=\left(a^{2}+b^{2}\right)+\left(b^{2}+1\right)+2 \geqslant \frac{2}{c}+2 b+2$$
Therefore,
$$\frac{1}{a^{2}+2 b^{2}+3} \leqslant \frac{c}{2(1+c+b c)}$$
Similarly, we have
$$\begin{array}{l}
\frac{1}{b^{2}+2 c^{2}+3} \leqslant \frac{a}{2(1+a+c a)} \\
\frac{1}{c^{2}+2 a^{2}+3} \leqslan... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "inequalities"
} | 122. ("Home of Math Olympiads" website, 2008. 04. 14, provided by polynasia) Let $a, b, c \in \mathbf{R}^{+}$, and $a b c=1$, then
$$\sum \frac{1}{a^{2}+2 b^{2}+3} \leqslant \frac{1}{2}$$ | [
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aops_80393 | Here's my Kalva version of a solution. :P (very brief)
[hide]Use arithmetic series sum formula (both versions). $\frac{n}{2}(29+2)=155\implies n=10$. Now, $\frac{10}{2}(2+29)=\frac{10}{2}(4+9d)\implies d=3$[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "In a given arithmetic sequence the first term is $2$, the last term is $29$, and the sum of all the terms is $155$. The common difference is:\r\n\r\n$\\text{(A)} \\ 3 \\qquad \\text{(B)} \\ 2 \\qquad \\tex... | In an arithmetic sequence the first term is \(2\), the last term is \(29\), and the sum of all the terms is \(155\). The common difference is:
\(\text{(A)}\ 3\qquad \text{(B)}\ 2\qquad \text{(C)}\ \dfrac{27}{19}\qquad \text{(D)}\ \dfrac{13}{9}\qquad \text{(E)}\ \dfrac{23}{38}\) | [
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aops_54962 | [quote="manuel"]Find n such that $\dfrac{1}{1+ \sqrt3} + \dfrac{1}{ \sqrt3 + \sqrt5}+...+ \dfrac{1}{\sqrt{2n-1}+\sqrt{2n+1}}=100$[/quote]
here is my idea:
\[ \dfrac{1}{1+ \sqrt{3}} + \dfrac{1}{ \sqrt{3} + \sqrt{5}}+...+ \dfrac{1}{\sqrt{2n-1}+\sqrt{2n+1}}=\dfrac{\sqrt{3}-1}{2}+\dfrac{\sqrt{5}-\sqrt{3}}{2}+...+\dfrac{\... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find n such that $\\dfrac{1}{1+ \\sqrt3} + \\dfrac{1}{ \\sqrt3 + \\sqrt5}+...+ \\dfrac{1}{\\sqrt{2n-1}+\\sqrt{2n+1}}=100$",
"content_html": "Find n such that <img src=\"//latex.artofproblemsolving.co... | Find \(n\) such that
\[
\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\cdots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}=100.
\] | [
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aops_1127658 |
Given the system 1 $\ \left\{\begin{array}{ll} x^{2}y - x^{2} - 16y = 56 \\ x^{2}y - 4x^{2} + y^{2} - 5y = 8 \end{array}\right.\ $
$\left\{\begin{array}{ll} x^{2}(y - 1) = 16 y + 56 \\ x^{2}(y - 4) = 8 +5y - y^{2}\end{array}\right.\ $
$x = 0$ is not a solution.
Dividing: $\frac{y-1}{y-4}=\frac{16y+56}{8 +5y - y^{2}}$... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Okay, so here I have two systems of equations to solve. I put them in a single post because they are very similar, and the methods to solve them are similar with one key difference (this is a hint to figur... | System 1:
\[
\begin{cases}
x^2y - x^2 - 16y = 56,\\[4pt]
x^2y - 4x^2 + y^2 - 5y = 8.
\end{cases}
\]
System 2:
\[
\begin{cases}
xy + 3x + y^2 + 2y = 5,\\[4pt]
x^2y + 4x^2 + 2xy^2 + 8xy + y^3 + 3y^2 - 3y = -10.
\end{cases}
\] | [
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... | null | null |
aops_290061 | I think this is a shorter way.
Without the current, he would have went 15 miles, but he only went 9. 15-9=6, and 6/3 for the mph of the current gives you the current is going at 2 mph.
Edit: Dope!!!! I forgot about that. GJ isabella. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Jim paddles a canoe upstream at a rate of 5 mph but travels a distance of only 9 miles in 3 hours. What was the rate of the current in miles per hour?",
"content_html": "Jim paddles a canoe upstream ... | Jim paddles a canoe upstream at a rate of 5 mph but travels a distance of only 9 miles in 3 hours. What was the rate of the current in miles per hour? | [
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... | null | null |
aops_2508673 | @Above I don't think that knowledge is necessary to solve the problem, unless I am missing smth.
[hide = My attempt]
Let $\cos A + \cos B + \cos C = x$ and $\sin A + \sin B + \sin C = y$.
The cosine difference identity and some factoring on the left side of the given equation yields
\begin{align*} \cos A \cos B+ \s... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $A, B$ and $C$ are real angles such that\n$$\\cos (B-C)+\\cos (C-A)+\\cos (A-B)=-3/2,$$\nfind\n$$\\cos (A)+\\cos (B)+\\cos (C)$$",
"content_html": "If <img src=\"//latex.artofproblemsolving.com/b/... | If \(A,B,C\) are real angles such that
\[
\cos(B-C)+\cos(C-A)+\cos(A-B)=-\tfrac{3}{2},
\]
find
\[
\cos A+\cos B+\cos C.
\] | [
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63714,
6855... | null | null |
aops_195017 | [quote="zaya_yc"]$ a,b,c\geq 0$
$ \frac {1 \plus{} a}{1 \plus{} b} \plus{} \frac {1 \plus{} b}{1 \plus{} c} \plus{} \frac {1 \plus{} c}{1 \plus{} a}\leq 3 \plus{} a^{2} \plus{} b^{2} \plus{} c^{2}$[/quote]
Let $ c \equal{} \max\{a,b,c\}.$ Then
$ \frac {1 \plus{} a}{1 \plus{} b} \plus{} \frac {1 \plus{} b}{1 \plus{... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$ a,b,c\\geq 0$ \r\n\r\n$ \\frac{1\\plus{}a}{1\\plus{}b}\\plus{}\\frac{1\\plus{}b}{1\\plus{}c}\\plus{}\\frac{1\\plus{}c}{1\\plus{}a}\\leq 3\\plus{}a^{2}\\plus{}b^{2}\\plus{}c^{2}$",
"content_html": ... | Let \(a,b,c\ge 0\). Prove
\[
\frac{1+a}{1+b}+\frac{1+b}{1+c}+\frac{1+c}{1+a}\le 3+a^{2}+b^{2}+c^{2}.
\] | [
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41443... | null | null |
numina_10098833 | Prove that for $y_{i}=\frac{1}{1+x_{i}}, i=1,2, \cdots, n+1$, then $\sum_{i=1}^{n+1} y_{i}=1$, let $s_{i}=\sum_{1 \leqslant j \leqslant n+1, j \neq i} y_{j}, p_{i}=$ $\prod_{1 \leqslant i \leqslant n+1, j \neq i} y_{j}$
By the arithmetic-geometric mean inequality, we have $\frac{1-y_{i}}{y_{i}}=\frac{s_{i}}{y_{i}} \geq... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 14 (1998 Canadian Olympiad Problem) Let $x_{1}, x_{2}, \cdots, x_{n+1}$ be positive real numbers satisfying the condition $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n+1}}=1$. Prove that: $x_{1} x_{2} \cdots x_{n+1} \geqslant n^{n-1}$. | [
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... | null | null |
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