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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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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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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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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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}
$$ | [
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aops_3202306 | [quote=williamxiao][hide=sol?]Subtract equation 1 from equation 2 to get $(y+x)(y-x) + (y-x)z = 1, (x+y+z)(y-x) = 1$
Same with 3 and 2, $(y+z)(y-z) + (y-z)x = 1$, $(y+z+x)(y-z) = 1$
So $(x+y+z)(y-x) = (x+y+z)(y-z)$. That means $(y-x) = (y-z)$, so $x = z$.
We then have $(2x+y)(y-x) = 1$. Next, we have by substituting x... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find $x^{2}+y^{2}+z^{2}$ given\n$$\\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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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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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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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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