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19. Let $x, y, z$ be positive real numbers, and $xyz=1$, prove: $\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+$ $\frac{z^{3}}{(1+x)(1+y)} \geqslant \frac{3}{4}$
(39th IMO Shortlist) | 19. From the generalization of Cauchy's inequality, we have
$$\begin{array}{l}
{\left[\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)}\right] \cdot} \\
{[(1+y)+(1+z)+(1+x)][(1+z)+(1+x)+(1+y)] \geqslant} \\
(x+y+z)^{3}
\end{array}$$
Therefore,
$$\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 732,654 |
20. Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ positive real numbers, and $a_{1} a_{2} \cdots a_{n}=1$, prove that: $\sum_{i=1}^{n} \frac{a_{i}^{n}\left(1+a_{i}\right)}{A} \geqslant$ $\frac{n}{2^{n-1}}$, where $A=\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)$. (A generalization of a problem from the... | 20. Similar to Question 19, let $u=a_{1}+a_{2}+\cdots+a_{n}$, by the AM-GM inequality we have
$$\begin{array}{r}
u=a_{1}+a_{2}+\cdots+a_{n} \geqslant n \sqrt[n]{a_{1} a_{2} \cdots a_{n}}=n \\
\sum_{i=1}^{n} \frac{a_{i}^{n}\left(1+a_{i}\right)}{A} \geqslant \frac{u^{n}}{(n+u)^{n-1}}
\end{array}$$
Let $f(u)=\frac{u^{n}}... | \frac{n}{2^{n-1}} | Inequalities | proof | Yes | Yes | inequalities | false | 732,655 |
21. If $a, b, c$ are positive real numbers, prove:
$$3(a+\sqrt{a b} \sqrt[3]{a b c}) \leqslant\left(8+\frac{2 \sqrt{a b}}{a+b}\right) \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}$$
(Generalization of Kiran - Kellaya Inequality) | 21. By Hölder's inequality, we have
$$\begin{array}{l}
a+\sqrt{a b}+\sqrt[3]{a b c}=a^{\frac{1}{3}} a^{\frac{1}{3}} a^{\frac{1}{3}}+a^{\frac{1}{3}}(\sqrt{a b})^{\frac{1}{3}} b^{\frac{1}{3}}+a^{\frac{1}{3}} b^{\frac{1}{3}} c^{\frac{1}{3}} \leqslant \\
(a+a+a)^{\frac{1}{3}}(a+\sqrt{a b}+b)^{\frac{1}{3}}(a+b+c)^{\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,656 |
22. For all positive real numbers $a, b, c, \lambda \geqslant 8$, prove: For positive numbers $a, b, c$, we have
$$\frac{a}{\sqrt{a^{2}+\lambda b c}}+\frac{b}{\sqrt{b^{2}+\lambda c a}}+\frac{c}{\sqrt{c^{2}+\lambda a b}} \geqslant \frac{3}{\sqrt{1+\lambda}}$$
(Strengthening of a problem from the 42nd IMO) | 22. By the generalization of Cauchy's inequality, we get
$$\begin{aligned}
\text { LHS }= & \sum \frac{a}{\sqrt{a^{2}+\lambda b c}}=\sum \frac{a^{\frac{3}{2}}}{\sqrt{a^{3}+\lambda a b c}} \geqslant \\
& \frac{\left(\sum a\right)^{\frac{3}{2}}}{\left[\sum\left(a^{3}+\lambda a b c\right)\right]^{\frac{1}{2}}}
\end{aligne... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,657 |
23. Given that $x, y, z$ are positive real numbers, and $x+y+z=1$, prove: $\left(\frac{1}{x^{2}}-x\right)\left(\frac{1}{y^{2}}-y\right)\left(\frac{1}{z^{2}}-z\right) \geqslant\left(\frac{26}{3}\right)^{3} \cdot$ (Mathematics in Middle School, Issue 3, 2006) | 23.
$$\begin{array}{l}
\left(\frac{1}{x^{2}}-x\right)\left(\frac{1}{y^{2}}-y\right)\left(\frac{1}{z^{2}}-z\right)= \\
\left(\frac{1-x}{x}\right)\left(\frac{1-y}{y}\right)\left(\frac{1-z}{z}\right)\left(\frac{1+x+x^{2}}{x}\right)\left(\frac{1+y+y^{2}}{y}\right)\left(\frac{1+z+z^{2}}{z}\right)= \\
\left(\frac{y+z}{x}\rig... | \left(\frac{26}{3}\right)^{3} | Inequalities | proof | Yes | Yes | inequalities | false | 732,658 |
24. Given that $a, b$ are positive real numbers, prove: $\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}} \leqslant \sqrt[3]{2\left(1+\frac{b}{a}\right)\left(1+\frac{b}{a}\right)}$.
$(2002$ Macao Mathematical Olympiad Problem) | $$\begin{array}{l}
\text { 24. } \sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}} \leqslant \sqrt[3]{2\left(1+\frac{b}{a}\right)\left(1+\frac{b}{a}\right)} \Leftrightarrow \sqrt[3]{a^{2}}+\sqrt[3]{b^{2}} \leqslant \\
\sqrt[3]{2(a+b)^{2}} \Leftrightarrow\left(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}\right)^{3} \leqslant 2(a+b)^{2}
\e... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,659 |
25. Given $x_{i}, y_{i}, z_{i}(i=1,2,3)$ are positive real numbers, $M=\left(x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+1\right)\left(y_{1}^{3}+y_{2}^{3}+\right.$ $\left.y_{3}^{3}+1\right)\left(z_{1}^{3}+z_{2}^{3}+z_{3}^{3}+1\right), N=A\left(x_{1}+y_{1}+z_{1}\right)\left(x_{2}+y_{2}+z_{2}\right)\left(x_{3}+y_{3}+z_{3}\right)$, the... | 25. From the generalization of the Cauchy inequality, we have
$$\begin{aligned}
M= & \left(x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+1\right)\left(y_{1}^{3}+y_{2}^{3}+y_{3}^{3}+1\right)\left(z_{1}^{3}+z_{2}^{3}+z_{3}^{3}+1\right)= \\
& \left(x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+... | \frac{3}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,660 |
26. Let $a, b, c$ be positive real numbers, prove that: $\frac{a+b+c}{3} \geqslant \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}} \geqslant$ $\frac{\sqrt{a b}+\sqrt{b c}+\sqrt{c a}}{3}$. (2004 China National Training Team Problem) | 26. By the mean inequality,
$$\frac{a+b+c}{3}=\frac{\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}}{2} \geqslant \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}$$
Using Hölder's inequality,
$$\begin{array}{l}
\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}=\sqrt[3]{\frac{\frac{a+b}{2}+a+b}{3} \cdot \frac{b+\frac{b+c}{2}+c}{3} \cdot \frac{a+c+\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,661 |
41. Given that $a, b, c$ are real numbers, prove the inequality: $a^{4}\left(b^{2}+c^{2}\right)+b^{4}\left(c^{2}+a^{2}\right)+c^{4}\left(a^{2}+\right.$ $\left.b^{2}\right)+2 a b c\left(a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b-a^{3}-b^{3}-c^{3}-3 a b c\right) \geqslant 2\left(a^{3} b^{3}+\right.$ $\left.b^{3} c^{... | 41. $P(a, b, c)=a^{4}\left(b^{2}+c^{2}\right)+b^{4}\left(c^{2}+a^{2}\right)+c^{4}\left(a^{2}+b^{2}\right)+2 a b c\left(a^{2} b+a^{2} c+\right.$ $\left.b^{2} a+b^{2} c+c^{2} a+c^{2} b-a^{3}-b^{3}-c^{3}-3 a b c\right)-2\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}\right)$ is symmetric in $a$, $b$, and $c$. When $a=b$, $b=c$,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,662 |
27. Let $a, b, c$ be positive real numbers, and $a+b+c \geqslant \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$, prove: $\frac{a^{3} c}{b(a+c)}+$ $\frac{b^{3} a}{c(a+b)}+\frac{c^{3} b}{a(b+c)} \geqslant \frac{3}{2} .(2005$ Romanian Mathematical Olympiad Problem) | 27. From the generalization of Cauchy's inequality, we have
$$\begin{array}{c}
\left(\frac{a^{3} c}{b(a+c)}+\frac{b^{3} a}{c(a+b)}+\frac{c^{3} b}{a(b+c)}\right)\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}\right) \\
{[(a+c)+(a+b)+(b+c)] \geqslant(a+b+c)^{3}} \\
\left(\frac{a^{3} c}{b(a+c)}+\frac{b^{3} a}{c(a+b)}+\frac{c^{3... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,663 |
28. Let $a, b, c$ be positive real numbers, prove: $3(a+b+c) \geqslant 8 \sqrt[3]{a b c}+\sqrt[3]{\frac{a^{3}+b^{3}+c^{3}}{3}}$. (2006 Austrian Mathematical Olympiad) | 28. From the generalization of Cauchy's inequality, we have
$$\begin{array}{l}
\left(a b c+a b c+\cdots+a b c+\frac{a^{3}+b^{3}+c^{3}}{3}\right) \cdot \\
(1+1+\cdots+1+1)(1+1+\cdots+1+1) \geqslant \\
\left(\sqrt[3]{a b c}+\sqrt[3]{a b c}+\cdots+\sqrt[3]{a b c}+\sqrt[3]{\frac{a^{3}+b^{3}+c^{3}}{3}}\right)^{3}
\end{array... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,664 |
29. In the right-angled $\triangle A B C$, find the largest positive real number $k$ such that the inequality $a^{3}+b^{3}+c^{3} \geqslant k(a+$ $b+c)^{3}$ holds. (2006 Iran Mathematical Olympiad) | 29. Let $c$ be the largest side, then $c=\sqrt{a^{2}+b^{2}}$, and
$$a^{3}+b^{3}+c^{3}=a^{3}+b^{3}+2 \sqrt{2}\left(\sqrt{\frac{a^{2}+b^{2}}{2}}\right)^{3}$$
By the weighted power mean inequality, we have
$$\begin{aligned}
\sqrt[3]{\frac{a^{3}+b^{3}+2 \sqrt{2}\left(\sqrt{\frac{a^{2}+b^{2}}{2}}\right)^{3}}{1+1+2 \sqrt{2}... | \frac{1}{\sqrt{2}(1+\sqrt{2})^{2}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,665 |
30. Let $a, b, c$ be positive real numbers, prove: $\frac{a^{4}}{a^{4}+\sqrt[3]{\left(a^{6}+b^{6}\right)\left(a^{3}+c^{3}\right)^{2}}}+$ $\frac{b^{4}}{b^{4}+\sqrt[3]{\left(b^{6}+c^{6}\right)\left(b^{3}+a^{3}\right)^{2}}}+\frac{c^{4}}{c^{4}+\sqrt[3]{\left(c^{6}+a^{6}\right)\left(c^{3}+b^{3}\right)^{2}}} \leqslant 1$. (2... | 30. By the generalization of Cauchy's inequality, we have
$$\begin{array}{l}
\left(a^{6}+b^{6}\right)\left(a^{3}+c^{3}\right)^{2}=\left(a^{6}+b^{6}\right)\left(c^{3}+a^{3}\right)\left(c^{3}+a^{3}\right) \geqslant \\
\left(a^{2} \cdot c \cdot c+b^{2} \cdot a \cdot a\right)^{3}=a^{6}\left(b^{2}+c^{2}\right)^{3}
\end{arra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,666 |
31. Given that $x, y, z$ are positive numbers, prove that $\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}} \geqslant \sqrt{\frac{3}{2}(x+y+z)}$. (2005 Serbian Mathematical Olympiad Problem) | 31. By the generalization of Cauchy's inequality, we have
$$\left(\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}}\right)\left(\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\right.$$
$$\begin{array}{r}
\left.\frac{z}{\sqrt{x+y}}\right)[x(y+z)+y(z+x)+z(x+y)] \geqslant(x+y+z)^{3} . \\
\text { Also, since }(x+y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,667 |
32. Let the positive integer $n \geqslant 2, a_{1}, a_{2}, \cdots, a_{n}$ be $n$ non-negative real numbers, prove the inequality: $\left(a_{1}^{3}+\right.$ 1) $\left(a_{2}^{3}+1\right) \cdots\left(a_{n}^{3}+1\right) \geqslant\left(a_{1}^{2} a_{2}+1\right)\left(a_{2}^{2} a_{3}+1\right) \cdots\left(a_{n}^{2} a_{1}+1\righ... | 32. From the generalization of Cauchy's inequality, we have
\[
\left(a_{k}^{3}+1\right)\left(a_{k}^{3}+1\right)\left(a_{k+1}^{3}+1\right) \geqslant\left(a_{k}^{2} a_{k+1}+1\right)^{3}, \quad k=1,2, \cdots, n,
\]
where \(a_{n+1}=a_{1}\).
Multiplying them together, we get
\[
\prod_{k=1}^{n}\left(a_{k}^{3}+1\right)^{3} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,668 |
33. Let $a, b, c, x, y, z$ be positive numbers, prove: $\sqrt[3]{a(b+1) y z}+\sqrt[3]{b(c+1) z x}+$ $\sqrt[3]{c(a+1) x y} \leqslant \sqrt[3]{(a+1)(b+1)(c+1)(x+1)(y+1)(z+1)}$. (2005 Ukrainian Mathematical Olympiad Problem) | 33. By H\"older's inequality, we have
$$\begin{array}{l}
\sqrt[3]{a(b+1) y z}+\sqrt[3]{b(c+1) z x}+\sqrt[3]{c(a+1) x y}= \\
\sqrt[3]{a z \cdot y \cdot(b+1)}+\sqrt[3]{z \cdot(c+1) \cdot b x}+\sqrt[3]{(a+1) \cdot c y \cdot x} \leqslant \\
\sqrt[3]{(a z+z+(a+1))(y+(c+1)+c y)((b+1)+b x+x)}= \\
\sqrt[3]{(a+1)(b+1)(c+1)(x+1)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,669 |
34. Let $a_{1}, a_{2}, \cdots, a_{n}>0$, and $\sum_{i=1}^{n} a_{i}^{3}=3, \sum_{i=1}^{n} a_{i}^{5}=5$, prove: $\sum_{i=1}^{n} a_{i}>\frac{3}{2}$. (2001 Baltic Way Competition Problem) | 34. By Hölder's inequality, we have
$$\begin{array}{c}
\sum_{i=1}^{n} a_{i}^{3}=\sum_{i=1}^{n}\left(a_{i} \cdot a_{i}^{2}\right) \leqslant\left(\sum_{i=1}^{n} a_{i}^{\frac{5}{3}}\right)^{\frac{3}{5}}\left(\sum_{i=1}^{n}\left(a_{i}^{2}\right)^{\frac{5}{2}}\right)^{\frac{2}{5}}= \\
\left(\sum_{i=1}^{n} a_{i}^{\frac{5}{3}... | \sum_{i=1}^{n} a_{i} > \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,670 |
35. Let $a_{1}, a_{2}, \cdots, a_{n}>0$, then $\left.\prod_{i=1}^{n} \prod_{j=1}^{n}\left(1+\frac{a_{i}}{a_{j}}\right)\right|^{\frac{1}{n}} \geqslant 2^{n}$. (Generalization of a 1988 Australian Mathematical Olympiad problem) | 35. Fix $i$, by the generalization of Cauchy-Schwarz inequality we get
$$\prod_{j=1}^{n}\left(1+\frac{a_{i}}{a_{j}}\right) \geqslant\left[1+\frac{a i}{\left(\prod_{j=1}^{n} a_{j}\right)^{\frac{1}{n}}}\right]^{n}$$
Therefore,
$$\left\{\prod_{i=1}^{n} \prod_{j=1}^{n}\left(1+\frac{a_{i}}{a_{j}}\right)\right\}^{\frac{1}{n... | 2^{n} | Inequalities | proof | Yes | Yes | inequalities | false | 732,671 |
36. Let $a, b, c>0$, and $a b+b c+c a=1$, prove: $\sqrt[3]{\frac{1}{a}+6 b}+\sqrt[3]{\frac{1}{b}+6 c}+$ $\sqrt[3]{\frac{1}{c}+6 a} \leqslant \frac{1}{a b c}$. (45th IMO Shortlist) | 36. From the generalization of Cauchy's inequality, we have
$$\begin{array}{l}
(1+1+1)(1+1+1)\left[\left(\frac{1}{a}+6 b\right)+\left(\frac{1}{b}+6 c\right)+\left(\frac{1}{c}+6 a\right)\right] \geqslant \\
\left(\sqrt[3]{\frac{1}{a}+6 b}+\sqrt[3]{\frac{1}{b}+6 c}+\sqrt[3]{\frac{1}{c}+6 a}\right)^{3}
\end{array}$$
That... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,672 |
42. Given real numbers $x, y, z$ satisfying $x y z = -1$, prove that: $x^{4} + y^{4} + z^{4} + 3(x + y + z) \geqslant \frac{x^{2}}{y} + \frac{x^{2}}{z} + \frac{y^{2}}{x} + \frac{y^{2}}{z} + \frac{z^{2}}{x} + \frac{z^{2}}{y}$. (2004 Iran Mathematical Olympiad Problem) | 42. Since $x y z=-1$, then
$$\begin{array}{l}
x^{4}+y^{4}+z^{4}+3(x+y+z)-\left(\frac{x^{2}}{y}+\frac{x^{2}}{z}+\frac{y^{2}}{x}+\frac{y^{2}}{z}+\frac{z^{2}}{x}+\frac{z^{2}}{y}\right)= \\
x^{4}+y^{4}+z^{4}-3(x+y+z) x y z-\frac{x^{2}(y+z)}{y z}-\frac{y^{2}(z+x)}{z x}-\frac{z^{2}(x+y)}{x y}= \\
x^{4}+y^{4}+z^{4}-3(x+y+z) x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,673 |
37. Given $x \geqslant 0, y \geqslant 0, z \geqslant 0$, prove the inequality: $8\left(x^{3}+y^{3}+z^{3}\right)^{2} \geqslant 9\left(x^{2}+y z\right) \left(y^{2}+z x\right)\left(z^{2}+x y\right) \cdot(1982$ German National Team Problem) | $$\begin{array}{l}
\text { 37. } 9\left(x^{2}+y z\right)\left(y^{2}+z x\right)\left(z^{2}+x y\right) \leqslant \\
\frac{9}{8}\left(2 x^{2}+y^{2}+z^{2}\right)\left(x^{2}+2 y^{2}+z^{2}\right)\left(x^{2}+y^{2}+2 z^{2}\right) \leqslant \\
\frac{9}{8}\left(\frac{4\left(x^{2}+y^{2}+z^{2}\right)}{3}\right)^{3}=9 \times 8\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,674 |
38. Let $a, b, c, d>0$, and $a^{2}+b^{2}=\left(c^{2}+d^{2}\right)^{3}$, prove: $\frac{c^{3}}{a}+\frac{d^{3}}{b} \geqslant 1$. (2000 Singapore Mathematical Olympiad Problem) | 38. From the generalization of the Cauchy-Schwarz inequality, we have $\left(\frac{c^{3}}{a}+\frac{d^{3}}{b}\right)\left(\frac{c^{3}}{a}+\frac{d^{3}}{b}\right)\left(a^{2}+b^{2}\right) \geqslant\left(c^{2}+d^{2}\right)^{3}$, and since $a^{2}+b^{2}=\left(c^{2}+d^{2}\right)^{3}$, it follows that $\frac{c^{3}}{a}+\frac{d^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,675 |
39. Given that $a, b, c$ are non-negative real numbers, prove that: $\left(\frac{a+2 b}{a+2 c}\right)^{3}+\left(\frac{b+2 c}{b+2 a}\right)^{3}+\left(\frac{c+2 a}{c+2 b}\right)^{3} \geqslant 3$. (2004 MOP Problem) | 39. By the generalization of Cauchy's inequality, we have
$$\begin{array}{l}
\left(1^{3}+1^{3}+1^{3}\right)\left(1^{3}+1^{3}+1^{3}\right)\left[\left(\frac{a+2 b}{a+2 c}\right)^{3}+\left(\frac{b+2 c}{b+2 a}\right)^{3}+\left(\frac{c+2 a}{c+2 b}\right)^{3}\right] \geqslant \\
\left(\frac{a+2 b}{a+2 c}+\frac{b+2 c}{b+2 a}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,676 |
40. Given that $a, b, c$ are positive numbers,
$$\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geqslant \frac{3}{\sqrt[3]{a b c}(1+\sqrt[3]{a b c})}$$
(2006 Balkan Mathematical Olympiad problem (generalization of Aassila's inequality)) | 40. Let $P=\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}$, by the inequality $(x+y+z)^{2} \geqslant$ $3(x y+y z+z x)$ we get
$$\begin{aligned}
P^{2} \geqslant & 3\left[\frac{1}{a b(1+b)(1+c)}+\frac{1}{b c(1+c)(1+a)}+\frac{1}{c a(1+a)(1+b)}\right]= \\
& \frac{3[a(1+b)+b(1+c)+c(1+a)]}{a b c(1+a)(1+b)(1+c)}= \\
& \fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,677 |
41. Given that $a, b, c$ are non-negative real numbers, prove that: $\sqrt[3]{a^{3}+7 a b c}+\sqrt[3]{b^{3}+7 a b c}+\sqrt[3]{c^{3}+7 a b c} \leqslant$ $2(a+b+c) \cdot$(2007 Poland and other countries' joint Mathematical Olympiad problem) | 41. By Hölder's inequality, we have
$$\begin{array}{l}
\sqrt[3]{a^{3}}+7 a b c+\sqrt[3]{b^{3}+7 a b c}+\sqrt[3]{c^{3}+7 a b c} \leqslant \\
\left(1^{3}+1^{3}+1^{3}\right)^{\frac{1}{3}}\left[\left(\sqrt[3]{a^{3}+7 a b c}\right)^{\frac{3}{2}}+\left(\sqrt[3]{b^{3}+7 a b c}\right)^{\frac{3}{2}}+\left(\sqrt[3]{c^{3}+7 a b c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,678 |
42. $\alpha, \beta, x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 1)$ are positive numbers, and $x_{1}+x_{2}+\cdots+x_{n}=1$, prove the inequality: $\frac{x_{1}^{3}}{\alpha x_{1}+\beta x_{2}}+\frac{x_{2}^{3}}{\alpha x_{2}+\beta x_{3}}+\cdots+\frac{x_{n}^{3}}{\alpha x_{n}+\beta x_{1}} \geqslant \frac{1}{n(\alpha+\beta)} \cdot... | 42. By the generalization of Cauchy's inequality, we have
$$\begin{array}{l}
\left(\frac{x_{1}^{3}}{\alpha x_{1}+\beta x_{2}}+\frac{x_{2}^{3}}{\alpha x_{2}+\beta x_{3}}+\cdots+\frac{x_{n}^{3}}{\alpha x_{n}+\beta x_{1}}\right) \\
{\left[\left(\alpha x_{1}+\beta x_{2}\right)+\left(\alpha x_{2}+\beta x_{3}\right)+\cdots+\... | \frac{1}{n(\alpha+\beta)} | Inequalities | proof | Yes | Yes | inequalities | false | 732,679 |
46. Given that $a, b, c$ are positive numbers, and $a b + b c + c a \leqslant 3 a b c$, prove: $\sqrt{\frac{a^{2}+b^{2}}{a+b}}+\sqrt{\frac{b^{2}+c^{2}}{b+c}}+$ $\sqrt{\frac{c^{2}+a^{2}}{c+a}}+3 \leqslant \sqrt{2(a+b)}+\sqrt{2(b+c)}+\sqrt{2(c+a)}$. (2009 IMO Shortlist, 2010 Iran National Training Team Problem) | 46. By Cauchy-Schwarz inequality (square mean is no less than arithmetic mean),
$$\begin{array}{l}
\sqrt{2} \sqrt{a+b}=2 \sqrt{\frac{a b}{a+b}} \sqrt{\frac{1}{2}\left(2+\frac{a^{2}+b^{2}}{a b}\right)} \geqslant \\
2 \sqrt{\frac{a b}{a+b}} \cdot \frac{1}{2}\left(\sqrt{2}+\sqrt{\frac{a^{2}+b^{2}}{a b}}\right)=\sqrt{\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,683 |
43. Given that $a, b, c$ are real numbers, prove: $a^{2}+b^{2}+c^{2}-(a b+b c+c a) \geqslant 3(b-c)(a-b)$. (1997 Spanish Mathematical Olympiad problem) | 43. Since $2\left(a^{2}+b^{2}+c^{2}-(a b+b c+c a)\right)-6(b-c)(a-b)=(a-b)^{2}+$ $(b-c)^{2}+(c-a)^{2}-6(b-c)(a-b)=(a-b)^{2}+(b-c)^{2}+[(a-b)+$ $(b-c)]^{2}-6(b-c)(a-b)=2\left[(a-b)^{2}+(b-c)^{2}-2(b-c)(a-b)\right]=$ $(a-2 b+c)^{2} \geqslant 0$. Therefore, the original inequality holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,685 |
48. Let $x, y, z$ be positive real numbers, prove: $\sqrt{x^{2}+y^{2}}+\sqrt{y^{2}+z^{2}}+\sqrt{z^{2}+x^{2}} \leqslant 3 \sqrt{2} \cdot \frac{x^{3}+y^{3}+z^{3}}{x^{2}+y^{2}+z^{2}}$ (2010 Czech and Slovak Mathematical Olympiad Problem) | 48. By the power mean inequality, we have $\sqrt{\frac{a^{2}+b^{2}}{2}} \leqslant \sqrt[3]{\frac{a^{3}+b^{3}}{2}}$, so $\left(\sqrt{\frac{a^{2}+b^{2}}{2}}\right)^{3} \leqslant \frac{a^{3}+b^{3}}{2}$. Therefore, by the mean inequality, we get
$$\begin{array}{l}
\sqrt{x^{2}+y^{2}}\left(x^{2}+y^{2}+z^{2}\right)=2 \sqrt{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,687 |
50. Let $a, b, c, d, e$ be positive real numbers, and $a^{3}+a b+b^{3}=c+d=1$, prove: $\sum_{a c}\left(a+\frac{1}{a}\right)^{3} \geqslant$
40. (2003 Greek Mathematical Olympiad Problem)
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result di... | 50. Since $c+d=1$, by the AM-GM inequality $(c+d)\left(\frac{l}{c}+\frac{1}{d}\right) \geqslant 4$, and by the power mean inequality we get
$$\begin{array}{l}
\left(c+\frac{1}{c}\right)^{3}+\left(d+\frac{1}{d}\right)^{3} \geqslant \frac{1}{4}\left[\left(c+\frac{1}{c}\right)+\left(d+\frac{1}{d}\right)^{2}\right]^{3}= \\... | 40 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,689 |
51. Given that $x, y, z$ are positive numbers, prove that $\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{z^{2}+x^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}>2$. (2005 Macau Mathematical Olympiad Problem) | 51. By Hölder's inequality, we have
$$\begin{array}{l}
\left(\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{z^{2}+x^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}\right)^{2}\left[x\left(y^{2}+z^{2}\right)+y\left(z^{2}+x^{2}\right)+z\left(x^{2}+y^{2}\right)\right] \geqslant \\
(x+y+z)^{3}
\end{array}$$
It suffices to prove that
$$\be... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,690 |
52. Given that $a, b, c$ are positive numbers, prove: $\left(\frac{2 a}{b+c}\right)^{\frac{2}{3}}+\left(\frac{2 b}{c+a}\right)^{\frac{2}{2}}+\left(\frac{2 c}{a+b}\right)^{\frac{2}{3}} \geqslant 3$. (2002 USA Math MOP Summer Camp Problem) | 52: By Hölder's inequality, we have
$$\begin{array}{l}
{\left[\left(\frac{2 a}{b+c}\right)^{\frac{2}{3}}+\left(\frac{2 b}{c+a}\right)^{\frac{2}{3}}+\left(\frac{2 c}{a+b}\right)^{\frac{2}{3}}\right]^{3} \cdot} \\
{\left[(2 a)^{2}(b+c)^{2}+(2 b)^{2}(c+a)^{2}+(2 c)^{2}(a+b)^{2}\right] \geqslant} \\
{[2(a+b+c)]^{4}}
\end{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,691 |
Example 1 Prove: For all positive numbers $a, b, c$, we have
$$\frac{1}{a^{3}+b^{3}+a b c}+\frac{1}{b^{3}+c^{3}+a b c}+\frac{1}{c^{3}+a^{3}+a b c} \leqslant \frac{1}{a b c}$$
(26th United States of America Mathematical Olympiad problem); | Prove that since $a, b, c$ are positive real numbers, we have
$$a^{3}+b^{3} \geqslant a^{2} b+a b^{2}, b^{3}+c^{3} \geqslant b^{2} c+b c^{2}, c^{3}+a^{3} \geqslant c^{2} a+c a^{2}$$
Therefore,
$$\begin{array}{l}
\frac{1}{a^{3}+b^{3}+a b c}+\frac{1}{b^{3}+c^{3}+a b c}+\frac{1}{c^{3}+a^{3}+a b c} \leqslant \\
\frac{1}{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,692 |
Example 2 $a, b, c$ are positive real numbers, and $a b c=1$, prove: $\frac{a b}{a^{5}+b^{5}+a b}+\frac{b c}{b^{5}+c^{5}+b c}+$ $\frac{c a}{c^{5}+a^{5}+c a} \leqslant 1$. (37th IMO Shortlist Problem) | Prove that since $a, b, c$ are positive real numbers, we have
$$\begin{array}{l}
a^{5}+b^{5} \geqslant a^{3} b^{2}+a^{2} b^{3} \\
b^{5}+c^{5} \geqslant b^{3} c^{2}+b^{2} c^{3} \\
c^{5}+a^{5} \geqslant c^{3} a^{2}+c^{2} a^{3}
\end{array}$$
Also, since $a b c=1$, we have
$$a^{5}+b^{5}+a b=a^{5}+b^{5}+a^{2} b^{2} c \geqs... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,693 |
Example 3 Given that $a, b$ are positive numbers, $n$ is a positive integer, prove: $\frac{a^{n}+b^{n}}{2} \geqslant\left(\frac{a+b}{2}\right)^{n}$. (1975 Soviet Union University Mathematics Competition Problem) | Prove that by the binomial theorem,
$$(a+b)^{n}=\sum_{k=0}^{n} a^{k} b^{n-k}=\sum_{k=0}^{n} \mathrm{C}_{n}^{k} a^{n-k} b^{k}$$
Therefore,
$$\begin{array}{l}
2(a+b)^{n}=\sum_{k=0}^{n} \mathrm{C}_{n}^{k}\left(a^{k} b^{n-k}+a^{n-k} b^{k}\right) \leqslant \\
\sum_{k=0}^{n} \mathrm{C}_{n}^{k}\left(a^{n}+b^{n}\right)=\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,694 |
44. Given that $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ are acute angles, prove the inequality: $\left(\frac{1}{\sin \alpha_{1}}+\frac{1}{\sin \alpha_{2}}+\cdots+\right.$
$$\left.\frac{1}{\sin \alpha_{n}}\right)\left(\frac{1}{\cos \alpha_{1}}+\frac{1}{\cos \alpha_{2}}+\cdots+\frac{1}{\cos \alpha_{n}}\right) \leqsla... | 44. Let $a_{i}=\sin \alpha_{i}, b_{i}=\cos \alpha_{i}(i=1,2, \cdots, n)$, then $a_{i}^{2}+b_{i}^{2}=1, i=1,2, \cdots, n$, $a_{i} b_{i}+a_{j} b_{i}=\sin \left(\alpha_{i}+\alpha_{j}\right) \leqslant 1$, from $\left(a_{i}-b_{i}\right)^{2} \geqslant 0$ we get $1 \geqslant 2 a_{i} b_{i}$, the original inequality is equivale... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,696 |
Example 5 A geometric sequence with the first term and common ratio both being positive numbers and an arithmetic sequence have equal first and last terms, respectively, then the sum of this geometric sequence is not greater than the sum of this arithmetic sequence. (1979 Shandong Province Mathematics Competition Quest... | Proof: Let the first term of a geometric sequence be $a$, the common ratio be $q$, and the number of terms be $n$, then the last term and the sum of this sequence are respectively
$$a_{n}=a q^{n-1}, S=a\left(1+q+q^{2}+\cdots+q^{n-1}\right)$$
For an arithmetic sequence with the first term $a$ and the common difference ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 732,697 |
Example 6 For any real numbers $a, b$, we have $\left(\frac{a+b}{2}\right)\left(\frac{a^{2}+b^{2}}{2}\right)\left(\frac{a^{3}+b^{3}}{2}\right)^{2} \leqslant \frac{a^{6}+b^{6}}{2}$. (1963 Polish Mathematical Competition Problem) | Prove that because $a^{6}+b^{6} \geqslant a^{4} b^{2}+a^{2} b^{4}$, we have $2\left(a^{6}+b^{6}\right) \geqslant\left(a^{4}+b^{4}\right)\left(a^{2}+b^{2}\right)$, so
$$\left(\frac{a^{2}+b^{2}}{2}\right)\left(\frac{a^{4}+b^{4}}{2}\right) \leqslant \frac{a^{6}+b^{6}}{2}$$
For any $a, b$, we have $a^{4}+b^{4} \geqslant a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,698 |
Example 7 Given positive real numbers $a, b, c, d$, prove:
$$\frac{a^{3}+b^{3}+c^{3}}{a+b+c}+\frac{b^{3}+c^{3}+d^{3}}{b+c+d}+\frac{c^{3}+d^{3}+a^{3}}{c+d+a}+\frac{d^{3}+a^{3}+b^{3}}{d+a+b} \geqslant a^{2}+b^{2}+c^{2}+d^{2}$$
(US College Mathematics Competition Problem) | Prove that from $a^{3}+b^{3} \geqslant a^{2} b+a b^{2}, b^{3}+c^{3} \geqslant b^{2} c+b c^{2}, c^{3}+a^{3} \geqslant c^{2} b+c a^{2}$, we get
$$\left(a^{3}+b^{3}+c^{3}\right)(1+1+1) \geqslant\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)$$
Thus
Similarly,
$$\frac{a^{3}+b^{3}+c^{3}}{a+b+c} \geqslant \frac{a^{2}+b^{2}+c^{2}}{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,699 |
Example 8 Given that $a, b, c$ are positive numbers, prove that $\frac{1}{a}+\frac{1}{b} \pm \frac{1}{c} \leqslant \frac{a^{8}+b^{8}+c^{8}}{a^{3} b^{3} c^{3}}$ (1967 HMO Preliminary Question) | Prove that because $a^{8}+b^{8} \geqslant a^{6} b^{2}+a^{2} b^{6} b^{8}+c^{8} \geqslant b^{6} c^{2}+b^{2} c^{6} c^{8}+a^{8} \geqslant c^{6} a^{2}+a^{2} c^{6}$, so
$$2\left(a^{8}+b^{8}+c^{8}\right) \geqslant a^{2}\left(b^{6}+c^{6}\right)+b^{2}\left(c^{6}+a^{6}\right)+c^{2}\left(a^{6}+b^{6}\right)$$
Adding $a^{8}+b^{8}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,700 |
Example 9 Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, $\gamma=\alpha+\bar{\beta}, \alpha \beta>0$, then
$$\frac{1}{n}\left(a_{1}^{\gamma}+a_{2}^{\gamma}+\cdots+a_{n}^{\gamma}\right) \geqslant \frac{1}{n}\left(a_{1}^{\alpha}+a_{2}^{\alpha}+\cdots+a_{n}^{\alpha}\right) \cdot \frac{1}{n}\left(a_{1}^{\beta}... | Prove that because $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and $\alpha, \beta$ have the same sign, so
$$\begin{array}{l}
a_{1}^{\alpha+\beta}+a_{j}^{\alpha+\beta} \geqslant a_{1}^{\alpha} a_{j}^{\beta}+a_{1}^{\beta} a_{j}^{\alpha}, j=2,3, \cdots, n \\
a_{2}^{\alpha+\beta}+a_{j}^{\alpha+\beta} \geqslant... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,701 |
Example 10 If $a_{1}, a_{2}, \cdots, a_{n}$ are all positive numbers, $k$ is a positive integer, and let $a_{n+1}=a_{1}$, then
$$\sum_{i=1}^{n} \frac{a_{i}^{k+1}}{a_{i}^{k}+a_{i}^{k-1} a_{i+1}+\cdots+a_{i} a_{i+1}^{k}+a_{i+1}^{k}} \geqslant \frac{1}{k+1} \sum_{i=1}^{n} a_{i}$$ | Proof: Let
$$\begin{array}{l}
M=\sum_{i=1}^{n} \frac{a_{i}^{k+1}}{a_{i}^{k}+a_{i}^{k-1} a_{i+1}+\cdots+a_{i} a_{i+1}^{k}+a_{i+1}^{k}} \\
N=\sum_{i=1}^{n} \frac{a_{i+1}^{k+1}}{a_{i}^{k}+a_{i}^{k-1} a_{i+1}+\cdots+a_{i} a_{i+1}^{k}+a_{i+1}^{k}}
\end{array}$$
Then
$$M-N=\sum_{i=1}^{n} \frac{a_{i}^{k+1}-a_{i+1}^{k+1}}{a_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,702 |
Example 11 Positive real numbers $x, y, z$ satisfy $x y z \geqslant 1$, prove: $\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+$ $\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \geqslant 0$. (46th IMO problem) | Prove that the original inequality is equivalent to
$$\begin{array}{l}
\frac{x^{5}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}}{z^{5}+x^{2}+y^{2}} \geqslant \\
\frac{x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{2}}{z^{5}+x^{2}+y^{2}}
\end{array}$$
From $\frac{a^{2}}{b} \geqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,703 |
Example 12 Let $a, b, c$ be positive real numbers, and satisfy $abc=1$. Try to prove: $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+$ $\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}$. (36th IMO Problem) | To prove the inequality, we homogenize both ends, which is equivalent to proving:
$$\frac{1}{a^{3}(b c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2(a b c)^{\frac{4}{3}}}$$
Let \(a=x^{3}, b=y^{3}, c=z^{3}\), substituting into the above inequality, we get:
$$\sum_{c y c} \frac{1}{x^{9}\left(y^{3}+z^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,704 |
Example 13 Let $x, y, z$ be positive real numbers, prove that: $(x y+y z+z x)\left[\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\right.$ $\left.\frac{1}{(z+x)^{2}}\right] \geqslant \frac{9}{4}$. (1996 Iran Mathematical Olympiad Problem) | Prove that
$$\begin{array}{l}
4(x y+y z+z x)\left[(x+y)^{2}(y+z)^{2}+(y+z)^{2}(z+x)^{2}+(z+x)^{2}(x+y)^{2}\right]- \\
9(x+y)^{2}(y+z)^{2}(z+x)^{2}=4\left(x^{5} y+x y^{5}+y^{5} z+y z^{5}+z^{5} x+z x^{5}\right)- \\
\left(x^{4} y^{2}+x^{2} y^{4}+y^{4} z^{2}+y^{2} z^{4}+z^{4} x^{2}+z^{2} x^{4}\right)+ \\
2\left(x^{4} y z+x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,705 |
Example 14 Let $x, y, z$ be non-negative real numbers satisfying $x y+y z+z x=1$, prove: $\frac{1}{x+y}+\frac{1}{y+z}+$ $\frac{1}{z+x} \geqslant \frac{5}{2} \cdot(2006$ National Training Team Test Question) | $$\begin{array}{l}
(x y+y z+z x)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)^{2} \geqslant\left(\frac{5}{2}\right)^{2} \\
4 \sum_{s y m} x^{5} y+\sum_{s y m} x^{4} y z+14 \sum_{s y m} x^{3} y^{2} z+38 x^{2} y^{2} z^{2} \geqslant \\
\sum_{s y m} x^{4} y^{2}+3 \sum_{s y m} x^{3} y^{3} \Leftrightarrow \\
\left(\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,706 |
45. Given $a, b, c \in\left[\frac{1}{2}, 1\right]$, prove the inequalities:
(1) $\frac{a}{b}+\frac{b}{a} \leqslant \frac{5}{2}$;
(2) $\frac{a b+b c}{a^{2}+2 b^{2}+c^{2}} \geqslant \frac{2}{5}$. (2007 Shaanxi Province Mathematics Competition Problem) | 45. (1) Since $a, b \in\left[\frac{1}{2}, 1\right]$, therefore, $\frac{1}{2} b \in\left[\frac{1}{4}, \frac{1}{2}\right], 2 b \in[1,2]$, so $a \leqslant 2 b, a \geqslant \frac{1}{2} b$, i.e., $2 a \geqslant b$, thus $(a-2 b)(2 a-b) \leqslant 0$, i.e., $2\left(a^{2}+b^{2}\right) \leqslant 5 a b$. Dividing both sides by $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,707 |
1. If $a, b$ are positive numbers, and $a^{3}+b^{3}=2$, prove: $a+b \leqslant 2$. | 1. Since $a^{3}+b^{3} \geqslant a^{2} b+a b^{2},(a+b)^{3}=a^{3}+b^{3}+3\left(a^{2} b+a b^{2}\right) \leqslant 4\left(a^{3}+\right.$ $b^{3}$ ), therefore $a+b \leqslant 2$. | a+b \leqslant 2 | Inequalities | proof | Yes | Yes | inequalities | false | 732,708 |
2. Given two positive term sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ which are arithmetic and geometric sequences respectively, and $a_{1}=b_{1}=$ $a, a_{2}: b_{2}=b$, prove that when $n \geqslant 3$, $a_{n} \leqslant b_{n}$. | 2. Since $b_{n}=b_{1}\left(\frac{b}{a}\right)^{n-1}=a\left(\frac{b}{a}\right)^{n-1}, a_{n}=a_{1}+(n-1) d=a+(n-1)(b-$
$a)$, mathematical induction can be used to prove it, using $a^{n+1}+b^{n+1} \geqslant a b^{n}+a^{n} b$ in the process. | proof | Algebra | proof | Yes | Yes | inequalities | false | 732,709 |
3. Let $a>1, n \in \mathrm{N}^{*}$, prove: $\frac{n\left(a^{2 n+1}+1\right)}{a^{2 n}-1}>\frac{a}{a-1}$. | 3. To prove $n\left(a^{2 n+1}+1\right)>a\left(a^{2 n-1}+a^{2 n-2}+\cdots+a+1\right)$, since $a^{2 n+1}+1 \geqslant$ $a^{k}+a^{2 n+1-k}(k=0,1, \cdots, 2 n+1)$, adding them up yields the result. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,710 |
4. (1) If $a, b, c$ are positive numbers, prove:
$$\frac{a^{3}}{a^{2}+a b+b^{2}}+\frac{b^{3}}{b^{2}+b c+c^{2}}+\frac{c^{3}}{c^{2}+c a+a^{2}} \geqslant \frac{a+b+c}{3}$$
(2003 Beijing High School Mathematics Competition (Grade 1) Re-test)
(2) If $a, b, c$ are positive numbers, and $n$ is a positive integer, prove: $\fra... | 4. Directly obtained from Example 10. Another proof:
$$\begin{array}{l}
\sum_{o c} \frac{a^{3}}{a^{2}+a b+b^{2}}=\sum_{o c}\left[a-\frac{a b(a+b)}{a^{2}+a b+b^{2}}\right] \geqslant \\
\sum_{o c}\left[a-\frac{a b(a+b)}{3 a b}\right]=\frac{a+b+c}{3}
\end{array}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,711 |
5. (1) If $x, y, z$ are positive numbers, and satisfy $x y z=1$, prove: $\frac{x^{9}+y^{9}}{x^{6}+x^{3} y^{3}+y^{6}}+$ $\frac{y^{9}+z^{9}}{y^{6}+y^{3} z^{3}+z^{6}}+\frac{z^{9}+x^{9}}{z^{6}+z^{3} x^{3}+x^{6}} \geqslant 2$. (1997 Romanian Mathematical Olympiad Problem) | 5. (1) Since
$$\begin{array}{c}
x^{9}+y^{9}=\left(x^{3}+y^{3}\right)\left(x^{6}-x^{3} y^{3}+y^{6}\right) \\
\frac{x^{6}-x^{3} y^{3}+y^{6}}{x^{6}+x^{3} y^{3}+y^{6}}=1-\frac{2 x^{3} y^{3}}{x^{6}+x^{3} y^{3}+y^{6}} \geqslant 1-\frac{2 x^{3} y^{3}}{2 x^{3} y^{3}+x^{3} y^{3}}=\frac{1}{3}
\end{array}$$
Therefore,
$$\frac{x^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,712 |
6. (1) For any real numbers $x, y$, we have $2 x^{4}+2 y^{4} \geqslant x y(x+y)^{2} \cdot(1994$ th Russian 21st Mathematical Olympiad problem)
(2) For any real numbers $x, y$, and $n$ is a positive integer, we have $2 x^{2 n+2}+2 y^{2 n+2} \geqslant x y(x+y)^{2 n}$. | 6. When $x y \leqslant 0$, the inequality obviously holds. When $x, y$ are both positive (or both negative, using $-x, -y$ to replace $x, y$ respectively), since $x^{4}+y^{4} \geqslant x^{3} y+x y^{3}, x^{4}+y^{4} \geqslant 2 x^{2} y^{2}$, adding the two inequalities yields $2 x^{4}+$ $2 y^{4} \geqslant x y(x+y)^{2}$. | 2 x^{4}+2 y^{4} \geqslant x y(x+y)^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,714 |
7. Given that $p, q, r$ are positive numbers, satisfying $p q=F$, prove that for all $n \in \mathbf{N}^{*}$, we have $\frac{1}{p^{n}+q^{n}+1}+$ $\frac{1}{q^{n}+r^{n}+1}+\frac{1}{r^{n}+p^{n}+1} \leqslant 1 .(2004$ Baltic Way Mathematical Contest Problem) | 7. Let $p^{n}=a^{3}, q^{n}=b^{3}, r^{n}=c^{3}$, then the inequality can be transformed into
$$\frac{1}{a^{3}+b^{3}+1}+\frac{1}{b^{3}+c^{3}+1}+\frac{1}{c^{3}+a^{3}+1} \leqslant 1$$
Since $a, b, c$ are positive real numbers, we have $a^{3}+b^{3} \geqslant a^{2} b + a b^{2}, b^{3}+c^{3} \geqslant b^{2} c + b c^{2}, c^{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,715 |
8. (1) Given that $a, b, c$ are positive numbers, prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}$. (1963 Moscow Mathematical Olympiad Problem) | 8. (1) It is easy to see that the inequality to be proved is equivalent to $2\left(a^{3}+b^{3}+c^{3}\right) \geqslant a^{2} b+a^{2} c+b^{2} a+b^{2} c+$ $c^{2} a+c^{2} b$. Since $a, b, c$ are positive real numbers, we have
$$a^{3}+b^{3} \geqslant a^{2} b+a b^{2}, b^{3}+c^{3} \geqslant b^{2} c+b c^{2}, c^{3}+a^{3} \geqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,716 |
(2) Given that $a, b, c$ are positive numbers, prove that: $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c}{a+b} \geqslant \frac{a+b+c}{2}$ (2nd World Friendship Cup Mathematics Competition Problem.) | (2) Since $a, b, c$ are positive numbers, we have $a^{3}+b^{3} \geqslant a^{2} b+a b^{2}, a^{2}+b^{2} \geqslant 2 a b$, hence
$$\begin{aligned}
\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}= & \frac{a^{2}(c+a)+b^{2}(b+c)}{(b+c)(c+a)}=\frac{a^{3}+b^{3}+c\left(a^{2}+b^{2}\right)}{(b+c)(c+a)} \geqslant \\
& \frac{a^{2} b+a b^{2}+c(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,717 |
46. Given that $a, b$ are positive real numbers, and $a^{5}+b^{5}=a^{3}+b^{3}$, prove: $a^{2}+b^{2} \leqslant 1+a b$. (2003 Polish Mathematical Olympiad Problem) | 46. Since
$$\begin{aligned}
1+a b-\left(a^{2}+b^{2}\right)= & \frac{a^{5}+b^{5}}{a^{3}+b^{3}}+a b-\left(a^{2}+b^{2}\right)= \\
& \frac{a^{4} b+a b^{4}-a^{3} b^{2}-a^{2} b^{3}}{a^{3}+b^{3}}= \\
& \frac{a b(a-b)^{2}(a+b)}{a^{3}+b^{3}} \geqslant 0
\end{aligned}$$
Therefore | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,718 |
9. Given that $a, b$ are positive numbers, and $\bar{a}+b=1$, prove: $\frac{a^{2}}{a+1}+\frac{b^{2}}{b+1} \geqslant \frac{1}{3}$ (1996 Hungarian Mathematical Olympiad problem) | 9. Since $a+b=1$, then $\frac{a^{2}}{a+1}+\frac{b^{2}}{b+1} \geqslant \frac{1}{3} \Leftrightarrow \frac{a^{2}}{a(a+b)+(a+b)^{2}}+$ $\frac{b^{2}}{b a(a+b)+(a+b)^{2}} \geqslant \frac{1}{3} \Leftrightarrow a^{3}+b^{3} \geqslant a^{2} b+a b^{2}$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,719 |
12. Given that $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are all positive numbers, determine the largest real number $C$ such that the inequality $C\left(x_{1}^{2005}+x_{2}^{2005}+x_{3}^{2005}+x_{4}^{2005}+x_{5}^{2005}\right) \geqslant x_{1} x_{2} x_{3} x_{4} x_{5}\left(x_{1}^{125}+x_{2}^{125}+x_{3}^{125}+x_{4}^{125}+x_{5}^{... | 12 . From Example 9, we get
$$\begin{array}{l}
5\left(x_{\mathrm{L}}^{2005}+x_{2}^{2005}+x_{3}^{2005}+x_{4}^{2005}+x_{5}^{2005}\right) \geqslant \\
\left(x_{1}^{5}+x_{2}^{5}+x_{3}^{5}+x_{4}^{5}+x_{5}^{5}\right)\left(x_{1}^{2000}+x_{2}^{2000}+x_{3}^{2000}+x_{4}^{2000}+x_{5}^{2000}\right)
\end{array}$$
By the AM-GM ineq... | 5^{15} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,722 |
13. (1) $a, b, c$ are positive real numbers, and $abc=1, n$ is a positive integer, prove $\frac{ab}{a^{3n+2}+b^{3n+2}+ab}+$ $\frac{bc}{b^{3n+2}+c^{3n+2}+bc}+\frac{ca}{c^{3n+2}+a^{3n+2}+ca} \leqslant 1$. (Generalization of a problem from the 37th IMO Shortlist) | 13. (1) Since $a, b, c$ are positive real numbers, we have $a^{3 n+2}+b^{3 n+2} \geqslant a^{2 n+1} b^{n+1}+a^{n+1} b^{2 n+1}$, and since $a^{n} b^{n} c^{n}=1$, it follows that
$$a^{5}+b^{5}+a b=a^{5}+b^{5}+a^{2} b^{2} c \geqslant a^{3} b^{2}+a^{2} b^{3}+a^{2} b^{2} c=a^{2} b^{2}(a+b+c)$$
Therefore,
$$\frac{a b}{a^{3 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,723 |
(2) $a, b, c$ are positive real numbers, and $a b c=1, f(\alpha)=\frac{a b}{a^{\alpha}+b^{\alpha}+a b}+\frac{b c}{b^{\alpha}+c^{\alpha}+b c}+$ $\frac{c a}{c^{\alpha}+a^{\alpha}+c a}$, then when $\alpha\frac{1}{2}$, $f(\alpha) \leqslant 1$; when $\alpha=-1$ or $\alpha=\frac{1}{2}$, $f(\alpha)=1$; when $-1<\alpha<\frac{1... | (2) From the identity
$$b^{\alpha}+c^{\alpha}=\left(\sqrt[3]{b^{\alpha+1}}-\sqrt[3]{c^{\alpha+1}}\right)\left(\sqrt[3]{b^{2 \alpha-1}}-\sqrt[3]{c^{2 \alpha-1}}\right)+\sqrt[3]{b^{\alpha+1} c^{\alpha+1}}\left(\sqrt[3]{b^{\alpha-2}}+\sqrt[3]{c^{\alpha-2}}\right)$$
we know that when $\alpha > \frac{1}{2}$,
$$b^{\alpha}+c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,724 |
14. Find the smallest real number $m$, such that for any positive real numbers $a, b, c$ satisfying $a+b+c=1$, the inequality $m\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1$ holds. (2006 China Southeast Mathematical Olympiad) | 14. When $a=b=c=\frac{1}{3}$, we have $m \geqslant 27$.
Below, we prove the inequality $27\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1$ for any positive real numbers $a, b, c$ satisfying $a+b+c=1$.
Since $a, b, c$ are positive real numbers, we have $a^{3}+b^{3} \geqslant a^{2} b+a b^{2},... | 27 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,725 |
15. Find the real numbers $a, b, c$ that satisfy the following inequalities: $4(ab + bc + ca) - 1 \geqslant a^2 + b^2 + c^2 \geqslant 3(a^3 + b^3 + c^3) \cdot$ (2005 Australian National Training Team Problem) | 15. Since $3\left(a^{3}+b^{3}+c^{3}\right) \geqslant\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)$, from $a^{2}+b^{2}+$ $c^{2} \geqslant 3\left(a^{3}+b^{3}+c^{3}\right)$ we get $a+b+c \leqslant 1$, thus $(a+b+c)^{2} \leqslant 1$, which means
$$a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \leqslant 1$$
Also,
$$4(a b+b c+c a) \geqslant a^{... | a=b=c=\frac{1}{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,726 |
(2) Let $a, b, c$ be positive real numbers, and $a^{2}+b^{2}+c^{2}=1$, prove:
$$\frac{a^{5}+b^{5}}{a b(a+b)}+\frac{b^{5}+c^{5}}{b c(b+c)}+\frac{c^{5}+a^{5}}{c a(c+a)} \geqslant 6-5(a b+b c+c a)$$ | (2) $\frac{a^{5}+b^{5}}{a b(a+b)}=\frac{a^{4}+b^{4}-a b\left(a^{2}+b^{2}\right)+a^{2} b^{2}}{a+b}=$
$\frac{(a-b)^{4}+4 a b\left(a^{2}+b^{2}\right)-6 a^{2} b^{2}-a b\left(a^{2}+b^{2}\right)+a^{2} b^{2}}{a b} \geqslant$
$$\frac{3 a b\left(a^{2}+b^{2}\right)-5 a^{2} b^{2}}{a b}=3\left(a^{2}+b^{2}\right)-5 a b$$
Similarly... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,731 |
19. Let $x, y$ be positive numbers, and $n$ be a positive integer, prove: $\frac{x^{n}}{1+x^{2}}+\frac{y^{n}}{1+y^{2}} \leqslant \frac{x^{n}+y^{n}}{1+x y}$. (2008 Shaanxi Province Mathematics Competition Problem) | 19. Since $x, y$ are positive numbers, and $n$ is a positive integer, we have $x^{n}+y^{n} \geqslant x^{n-1} y+y^{n-1} x$. By the Cauchy-Schwarz inequality, we get $\left(1+x^{2}\right)\left(1+y^{2}\right) \geqslant(1+x y)^{2}$.
Thus,
$$\begin{aligned}
\frac{x^{n}}{1+x^{2}}+\frac{y^{n}}{1+y^{2}}= & \frac{x^{n}\left(1+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,732 |
20. Let $a, b, c$ be positive real numbers, and $abc=1$, prove: $a^{3}+b^{3}+c^{3} \geqslant ab+bc+ca$. (2005 Georgia Training Team Problem) | 20. $\begin{aligned} 3\left(a^{3}+b^{3}+c^{3}\right) \geqslant & (a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \geqslant \\ & 3 \sqrt[3]{a b c}(a b+b c+c a) \geqslant 3(a b+b c+c a)\end{aligned}$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,733 |
21. Given that $a, b, c$ are positive real numbers, and satisfy $a b c=1$, prove:
$$\frac{1}{a^{5}(b+2 c)^{2}}+\frac{1}{b^{5}(c+a)^{2}}+\frac{1}{c^{5}(a+b)^{2}} \geqslant \frac{1}{3}$$
(2010 USA National Training Team Problem) | 21. Let $a=x^{6}, b=y^{6}, c=z^{6}$, then $x y z=1$. By the Cauchy-Schwarz inequality, we have
$$\sum_{c y c} \frac{1}{a^{5}(b+2 c)^{2}}=\sum_{g c} \frac{y^{30} z^{30}}{\left(y^{6}+2 z^{6}\right)^{2}} \geqslant \frac{\left(\sum_{9 c} y^{15} z^{15}\right)^{2}}{\sum_{9 c}\left(y^{6}+2 z^{6}\right)^{2}}=\frac{\sum_{c y} y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,734 |
Example 1 Let $a, b, c$ be the sides of a triangle, prove that; $a^{2}(b+c-a)+b^{2}(c+a-b)+$ $c^{2}(a+b-c) \leqslant 3 a b c$. (6th IMO problem.) | Prove that by expanding both sides of the inequality, rearranging terms, and using transformation I, we can obtain.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | null | Inequalities | proof | Yes | Yes | inequalities | false | 732,735 |
Example 2 For $x, y, z \geqslant 0$, prove: the inequality $x(x-z)^{2}+y(y-z)^{2} \geqslant(x-z)(y-z)(x+y+z)$. (1992 Canadian Mathematical Olympiad) | Prove that by expanding both sides of the inequality, moving terms, and organizing, we get
$$x^{3}+y^{3}+z^{3}-\left(x^{2} y+x y^{2}+x^{2} z+x z^{2}+y^{2} z+y z^{2}\right)+3 x y z \geqslant 0$$
This is precisely the transformed form I of Schur's Inequality. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,736 |
Example 3 Let $x, y, z$ be positive numbers, and $x+y+z=xyz$. Prove: $x^{2}+y^{2}+z^{2}-2(xy+yz+$ $zx)+9 \geqslant 0 .(1993$ March "Mathematics Bulletin"/ Problem $)$ | Prove that since $x, y, z > 0$, and $x + y + z = xyz$, we have
$$\begin{array}{l}
x^{2} + y^{2} + z^{2} - 2(xy + yz + zx) + 9 \geqslant 0 \Leftrightarrow \\
\left(x^{2} + y^{2} + z^{2}\right)(x + y + z) - 2(xy + yz + zx)(x + y + z) + 9xyz \geqslant 0 \Leftrightarrow \\
x^{3} + y^{3} + z^{3} - \left(x^{2}y + xy^{2} + x^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,737 |
Example 6 In $\triangle A B C$, prove: $\frac{\sin ^{2} A}{a}+\frac{\sin ^{2} B}{b}+\frac{\sin ^{2} C}{c} \leqslant \frac{s^{2}}{a b c}$, where $s=\frac{a+b+c}{2}$. (2006 Taiwan Mathematical Olympiad Training Team Problem) | Proof from the Law of Sines:
$$\begin{array}{l}
\frac{\sin ^{2} A}{a}+\frac{\sin ^{2} B}{b}+\frac{\sin ^{2} C}{c} \leqslant \frac{s^{2}}{a b c} \Leftrightarrow \\
a b c\left(\frac{\sin ^{2} A}{a}+\frac{\sin ^{2} B}{b}+\frac{\sin ^{2} C}{c}\right) \leqslant \frac{1}{4}(a+b+c)^{2} \Leftrightarrow \\
a b c\left(\frac{a}{4... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,741 |
Example 8 Let $x, y, z$ be positive real numbers, and $x+y+z=1$, prove: $x^{2}+y^{2}+z^{2}+9 x y z \geqslant$ $2(xy+yz+zx)$. (2004 Nanchang City High School Mathematics Competition Problem) | Proof
Because
$$\begin{array}{l}
2(xy + yz + zx) = 2(xy + yz + zx)(x + y + z) = \\
\mathbf{6xyz} + 2x^2(y + z) + 2y^2(z + x) + 2z^2(x + y) \\
x^2 + y^2 + z^2 = (x^2 + y^2 + z^2)(x + y + z) = \\
x^3 + y^3 + z^3 + x^2(y + z) + \\
y^2(z + x) + z^2(x + y)
\end{array}$$
To prove
$$x^2 + y^2 + z^2 + 9xyz \geqslant 2(xy + y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,743 |
Example 10 Given that $a, b, c$ are non-negative real numbers, prove: $\frac{1}{3}\left[(a-b)^{2}+(b-c)^{2}+(c-\right.$ $\left.a)^{2}\right] \leqslant a^{2}+b^{2}+c^{2}-3 \sqrt[3]{a^{2} b^{2} c^{2}} \leqslant(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$. (2005 Irish Mathematical Olympiad) | Prove that by the AM-GM inequality, $a^{2}+b^{2}+c^{2}-3 \sqrt[3]{a^{2} b^{2} c^{2}} \geqslant 0$, then
$$\begin{array}{l}
\frac{1}{3}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]= \\
\frac{2}{3}\left(a^{2}+b^{2}+c^{2}\right)-\frac{2}{3}(a b+b c+c a) \leqslant \\
\frac{2}{3}\left(a^{2}+b^{2}+c^{2}\right)-\frac{2}{3} \cdot... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,745 |
Example 11 Let $a, b, c$ be positive real numbers, prove that: $\sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c})+(a+b+c)^{2} \geqslant$ $4 \sqrt{3 a b c(a+b+c)} \cdot$ (2004 China National Training Team Problem) | Prove that by substituting $x=\sqrt{a}, y=\sqrt{b}, z=\sqrt{c}$, the original inequality becomes
$$x y z(x+y+z)+\left(x^{2}+y^{2}+z^{2}\right)^{2} \geqslant 4 x y z \sqrt{3\left(x^{2}+y^{2}+z^{2}\right)}$$
Expanding, it suffices to prove
$$\begin{array}{l}
x^{4}+y^{4}+z^{4}+2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,746 |
Example 12 Proof: For any positive real numbers $a, b, c$, we have $\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geqslant$ $9(a b+b c+c a) .(2004$ Asia Pacific Mathematical Olympiad Problem) | To prove
$$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geqslant 9(a b+b c+c a)$$
It suffices to prove
$$a^{2} b^{2} c^{2}+2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+4\left(a^{2}+b^{2}+c^{2}\right)+8 \geqslant 9(a b+b c+c a)$$
By the AM-GM inequality, we have
$$a^{2}+b^{2} \geqslant 2 a b, \qu... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,747 |
Example 13 Let $x, y, z$ be positive real numbers, prove that: $\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}>2 \sqrt[3]{x^{3}+y^{3}+z^{3}}$. (2008 China National Training Team Problem) | Proof: Let $\frac{x y}{z} a^{2}, \frac{y z}{x} b^{2}, \frac{z x}{y} c^{2}$. Since $x, y, z$ are positive real numbers, we have $x=c a, y=a b, z=b c$.
The original inequality is transformed into proving
$$a^{2}+b^{2}+c^{2}>2 \sqrt[3]{a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}}$$
which is equivalent to proving
$$\begin{array}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,748 |
Example 14 Let $x, y, z$ be positive real numbers, prove that: $(x y+y z+z x)\left[\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\right.$ $\left.\frac{1}{(z+x)^{2}}\right] \geqslant \frac{9}{4} .$(1996 Iranian Mathematical Olympiad) | $$\begin{array}{l}
4(x y+y z+z x)\left[(x+y)^{2}(y+z)^{2}+(y+z)^{2}(z+x)^{2}+(z+x)^{2}(x+y)^{2}\right]= \\
4(x y+y z+z x)\left[\left(y^{2}+x y+y z+z x\right)^{2}+ \\
\left(z^{2}+x y+y z+z x\right)^{2}+\left(x^{2}+x y+y z+z x\right)^{2}\right]= \\
4(x y+y z+z x)\left[\left(x^{4}+y^{4}+z^{4}\right)+ \\
2\left(x^{2}+y^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,749 |
Example 15 For any positive real numbers $a, b, c$, we have
$$\frac{a^{3}}{b^{2}-b c+c^{2}}+\frac{b^{3}}{c^{2}-c a+a^{2}}+\frac{c^{3}}{a^{2}-a b+b^{2}} \geqslant a+b+c$$
(2006 Balkan Mathematical Olympiad) | $$\begin{array}{l}
\frac{a^{3}}{b^{2}-b c+c^{2}}+\frac{b^{3}}{c^{2}-c a+a^{2}}+\frac{c^{3}}{a^{2}-a b+b^{2}}= \\
\frac{a^{4}}{a\left(b^{2}-b c+c^{2}\right)}+\frac{b^{4}}{b\left(c^{2}-c a+a^{2}\right)}+\frac{c^{4}}{c\left(a^{2}-a b+b^{2}\right)} \geqslant \\
\frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a\left(b^{2}-b c+c^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,750 |
49. Let $x, y, z$ be positive real numbers, and $x+y+z=1$, prove: $x y+y z+z x \geqslant 4\left(x^{2} y^{2}+\right.$ $\left.y^{2} z^{2}+z^{2} x^{2}\right)+5 x y z$. And determine the condition for equality. (2006 Serbia Mathematical Olympiad Problem) | 49. Homogenization on both sides
$$\begin{array}{l}
x y+y z+z x \geqslant 4\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+5 x y z \Leftrightarrow \\
(x+y+z)^{2}(x y+y z+z x)-4\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)-5 x y(x+y+z) \geqslant 0 \Leftrightarrow \\
\left.x^{3} y+x y^{3}+y^{3} z+y z^{3}+z x+z x^{3}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,751 |
Example 16 Let $a, b, c$ be positive real numbers. When $a^{2}+b^{2}+c^{2}+4 a b c=4$, prove: $a+b+c \leqslant$
3. (2003 Iran Mathematical Olympiad) | Proof: We use proof by contradiction to prove: If $a+b+c>3$, then $a^{2}+b^{2}+c^{2}+4 a b c>4$.
By Schur's inequality, we have
$$\begin{array}{l}
2(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)+9 a b c-(a+b+c)^{3}= \\
a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b) \geqslant 0
\end{array}$$
Therefore,
$$2(a+b+c)\left(a^{2}+b^{2}+c^{2}\ri... | a+b+c \leqslant 3 | Inequalities | proof | Yes | Yes | inequalities | false | 732,752 |
(2) Let $x, y, z$ be non-negative real numbers, and $xy + yz + zx + xyz = 4$. Prove:
$$\frac{x}{\sqrt{y+z}} + \frac{y}{\sqrt{z+x}} + \frac{z}{\sqrt{x+y}} \geqslant \frac{\sqrt{2}}{2}(x+y+z)$$
(2007 Austrian-Polish Mathematical Olympiad Problem) | (2) From (1) we get
$$x+y+z \geqslant x y+y z+z x$$
By Cauchy-Schwarz inequality, we have
$$(x \sqrt{y+z}+y \sqrt{z+x}+z \sqrt{x+y})\left(\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}}\right) \geqslant(x+y+z)^{2}$$
By Cauchy-Schwarz inequality again, we get
$$\begin{array}{l}
x \sqrt{y+z}+y \sqrt{z+x}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,754 |
Example 18 Given that $a, b, c$ are all positive numbers, and $ab + bc + ca = 1$, prove: $\sqrt{a^{3}+a}+\sqrt{b^{3}+b}+$ $\sqrt{c^{3}+c} \geqslant 2 \sqrt{a+b+c}$. (2008 Iran National Team Selection Test) | Prove that because $ab + bc + ca = 1$, we have
$$\begin{array}{c}
a^{3} + a = a^{3} + a(ab + bc + ca) = a(a + b)(c + a) \\
b^{3} + b = b(a + b)(b + c) \\
c^{3} + c = c(c + a)(b + c)
\end{array}$$
The inequality is homogeneous, and the equivalent inequality is
$$\begin{array}{l}
\sqrt{a(a + b)(c + a)} + \sqrt{b(a + b)(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,755 |
1. Given that $a, b, c$ are the three sides of $\triangle ABC$, prove:
$$2<\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}-\frac{a^{3}+b^{3}+c^{3}}{a b c} \leqslant 3$$
(2001 Austria-Poland Mathematical Olympiad) | 1. Left side is equivalent to $(b+c-a)(c+a-b)(a+b-c)>0$, right side is equivalent to $(b+c-$ a) $(c+a-b)(a+b-c) \leqslant a b c$ (Schur's inequality). | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,756 |
2. In $\triangle A B C$, prove: $\frac{3\left(a^{4}+b^{4}+c^{4}\right)}{\left(a^{2}+b^{2}+c^{2}\right)^{2}}+\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \geqslant 2$. (2006 Costa Rican Mathematical Olympiad Problem- | 2. Eliminate the denominator
$$\begin{array}{l}
3\left(a^{4}+b^{4}+c^{4}\right)+\left(a^{2}+b^{2}+c^{2}\right)(a b+b c+c a)-2\left(a^{2}+b^{2}+c^{2}\right)^{2}= \\
a^{4}+b^{4}+c^{4}+\left(a^{3} b+a b^{3}+b^{3} c+b c^{3}+c^{3} a+c a^{3}\right)+ \\
a b c(a+b+c)-4\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)= \\
a^{4}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,757 |
3. Given that $a, b, c$ are positive numbers, and $a^{4}+b^{4}+c^{4}=3$, prove: $\frac{1}{4-a b}+\frac{1}{4-b c}+\frac{1}{4-c a} \leqslant$
1. (2005 Moldova Mathematical Olympiad Problem) | 3. From the known facts, we have
$$\begin{array}{c}
a b c \leqslant 1, a+b+c \leqslant 3 \\
a^{4}+b^{4}+c^{4} \geqslant \frac{1}{3}\left(a^{3}+b^{3}+c^{3}\right)(a+b+c) \\
a^{3}+b^{3}+c^{3}+6 a b c \geqslant a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}+3 a b c= \\
(a+b+c)(a b+b c+c a)
\end{array}$$
Inequality (1) i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,758 |
4. Given that $a, b, c$ are positive numbers, prove: $\sqrt{\frac{2}{3}+\frac{a b c}{a^{3}+b^{3}+c^{3}}}+\sqrt{\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a}} \geqslant 2$. (2006 Vietnam National Training Team Problem) | 4.
$$\begin{array}{l}
\sqrt{\frac{2}{3}+\frac{a b c}{a^{3}+b^{3}+c^{3}}}+\sqrt{\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a}} \geqslant 2 \Leftrightarrow \\
\sqrt{\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a}}-1 \geqslant 1-\sqrt{\frac{2}{3}+\frac{a b c}{a^{3}+b^{3}+c^{3}}} \Leftrightarrow \\
\left(\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,759 |
6. Let $a, b, c$ be positive numbers. Prove that: $\frac{a+b+c}{3}-\sqrt[3]{a b c} \leqslant \max (\sqrt{a}-\sqrt{b})^{2},(\sqrt{b}-$ $\left.\sqrt{c})^{2},(\sqrt{c}-\sqrt{a})^{2}\right)(2002$ US National Training Team Exam Question) | 6. To prove
$$\frac{a+b+c}{3}-\sqrt[3]{a b c} \leqslant \max (\sqrt{a}-\sqrt{b})^{2},(\sqrt{b}-\sqrt{c})^{2},(\sqrt{c}-\sqrt{a})^{2}$$
it suffices to prove $\square$
$$\frac{a+b+c}{3}-\sqrt[3]{a b c} \leqslant \frac{(\sqrt{a}-\sqrt{b})^{2}+(\sqrt{b}-\sqrt{c})^{2}+(\sqrt{c}-\sqrt{a})^{2}}{3}$$
which is equivalent to p... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,761 |
7. Let $\alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$, prove the inequality: $\frac{\sin \alpha \sin (\alpha-\beta) \sin (\alpha-\gamma)}{\sin (\beta+\gamma)}+$ $\frac{\sin \beta \sin (\beta-\alpha) \sin (\beta-\gamma)}{\sin (\gamma+\alpha)^{-}}+\frac{\sin \gamma \sin (\gamma-\alpha) \sin (\gamma-\beta)}{\sin... | $$\begin{array}{l}
\text { 7. Since } \sin (x+y) \sin (x-y)=(\sin x \cos y+\cos x \sin y)(\sin x \cos y- \\
\cos x \sin y)=\sin ^{2} x \cos ^{2} y-\sin ^{2} y \cos ^{2} x=\sin ^{2} x\left(1-\sin ^{2} y\right)-\sin ^{2} y\left(1-\sin ^{2} x\right)= \\
\sin ^{2} x-\sin ^{2} y
\end{array}$$
Therefore, the original inequa... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,763 |
8. Let $a, b, c$ be positive real numbers. Prove that: $\frac{(2 a+b+c)^{2}}{\left.2 a^{2}+(b+c)\right)^{2}}+\frac{(a+2 b+c)^{2}}{2 b^{2}+(c+a)^{2}}+$ $\frac{(a+b+2 c)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8 . \quad(2003$ USA Mathematical Olympiad Problem) | 8. For a function $f$ of $n$ variables, define its symmetric sum as
$$\sum_{s, m} f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum f\left(x_{\sigma(1)}, x_{\sigma(2)}, \cdots, x_{\sigma(n)}\right)$$
Here $\boldsymbol{\sigma}$ is a permutation of $1,2, \cdots, n$, and $s y m$ denotes the symmetric sum. For example, denot... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,764 |
9. $I$ is the incenter of $\triangle A B C$, prove: $I A^{2}+I B^{2}+I C^{2} \geqslant \frac{B C^{2}+C A^{2}+A B^{2}}{3}$ (1998 IMO problem) | 9 .
$$\begin{array}{c}
I A=r \csc \frac{A}{2}, I B=r \csc \frac{B}{2}, I C=r \csc \frac{C}{2} \\
B C=r\left(\cot \frac{B}{2}+\cot \frac{C}{2}\right) \\
C A=r\left(\cot \frac{C}{2}+\cot \frac{A}{2}\right) \\
A B=r\left(\cot \frac{A}{2}+\cot \frac{B}{2}\right)
\end{array}$$
$$\begin{array}{l}
I A^{2}+I B^{2}+I C^{2} \geq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,765 |
(2) $\frac{K A}{B C}, \frac{K B}{C A}, \frac{K C}{A B}$ among these, at least one is not less than $\frac{1}{\sqrt{3}}$. (2003 Vietnam National Team Selection Exam Problem) | (-2) By the median length formula $P K=\frac{1}{2} \sqrt{2\left(A K^{2}+B K^{2}\right)-A B^{2}}$, let the inradius of $\triangle A B C$ be $r$, then $P K=\frac{r}{\sin \frac{C}{2}}$, thus
$$2\left(A K^{2}+B K^{2}\right)=A B^{2}+4 P K^{2}=c^{2}+r^{2}+\frac{1}{4}(a+b-c)^{2}$$
Similarly,
$$\begin{array}{l}
2\left(B K^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,767 |
12. Given that $a, b ; c$ are positive numbers, prove: $3\left(a^{3}+b^{3}+c^{3}+a b c\right) \geqslant 4\left(a^{2} b+b^{2} c+c^{2} a\right)$ (2006 Ukrainian Mathematical Olympiad Problem) | 12. By Schur's inequality, we have
$$a^{3}+b^{3}+c^{3}+3 a b c \geqslant a^{2} b+a b^{2}+b^{2} c+b c^{2}+\dot{c}^{2} a+c a^{2}$$
By the AM-GM inequality, we have
$$\frac{a^{3}+a^{3}+b^{3}}{3} \geqslant a^{2} b, \frac{b^{3}+b^{3}+c^{3}}{3} \geqslant b^{2} c, \frac{c^{3}+c^{3}+a^{3}}{3} \geqslant c^{2} a$$
Adding these... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,769 |
13. Given that $x, y, z$ are non-negative numbers, and $x+y+z=2$, prove the inequality: $x^{2} y^{2}+y^{2} z^{2}+$ $z^{2} x^{2}+x y z \leqslant$ E $(2009$ Greek Mathematical Olympiad problem) | 13. Homogenization on both sides, equivalent proof
$$(x+y+z)^{4} \geqslant 16\left(-x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+8 x y z(x+y+z)$$
Because
$$\begin{aligned}
(x+y+z)^{4}= & x^{4}+y^{4}+z^{4}+4\left(x^{3} y+x y^{3}+y^{3} z+y z^{3}+z^{3} x+z x^{3}\right)+ \\
& 6\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,770 |
1 14. Given that $a, b, c$ are all positive numbers, $A=\frac{a+b+c}{3}, G=\sqrt[3]{a b c}, H=\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$, prove the inequality: $\left(\frac{A}{G}\right)^{3} \geqslant \frac{1}{4}+\frac{3}{4} \cdot \frac{A}{H} \cdot$ (1992 Polish Mathematical Olympiad Problem) | 14. Since $a, b, c$ are all positive numbers, $A=\frac{a+b+c}{3}, G=\sqrt[3]{a b c}, H=\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$, so
the inequality
$$\begin{array}{l}
\left(\frac{A}{G}\right)^{3} \geqslant \frac{1}{4}+\frac{3}{4} \cdot \frac{A}{H} \Leftrightarrow \\
\frac{(a+b+c)^{3}}{27 a b c} \geqslant \frac{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,771 |
15. Let $a, b, c$ be positive real numbers, prove: $a+b+c \leqslant \frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}+\frac{1}{2}\left(\frac{a b}{c}+\right.$ $\left.\frac{b c}{a}+\frac{c a}{b}\right) .(2009$ Oliforum Mathematical Olympiad Problem) | 15. Let $a=x y, b=y z, c=z x$, then the original inequality becomes
$$\frac{1}{2}\left(x^{2}+y^{2}+z^{2}\right)+x y z\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right) \geqslant x y+y z+z x$$
By the Cauchy-Schwarz inequality, we have
$$\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \geqslant \frac{9}{2(x+y+z)}$$
It su... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,772 |
51. Let $x, y, z$ be positive real numbers, $\sqrt{a}=x(y-z)^{2}, \sqrt{b}=y(z-x)^{2}, \sqrt{c}=z(x-y)^{2}$, prove that: $a^{2}+b^{2}+c^{2} \geqslant 2(a b+b c+c a) \cdot(2009$ China Southeast Mathematical Olympiad Problem) | 51. First, prove that $\sqrt{a}, \sqrt{b}, \sqrt{c}$ can form the sides of a triangle. The conditions are:
$$\begin{array}{l}
\sqrt{b}+\sqrt{c}-\sqrt{a}=-(y+z)(z-x)(x-y) \\
\sqrt{c}+\sqrt{a}-\sqrt{b}=-(z+x)(x-y)(y-z) \\
\sqrt{a}+\sqrt{b}-\sqrt{c}=-(x+y)(y-z)(z-x)
\end{array}$$
Therefore,
$$\begin{array}{l}
(\sqrt{b}+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,773 |
16: Let $a, b, c$ be positive real numbers, prove: $(b+c-a)(c+a-b)+(c+a-b)(-a+b-c)+(a+b-c)(b+c-a) \leqslant \sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c})$. (2001 Romanian National Training Team Problem) | 16. Since
$$\begin{array}{l}
(b+c-a)(c+a-b)+(c+a-b)(a+b-c)+(a+b-c)(b+c-a)= \\
2(a b+b c+c a)-\left(a^{2}+b^{2}+c^{2}\right)
\end{array}$$
By Schur's inequality and the AM-GM inequality,
$$2(a b+b c+c a)-\left(a^{2}+b^{2}+c^{2}\right) \leqslant \frac{9 a b c}{a+b+c} \leqslant 3 \sqrt[3]{a^{2} b^{2} c^{2}}$$
By the AM-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,774 |
17. Let $x, y, z$ be positive real numbers, prove that $\frac{x^{3}+y^{3}+z^{3}}{3 x y z}+\frac{3 \sqrt[3]{x y z}}{x+y+z} \geqslant 2$. (Mircea Lasscu Inequality) | 17. From the proof in question 6, we know that
$$x+y+z+3 \sqrt[3]{x y z} \geqslant 2(\sqrt{x y}+\sqrt{y z}+\sqrt{z x})$$
Thus,
$$3 \sqrt[3]{x y z} \geqslant 2(\sqrt{x y}+\sqrt{y z}+\sqrt{z x})-(x+y+z)$$
We have
$$\frac{3 \sqrt[3]{x y z}}{x+y+z} \geqslant \frac{2(\sqrt{x y}+\sqrt{y z}+\sqrt{z x})-(x+y+z)}{x+y+z}$$
Th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,775 |
18. Let $a, b, c$ be positive real numbers, find the value of $k$ such that
$$\left(k+\frac{a}{b+c}\right)\left(k+\frac{b}{c+a}\right)\left(k+\frac{c}{a+b}\right) \geqslant\left(k+\frac{1}{2}\right)^{3}$$
(2009 Vietnam National Team Selection Exam Problem) | 18. When $a=b, c \rightarrow 0$, the original inequality transforms to
$$(k+1)^{2} k \geqslant\left(k+\frac{1}{2}\right)^{3} \Rightarrow 2 k^{2}+k \geqslant \frac{3}{2} k^{2}+\frac{3}{4} k+\frac{1}{8} \Rightarrow 4 k^{2}+2 k-1 \geqslant 0$$
Solving, we get $k \geqslant \frac{\sqrt{5}-1}{4}$ or $k \leqslant-\frac{\sqrt... | \frac{\sqrt{5}-1}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,776 |
19. Given that $a, b, c$ are positive numbers, and $a+b+c=1$, prove:
$$\frac{1}{b c+a+\frac{1}{a}}+\frac{1}{c a+b+\frac{1}{b}}+\frac{1}{a b+c+\frac{1}{c}} \leqslant \frac{27}{31}$$
(2008 Serbian Mathematical Olympiad Problem) | 19. Since $a+b+c=1$, we have
$$
\frac{1}{b c+a+\frac{1}{a}}+\frac{1}{c a+b+\frac{1}{b}}+\frac{1}{a b+c+\frac{1}{c}} \leqslant \frac{27}{31} \Leftrightarrow
$$
$$
\frac{a}{a b c+a^{2}+1}+\frac{b}{a b c+b^{2}+1}+\frac{c}{a b c+c^{2}+1} \leqslant \frac{27}{31} \Leftrightarrow
$$
$$
\frac{a}{a b c+a^{2}+1}-a+\frac{b}{a b c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,777 |
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