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53. Given that $a, b, c$ are positive numbers, and $a^{2}+b^{2}+c^{2}=1$, prove: $\frac{1}{1-a b}+\frac{1}{1-b c}+\frac{1}{1-c a} \leqslant \frac{9}{2} \cdot($ Canada Crux) | $$\begin{array}{l}
\text { 53. } \frac{1}{1-a b}+\frac{1}{1-b c}+\frac{1}{1-c a} \leqslant \frac{9}{2} \Leftrightarrow 2((1-b c)(1-c a)+(1-c a)(1-a b)+ \\
(1-a b)(1-b c)] \leqslant 9(1-a b)(1-b c)(1-c a) \Leftrightarrow 3-5(a b+b c+c a)+7 a b c(a+ \\
b+c)-9 a^{2} b^{2} c^{2} \geqslant 0
\end{array}$$
$$a+b+c=(a+b+c)\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,217 |
55. Prove that for any positive real numbers $a, b, c$, we have $\sqrt{a^{2}+b^{2}-a b}+\sqrt{b^{2}+c^{2}-b c} \geqslant$ $\sqrt{a^{2}+c^{2}+a c}$. (31st IMO National Training Team Problem) | 55. It is sufficient to prove
$$\sqrt{a^{2}+b^{2}-a b} \geqslant \sqrt{a^{2}+c^{2}+a c}-\sqrt{b^{2}+c^{2}-b c}$$
Obviously,
$$(1) \Leftrightarrow a^{2}+b^{2}-a b \geqslant a^{2}+c^{2}+a c+b^{2}+c^{2}-b c-2 \sqrt{a^{2}+c^{2}+a c} \sqrt{b^{2}+c^{2}-b c}$$
That is $\square$
$$4\left(a^{2}+c^{2}+a c\right)\left(b^{2}+c^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,219 |
56. Given that $a, b, c$ are positive numbers, and $a b+b c+c a+2 a b c=1$, prove: $\frac{1}{4 a+1}+\frac{1}{4 b+1}+$ $\frac{1}{4 c+1} \geqslant 1$. (2005 Morocco Mathematical Olympiad Problem) | 56. $\frac{1}{4 a+1}+\frac{1}{4 b+1}+\frac{1}{4 c+1} \geqslant 1 \Leftrightarrow 1+2(a+b+c) \geqslant 32 a b c$.
From the given condition
$$a b+b c+c a+2 a b c=1 \geqslant 3 \sqrt[3]{a b \cdot b c \cdot c a}+2 a b c=3 \sqrt[3]{(a b c)^{2}}+2 a b c$$
Let $x=\sqrt[3]{a b c}$, then $3 x^{2}+2 x^{3}-1=(2 x-1)(x+1)^{2} \l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,220 |
57. Let $x, y, z$ be real numbers, prove that: $\frac{x^{2}-y^{2}}{2 x^{2}+1}+\frac{y^{2}-z^{2}}{2 y^{2}+1}+\frac{z^{2}-x^{2}}{2 z^{2}+1} \leqslant 0$. | 57. Let $a=x^{2}, b=y^{2}, c=z^{2}$, then the problem reduces to proving under the condition $a, b, c \geqslant 0$
$$\begin{array}{c}
\frac{a}{2 a+1}+\frac{b}{2 b+1}+\frac{c}{2 c+1} \leqslant \frac{b}{2 a+1}+\frac{c}{2 b+1}+\frac{a}{2 c+1} \\
(1) \Leftrightarrow 12 a b c+2(a b+b c+c a) \leqslant 4\left(a^{2} b+b^{2} c+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,221 |
3. Let $m$ be a positive integer, $x, y, z$ be positive numbers, and $xyz=1$. Prove that: $\frac{x^{m}}{(1+y)(1+z)}+$ $\frac{y^{m}}{(1+z)(1+x)}+\frac{z^{m}}{(1+x)(1+y)} \geqslant \frac{3}{4} \cdot(1999$ IMO Shortlist $)$ | 3. (1) When $m=1$, let $t=x+y+z$, then $x^{2}+y^{2}+z^{2} \geqslant \frac{1}{3}(x+y+z)^{2}=\frac{1}{3} t^{2}$, $x y+y z+z x \leqslant \frac{1}{3}(x+y+z)^{2}=\frac{1}{3} t^{2}$, the left side of the inequality $=\frac{x}{(1+y)(1+z)}+$ $\frac{y}{(1+z)(1+x)}+\frac{z}{(1+x)(1+y)}=\frac{(x+y+z)+\left(x^{2}+y^{2}+z^{2}\right... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 733,222 |
58. Given that $a, b, c$ are positive numbers, and $a b+b c+c a+2 a b c=1$, prove the inequality: $2(a+b+c)+1 \geqslant 32 a b c$. (2004 Hungarian - Mediterranean Mathematical Olympiad problem) | 58. By the AM-GM inequality,
$$a b+b c+c a+2 a b c=1 \geqslant 4 \sqrt[4]{a b \cdot b c \cdot c a \cdot 2 a b c}=4 \sqrt[4]{2(a b c)^{3}}$$
Therefore, $a b c \leqslant \frac{1}{8}, a b+b c+c a=1-2 a b c \geqslant 1-2 \times \frac{1}{8}=\frac{3}{4}, 32 a b c \leqslant 4$, to prove $2(a+b+c)+1 \geqslant 32 a b c$, it su... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,223 |
59. Find the largest positive real number $a$, such that $\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{z^{2}+x^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}>a$ holds for all positive real numbers $x, y, z$. (1994 Romanian National Training Team Problem) | 59. The maximum positive number $a=2$.
On the one hand, let $y=z, x \rightarrow 0^{+}$, then we know $\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{z^{2}+x^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}} \rightarrow 2$. Therefore, $a \leqslant 2$.
On the other hand, if $a=2$, we will prove that the original inequality holds. Therefo... | 2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,224 |
60. Given that $a, b, c, d$ are positive numbers, and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=4$, prove that $\sqrt[3]{\frac{a^{3}+b^{3}}{2}}+$ $\sqrt[3]{\frac{b^{3}+c^{3}}{2}}+\sqrt[3]{\frac{c^{3}+d^{3}}{2}}+\sqrt[3]{\frac{d^{3}+a^{3}}{2}} \leqslant 2(a+b+c+d)-4 .(2007$ Polish Mathematics | 60. First, prove a lemma using the analytical method: If $a, b$ are positive numbers, then
$$\sqrt[3]{\frac{a^{3}+b^{3}}{2}} \leqslant \frac{a^{2}+b^{2}}{a+b}$$
In fact,
$$\begin{array}{l}
\sqrt[3]{\frac{a^{3}+b^{3}}{2}} \leqslant \frac{a^{2}+b^{2}}{a+b} \Leftrightarrow 2\left(a^{2}+b^{2}\right)^{3} \geqslant\left(a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,225 |
61. Given real numbers $a, b, c, x, y, z$ satisfy $(a+b+c)(x+y+z)=3,\left(a^{2}+b^{2}+c^{2}\right)$ $\left(x^{2}+y^{2}+z^{2}\right)=4$, prove that: $a x+b y+c z \geqslant 0 \leqslant(2004$ National Training Team Training Problem) | 61. Let $A=a x+b y+c z, B=a y+b z+c x, C=a z+b x+c y$. By symmetry, we can conjecture that $A \geqslant 0, B \geqslant 0, C \geqslant 0$.
Considering $A+B+C=(a+b+c)(x+y+z)=3$, if we can prove $A^{2}+B^{2}+C^{2} \leqslant \frac{9}{2}$, then it must be that $A \geqslant 0, B \geqslant 0, C \geqslant 0$.
In fact, if two ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,226 |
64. Given $a, b \geqslant 0$, prove $\left(\frac{\sqrt{a}+\sqrt{b}}{2}\right)^{2} \leqslant \frac{a+\sqrt[3]{a^{2} b}+\sqrt[3]{a b^{2}}+b}{4} \leqslant$ $\frac{a+\sqrt{a b}+b}{3} \leqslant \sqrt{\left(\frac{\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}}{2}\right)^{3}} \cdot(1993$ Austrian-Polish Mathematical Olympiad problem) | 64. Let $A=\sqrt[6]{a}, B=\sqrt[6]{b}$, then
$$\begin{array}{l}
\left(\frac{\sqrt{a}+\sqrt{b}}{2}\right)^{2} \leqslant \frac{a+\sqrt[3]{a^{2} b}+\sqrt[3]{a b^{2}}+b}{4} \Leftrightarrow \\
\sqrt{a}+\sqrt{b} \leqslant(\sqrt[3]{a}+\sqrt[3]{b})\left(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}\right) \Leftrightarrow \\
\left(A^{3}+B^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,229 |
65. Given $a, b, c>0$, prove: $\frac{a b}{3 a+b}+\frac{b c}{b+2 c}+\frac{c a}{c+2 a} \leqslant \frac{2 a+20 b+27 c}{49}$. | 65. Since $a, b>0$, we can use the method of analysis to prove $\frac{a b}{3 a+b} \leqslant \frac{a+12 b}{49}, \frac{b c}{b+2 c} \leqslant$ $\frac{8 b+9 c}{49}, \frac{c a}{c+2 a} \leqslant \frac{a+18 c}{49}$, adding the three inequalities yields $\frac{a b}{3 a+b}+\frac{b c}{b+2 c}+\frac{c a}{c+2 a} \leqslant$ $\frac{2... | \frac{2 a+20 b+27 c}{49} | Inequalities | proof | Yes | Yes | inequalities | false | 733,230 |
67. Given $a, b, c > 0$, prove: $3 + a + b + c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geqslant$ $\frac{3(a+1)(b+1)(c+1)}{a b c + 1} \cdot(1988$ Kvant Mathematical Olympiad Problem) | 67.
$$\begin{array}{l}
3+a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant \frac{3(a+1)(b+1)(c+1)}{a b c+1} \Leftrightarrow \\
3+a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+ \\
3 a b c+a^{2} b c+a b^{2} c+a b c^{2}+b c^{2}+a b+b c+c a+a^{2} c+a b^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,232 |
4. Let non-negative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy $\sum_{i=1}^{5} \frac{1}{1+x_{i}}=1$, prove that: $\sum_{i=1}^{5} \frac{x_{i}}{4+x_{i}^{2}} \leqslant 1$. | 4. Let $y_{i}=\frac{1}{1+x_{i}}, i=1,2,3,4,5$, then $x_{i}=\frac{1-y_{i}}{y_{i}}, i=1,2,3,4,5$, and $y_{1}+y_{2}+y_{3}$ $+y_{4}+y_{5}=1$, thus,
$$\begin{array}{l}
\sum_{i=1}^{5} \frac{x_{i}}{4+x_{i}^{2}} \leqslant 1 \Leftrightarrow \sum_{i=1}^{5} \frac{-y_{i}^{2}+y_{i}}{5 y_{i}^{2}-2 y_{i}+1} \leqslant 1 \Leftrightarro... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,233 |
69. Let $a, b, c$ be positive real numbers, and $a+b+c=1$, prove: $\frac{1}{a b+2 c^{2}+2 c}+$ $\frac{1}{b c+2 a^{2}+2 a}+\frac{1}{c a+2 b^{2}+2 b} \geqslant \frac{1}{a b+b c+c a} \cdot$(2007 Turkish National Training Team Test | 69. The original inequality is equivalent to proving
$$\begin{array}{l}
I=\frac{a b+b c+c a}{a b+2 c^{2}+2 c}+\frac{a b+b c+c a}{b c+2 a^{2}+2 a}+\frac{a b+b c+c a}{c a+2 b^{2}+2 b} \geqslant 1 \Leftrightarrow \\
\left(1-\frac{a b+b c+c a}{a b+2 c^{2}+2 c}\right)+\left(1-\frac{a b+b c+c a}{b c+2 a^{2}+2 a}\right)+\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,235 |
71. Let $a, b, c$ be positive real numbers, and $a b c \geqslant 1$, prove:
(1) $\left(a+\frac{1}{a+1}\right)\left(b+\frac{1}{b+1}\right)\left(c+\frac{1}{c+1}\right) \geqslant \frac{27}{8}$;
(2) $27\left(a^{3}+a^{2}+a+1\right)\left(b^{3}+b^{2}+b+1\right)\left(c^{3}+c^{2}+c+1\right) \geqslant 64\left(a^{2}+a+1\right)\le... | 71. (1) $a^{2}+a+1 \geqslant \frac{3}{4}(a+1)^{2} \Leftrightarrow (a-1)^{2} \geqslant 0$, so,
$$a+\frac{1}{a+1} \geqslant \frac{3}{4}(a+1)$$
Similarly,
$$\begin{array}{l}
b+\frac{1}{b+1} \geqslant \frac{3}{4}(b+1) \\
c+\frac{1}{c+1} \geqslant \frac{3}{4}(c+1)
\end{array}$$
Multiplying these inequalities, we get
$$\be... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,237 |
74. Given positive integers $n \geqslant 2, a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$ or $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$, and $a_{1} \neq a_{n}$, positive numbers $x, y$ satisfy $\frac{x}{y} \geqslant \frac{a_{1}-a_{2}}{a_{1}-a_{n}}$, prove the inequality: $\frac{a_{1}}{a_{2} x+a_... | 74. Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$, the original inequality is equivalent to
$$(x+y)\left(\frac{a_{1}}{a_{2} x+a_{3} y}+\frac{a_{2}}{a_{3} x+a_{4} y}+\cdots+\frac{a_{n-1}}{a_{n} x+a_{1} y}+\frac{a_{n}}{a_{1} x+a_{2} y} \geqslant n\right.$$
Let $a_{n+1}=a_{1}, a_{n+2}=a_{2}$, by the AM-GM ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,239 |
76. Let $a, b, c$ be positive numbers, prove that: $\left(1+\frac{4 a}{b+c}\right)\left(1+\frac{4 b}{c+a}\right)\left(1+\frac{4 c}{a+b}\right)>25$. | $$\begin{array}{l}
\quad \text { 76. }\left(1+\frac{4 a}{b+c}\right)\left(1+\frac{4 b}{c+a}\right)\left(1+\frac{4 c}{a+b}\right)>25 \Leftrightarrow(b+c+4 a)(c+a+4 b) \\
(a+b+4 c)>25(a+b)(b+c)(c+a) \Leftrightarrow a^{3}+b^{3}+c^{3}+7 a b c>a^{2} b+a b^{2}+ \\
b^{2} c+b c^{2}+c^{2} a+a c^{2} .
\end{array}$$
By Schur's i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,240 |
79. Let $x, y$ be positive numbers, prove: $\frac{1}{(1+\sqrt{x})^{2}}+\frac{1}{(1+\sqrt{y})^{2}} \geqslant \frac{2}{x+y+2}$. | $$\begin{array}{l}
\text { 79. } \frac{1}{(1+\sqrt{x})^{2}}+\frac{1}{(1+\sqrt{y})^{2}} \geqslant \frac{2}{x+y+2} \Leftrightarrow(x+y+2)\left[(1+\sqrt{x})^{2}+(1+\right. \\
\left.\sqrt{y})^{2}\right]-2[(1+\sqrt{x})(1+\sqrt{y})]^{2} \geqslant 0 \Leftrightarrow x^{2}+y^{2}+2 x+2 y+2+2 x \sqrt{x}+2 y \sqrt{y}- \\
2 y \sqrt... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,243 |
5. Let $a, b, c$ be positive real numbers, prove that: $\frac{(b+c-a)^{2}}{a^{2}+(b+c)^{2}}+\frac{(c+a-b)^{2}}{b^{2}+(c+a)^{2}}+$ $\frac{(a+b-c)^{2}}{c^{2}+(a+b)^{2}} \geqslant \frac{3}{5} . \quad(1997$ Japan Mathematical Olympiad Problem) | 5. Replacing $a, b, c$ with $\frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c}$ respectively, the original inequality remains unchanged, so we can assume without loss of generality that $0<a, b, c<1, a+b+c=1$, then
$$\frac{(b+c-a)^{2}}{a^{2}+(b+c)^{2}}=\frac{(1-2 a)^{2}}{a^{2}+(1-a)^{2}}=2-\frac{2}{1+(1-2 a)^{2}}$$
Si... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,244 |
81. Given that $a, b, c$ are positive numbers, and $abc=8$, prove: $\frac{a-2}{a+1}+\frac{b-2}{b+1}+\frac{c-2}{c+1} \leqslant 0$. | 81. $\frac{a-2}{a+1}+\frac{b-2}{b+1}+\frac{c-2}{c+1} \leqslant 0 \Leftrightarrow 3-3\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right) \leqslant 0 \Leftrightarrow \frac{1}{a+1}+$ $\frac{1}{b+1}+\frac{1}{c+1} \geqslant 1 \Leftrightarrow a+b+c \geqslant a b c-2$, since $a b c=8$, so $a+b+c \geqslant a b c-$ $2 \Leftr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,248 |
83. Given non-negative real numbers $a, b, c$ satisfying $a+b+c=1$, prove: $2 \leqslant\left(1-a^{2}\right)^{2}+(1-$ $\left.b^{2}\right)^{2}+\left(1-c^{2}\right)^{2} \leqslant(1+a)(1+b)(1+c)$ and find the conditions under which equality holds. (2000 Year | 83. Proof 1 Since $a+b+c=1$, we have $\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2} \geqslant 2 \Leftrightarrow (b+c)^{2}(2 a+b+c)^{2}+(c+a)^{2}(2 b+c+a)^{2}+(a+b)^{2}(2 c+a+b)^{2} \geqslant 2(a+b+c)^{4} \Leftrightarrow 2\left(a^{4}+b^{4}+c^{4}\right)+14\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,250 |
86. Given that $a, b$ are positive numbers, prove: $\frac{(a-b)^{2}}{2(a+b)} \leqslant \sqrt{\frac{a^{2}+b^{2}}{2}}-\sqrt{a b} \leqslant \frac{(a-b)^{2}}{4 \sqrt{a b}}$. | $$\begin{array}{l}
\text { 86. } \frac{(a-b)^{2}}{2(a+b)} \leqslant \sqrt{\frac{a^{2}+b^{2}}{2}}-\sqrt{a b} \Leftrightarrow \frac{(a-b)^{2}}{2(a+b)} \leqslant \frac{\frac{a^{2}+b^{2}}{2}-a b}{\sqrt{\frac{a^{2}+b^{2}}{2}}+\sqrt{a b}}= \\
\frac{(a-b)^{2}}{2\left(\sqrt{\frac{a^{2}+b^{2}}{2}}+\sqrt{a b}\right)} \Leftrighta... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,253 |
6. Let $a, b, c$ be positive real numbers, and $a+b+c=3$, prove that: $\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+$ $\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}}+\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \leqslant 5 \quad(2006$ Northern China Mathematical Olympiad Problem) | 6. From $a+b+c=3$, we get
$$\begin{array}{l}
\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}=\frac{a^{2}+9}{2 a^{2}+(3-a)^{2}}=\frac{1}{3} \cdot \frac{a^{2}+9}{a^{2}-2 a+3}=\frac{1}{3} \cdot \frac{a^{2}+9}{(a-1)^{2}+2}= \\
\frac{1}{3} \cdot\left(1+\frac{2 a+6}{(a-1)^{2}+2}\right) \leqslant \frac{1}{3} \cdot\left(1+\frac{2 a+6}{2}\ri... | 5 | Inequalities | proof | Yes | Yes | inequalities | false | 733,255 |
94. Given that $x, y, z$ are positive numbers, and $x y + y z + z x = 1$, prove: $3 - \sqrt{3} + \frac{x^{2}}{y} + \frac{y^{2}}{z} + \frac{z^{2}}{x} \geqslant (x + y + z)^{2}$. (2010 Iran Mathematical Olympiad Summer Camp Problem) | 94. First, prove a lemma:
$$\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x} \geqslant \frac{\left(x^{2}+y^{2}+z^{2}\right)(x+y+z)}{x y+y z+z x}$$
$$\begin{aligned}
(1) \Leftrightarrow & \left(\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x}\right)(x y+y z+z x) \geqslant\left(x^{2}+y^{2}+z^{2}\right)(x+y+z) \Leftrightarr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,262 |
95. Let real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $\sum_{k=1}^{n} x_{k}^{2}=1, n \geqslant 2$, prove that: $\sum_{k=1}^{n}\left(1-\frac{k}{\sum_{i=1}^{n} i x_{i}^{2}}\right) \frac{x_{k}^{2}}{k} \leqslant$ $\left(\frac{n-1}{n+1}\right)^{2} \sum_{k=1}^{n} \frac{x_{k}^{2}}{k}$. And determine the condition for equ... | \begin{array}{l}\text { 95. } \sum_{k=1}^{n}\left(1-\frac{k}{\sum_{i=1}^{n} i x_{i}^{2}}\right) \frac{x_{k}^{2}}{k} \leqslant\left(\frac{n-1}{n+1}\right)^{2} \sum_{k=1}^{n} \frac{x_{k}^{2}}{k} \Leftrightarrow(n+1)^{2} \sum_{k=1}^{n}\left(1-\frac{k}{\sum_{i=1}^{n} i x_{i}^{2}}\right) \frac{x_{k}^{2}}{k} \leqslant \\ (n-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,263 |
98. Prove: For any positive real numbers $a, b, c, d$, we have $\frac{(a-b)(a-c)}{a+b+c}+$ $\frac{(b-c)(b-d)}{b+c+d}+\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b} \geqslant 0$, and determine the conditions for equality. (49th IMO Shortlist) | 98. The left side of the inequality
$$\begin{aligned}
L= & \frac{(a-c)^{2}}{a+b+c}+\frac{(b-d)^{2}}{b+c+d}+(a-c)(b-d)\left(\frac{2 d+b}{(b+c+d)(d+a+b)}-\right. \\
& \left.\frac{2 c+a}{(a+b+c)(c+d+a)}\right)
\end{aligned}$$
To prove \( L \geqslant 0 \), it suffices to prove
$$\begin{array}{l}
(a-c)(b-d)\left(\frac{2 d+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,267 |
99. Let $a, b, c$ be positive real numbers, and $a^{4}+b^{4}+c^{4} \geqslant a^{3}+b^{3}+c^{3}$, prove: $\frac{a^{3}}{\sqrt{b^{4}+b^{2} c^{2}+c^{4}}}+$ $\frac{b^{3}}{\sqrt{c^{4}+c^{2} a^{2}+a^{4}}}+\frac{c^{3}}{\sqrt{a^{4}+a^{2} b^{2}+b^{4}}} \geqslant \sqrt{3}$. (62nd Polish Mathematical Olympiad Problem) | 99. We prove the local inequality
$$\begin{array}{l}
\frac{a^{3}}{\sqrt{b^{4}+b^{2} c^{2}+c^{4}}} \geqslant \frac{\sqrt{3} a^{4}}{a^{3}+b^{3}+c^{3}} \\
\frac{a^{3}}{\sqrt{b^{4}+b^{2} c^{2}+c^{4}}} \geqslant \frac{\sqrt{3} a^{4}}{a^{3}+b^{3}+c^{3}} \Leftrightarrow \\
\frac{F}{\sqrt{b^{4}+b^{2} c^{2}+c^{4}}} \geqslant \f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,268 |
101. Let $x, y, z \geqslant 0$, and at least two of them are not 0, prove: $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant$ $\sqrt{4-\frac{14 x y z}{(x+y)(y+z)(z+x)}} \cdot(2008$ Romanian Mathematical Olympiad problem) | 101. When $x, y, z$ has one being 0, the inequality obviously holds. Assuming $x, y, z$ are all greater than 0, the original inequality is equivalent to $\left(\frac{x}{y+z}\right)^{2}+\left(\frac{y}{z+x}\right)^{2}+\left(\frac{z}{x+y}\right)^{2}+\frac{10 x y z+2(x+y)(y+z)(z+x)}{(x+y)(y+z)(z+x)} \geqslant$ $4 \Leftrigh... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,270 |
Example 1
(1) Let $x, y, z$ be positive numbers, then $\frac{y^{2}-x^{2}}{z+x}+\frac{z^{2}-y^{2}}{x+y}+$ $\frac{x^{2}-z^{2}}{y+z} \geqslant 0$.
( W. Janous Conjecture)
(2) Let $x, y, z$ be positive numbers, then $\frac{y^{2}-z x}{z+x}+\frac{z^{2}-x y}{x+y}+\frac{x^{2}-y z}{y+z} \geqslant 0$. | $$\begin{array}{l}
\quad \text { Proof. (1) Let } z+x=a, x+y=b, y+z=c, \text { then } x+y+z \\
=\frac{1}{2}(a+b+c) \\
x=\frac{1}{2}(a+b-c) \\
y=\frac{1}{2}(b+c-a) \\
z=\frac{1}{2}(c+a-b)
\end{array}$$
Therefore, the original inequality can be transformed into
$$\frac{b c}{a}+\frac{c a}{b}+\frac{a b}{c} \geqslant a+b+c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,272 |
Example 2 For real numbers $x_{1}, y_{1}, z_{1}, x_{2}, y_{2}, z_{2}$ satisfying $x_{i}>0, y_{i}>0, x_{i} y_{i}-z_{i}^{2}>0(i=1 ; 2)$, the following inequality holds: $\frac{8}{\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(z_{1}+z_{2}\right)^{2}} \leqslant \frac{1}{x_{1} y_{1}-z_{1}^{2}}+\frac{1}{x_{2} y_{2}-z... | Proof: Let $\boldsymbol{a}=x_{1} y_{1}-z_{1}^{2}, b=x_{2} y_{2}-z_{2}^{2}$; then
$$\begin{array}{l}
\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(z_{1}+z_{2}\right)^{2}=a+b+x_{1} y_{2}+x_{2} y_{1}-2 z_{1} z_{2}= \\
a+b+2 \sqrt{a b}+\left(\frac{x_{1}}{x_{2}} b+\frac{x_{2}}{x_{1}} a-2 \sqrt{a b}\right)+\frac{x_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,273 |
Example 3 If $x_{i}>0(i=1,2, \cdots, n), n \geqslant 2$, and $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n}}=$
1, prove that: $x_{1} x_{2} \cdots x_{n} \geqslant(n-1)^{n}$ | $$\begin{array}{l}
\text { Proof: Let } \frac{1}{1+x_{1}}=\frac{a_{1}}{a_{1}+a_{2}+\cdots+a_{n}}, \frac{1}{1+x_{2}}=\frac{a_{2}}{a_{1}+a_{2}+\cdots+a_{n}}, \cdots, \\
\frac{1}{1+x_{n}}=\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n}}, \text { where } a_{1}, a_{2}, \cdots, a_{n} \text { are positive numbers, then } \\
x_{1}=\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,274 |
Example 4 Let positive real numbers $x, y, z$ satisfy $x+y+z=xyz$, find the minimum value of $x^{7}(yz-1)+y^{7}(zx-1)+z^{7}(xy-1)$ (2003 China National Training Team Problem) | Given the known conditions, we have $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$. Therefore, let $x=\frac{a+b+c}{a}, y=\frac{a+b+c}{b}, z=\frac{a+b+c}{c}$; where $a, b, c$ are positive real numbers. Then
$$\begin{array}{l}
x^{7}(y z-1)+y^{7}(z x-1)+z^{7}(x y-1)= \\
\left(\frac{a+b+c}{a}\right)^{7}\left[\frac{(a+b+c)^{2}}{b ... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,275 |
Example 5 Let $x, y, z$ be non-negative real numbers, and $x+y+z=1$, prove: $0 \leqslant y z+z x+x y-2 x y z \leqslant \frac{7}{27}$. (25th IMO Problem) | Assume without loss of generality that $x \geqslant y \geqslant z \geqslant 0$, from $x+y+z=1$ we know $z \leqslant \frac{1}{3}, x+y \geqslant \frac{2}{3}$, thus $2 x y z \leqslant \frac{2}{3} x y \leqslant x y$, so $y z+z x+x y-2 x y z \geqslant 0$. To prove the right inequality, let $x+y=\frac{2}{3}+\delta$, then $z=... | \frac{7}{27} | Inequalities | proof | Yes | Yes | inequalities | false | 733,276 |
8. Let $a, b, A, B$ be known real numbers, and for any real number $x$ the inequality $A \cos 2 x + B \sin 2 x + a \cos x + b \sin x \leqslant 1$ always holds. Prove: $a^{2} + b^{2} \leqslant 2, A^{2} + B^{2} \leqslant 1$ (19th IMO Problem) | 8. Construct the function $f(x)=1-A \cos 2 x-B \sin 2 x-a \cos x-b \sin x$. Then it can be transformed into
$$f(x)=1-\sqrt{A^{2}+B^{2}} \cos 2(x-\varphi)+\sqrt{a^{2}+b^{2}} \cos (x-\theta)$$
According to the problem, we have
$$\begin{array}{l}
f\left(\theta+\frac{\pi}{4}\right)=1-\sqrt{A^{2}+B^{2}} \cos 2\left(\theta-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,277 |
Example $6 \quad 0 \leqslant \alpha \leqslant 1,0 \leqslant x \leqslant \pi$, prove that $(2 \alpha-1) \sin x+(1-\alpha) \sin (1-$ $\alpha) x \geqslant \theta$ (1983 Swiss Mathematical Olympiad Problem) | Prove that when $\alpha=0,1$ and $x=0$, the inequality obviously holds.
When $0<\alpha<1$ and $x>0$, then
$$\frac{\sin x}{x} \leqslant \frac{\sin (1-\alpha) x}{(1-\alpha) x}$$
That is
$$\begin{array}{l}
(1-\alpha) \sin x \leqslant \sin (1-\alpha) x \\
(1-\alpha) \sin (1-\alpha) x \geqslant(1-\alpha)^{2} \sin x= \\
\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,278 |
Example 7 Given that $a, b, c$ are all positive numbers, and satisfy $\frac{a^{2}}{1+a^{2}}+\frac{b^{2}}{1+b^{2}}+\frac{c^{2}}{1+c^{2}}=1$, prove: $a b c \leqslant \frac{\sqrt{2}}{4}$. (31st IMO National Training Team Problem) | Let $a=\tan \alpha, b=\tan \beta, c=\tan \gamma, \alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$, then the given condition can be transformed into $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=1$.
So
$$\cos ^{2} \alpha=\sin ^{2} \beta+\sin ^{2} \gamma \geqslant 2 \sin \beta \sin \gamma$$
Similarly,
$$\beg... | a b c \leqslant \frac{\sqrt{2}}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 733,279 |
Example 8 Given that $a, b$ are positive real numbers, and $\frac{1}{a}+\frac{1}{b}=1$, prove: for every $n \in \mathbf{N}^{*}$, $(a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1}$. (1988 National High School Mathematics League Question) | Prove that for $a=\sec ^{2} \theta, b=\csc ^{2} \theta, 0<\theta<\frac{\pi}{2}$, we have $\frac{1}{a}+\frac{1}{b}=1$. The original inequality is equivalent to
$$\left(a^{n}-1\right)\left(b^{n}-1\right) \geqslant\left(2^{n}-1\right)^{2}$$
which is
$$\left(\sec ^{2 n} \theta-1\right)\left(\csc ^{2 n} \theta-1\right) \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,280 |
Example 9 Given real numbers $a, b, c, d, e$ satisfy the equations $a+b+c+d+e=8, a^{2}+b^{2}+c^{2}+$ $d^{2}+e^{2}=16$, prove that $0 \leqslant e \leqslant \frac{16}{5}$. (7th US Mathematical Olympiad problem)- | Prove that regarding $e$ as a constant, let $a=\frac{8-e}{4}+t_{1}, b=\frac{8-e}{4}+t_{2}, c=\frac{8-e}{4}+t_{3}, d=$ $\frac{8-e}{4}+t_{4}$, where $t_{1}+t_{2}+t_{3}+t_{4}=0$, then
$$16-e^{2}=a^{2}+b^{2}+c^{2}+d^{2}=\frac{(8-e)^{2}}{4}+t_{1}^{2}+t_{2}^{2} + t_{3}^{2}+t_{4}^{2} \geqslant \frac{(8-e)^{2}}{4}$$
That is
$... | 0 \leqslant e \leqslant \frac{16}{5} | Algebra | proof | Yes | Yes | inequalities | false | 733,281 |
Example 10 Let $x, y, z$ be non-negative real numbers, and $x+y+z=E$. Prove: $0 \leqslant y z+z x+x y-2 x y z$ $\leqslant \frac{7}{27}$. (25th IMO Problem) | Proof: By symmetry, we may assume $x \geqslant y \geqslant z$, thus $1=x+y+z \geqslant 3 z$, which means $z \leqslant \frac{1}{3}$. Therefore, $2 x y z \leqslant \frac{2}{3} x y \leqslant x y$, so $y z+z x+x y-2 x y z \geqslant 0$. For the right inequality, we can consider using the mean substitution: Let $x+y=\frac{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,282 |
Example 11 Given that $a, b, e$ are positive numbers, then
$$\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geqslant \frac{3}{\sqrt[3]{a b c}(1+\sqrt[3]{a b c})}$$ | Proof: Let $\sqrt[3]{a b c}=k(k>0)$, then $a b c=k^{3}$, so we can set $a=k \frac{a_{2}}{a_{1}}, b=k \frac{a_{3}}{a_{2}}, c=k \frac{a_{1}}{a_{3}}$, $\left(a_{1}, a_{2}, a_{3}>0\right)$, substituting into (1), we only need to prove
$$\frac{1}{k \frac{a_{2}}{a_{1}}+k^{2} \frac{a_{3}}{a_{1}}}+\frac{1}{k \frac{a_{3}}{a_{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,283 |
Example 12 (1) Let real numbers $x, y, z$ all be not equal to 1, and satisfy $xyz=1$. Prove: $\frac{x^{2}}{(x-1)^{2}}+$ $\frac{y^{2}}{(y-1)^{2}}+\frac{z^{2}}{(z-1)^{2}} \geqslant 1$; (31st IMO Problem)
(2) Prove: There exist infinitely many sets of rational triples $(x, y, z)$, where $x, y, z$ are all not equal to 1, a... | Prove (1) Let $\frac{x}{x-1}=a, \frac{y}{y-1}=b, \frac{z}{z-1}=c$, then $x=\frac{a}{a-1}, y=\frac{b}{b-1}$,
$$z=\frac{c}{c-1}$$
From
$$\begin{aligned}
x y z=1 \Rightarrow & a b c=(a-1)(b-1)(c-1) \Rightarrow a+b+c-1=a b+b c+c a \Rightarrow \\
& a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a)= \\
& (a+b+c)^{2}-2(a+b+c-1)= ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,284 |
Example 13 Proof: For any positive real numbers $a, b, c$, we have
$$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geqslant 9(a b+b c+c a)$$ | Let $a=\sqrt{2 \tan A}, b=\sqrt{2 \tan B}, c=\sqrt{2 \tan C}$, where $A, B, C \in\left(0, \frac{\pi}{2}\right)$. Since $1+\tan ^{2} \theta=\sec ^{2} \theta$, (1) is equivalent to $\cos A \cos B \cos C(\cos A \sin B \sin C+\sin A \cos B \sin C+\sin A \sin B \cos C) \leqslant \frac{4}{9}$.
Because $\cos (A+B+C)=\cos A \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,285 |
9. Let $a, b, c, d$ be positive real numbers, and satisfy $a+b+c+d=1$. Prove that: $6\left(a^{3}+b^{3}+c^{3}+d^{3}\right) \geqslant$ $\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+\frac{1}{8}$. (8th Hong Kong Mathematical Olympiad Problem) | 9. Clearly, $0<a, b, c, d<1$, we first prove that when $0<x<1$, we have
$$f(x)=6 x^{3}-x^{2} \geqslant \frac{5}{8} x-\frac{1}{8}$$
Multiplying both sides by 8 and rearranging, we know that the above inequality is equivalent to $48 x^{3}-8 x^{2} \geqslant 5 x-1 \Leftrightarrow (4 x-1)^{2}(3 x+1) \geqslant 0$, which is ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,288 |
1. For $x, y, z \geqslant 0$, prove the inequality $x(x-z)^{2}+y(y-z)^{2} \geqslant(x-z)(y-z)(x+y-z) . \quad(1992$ Canadian Mathematical Olympiad problem) | 1. The inequality to be proved is equivalent to $x^{3}+y^{3}+z^{3}+3 x y z-x^{2} y-x^{2} z-y^{2} z-y^{2} x-$ $z^{2} x-z^{2} y \geqslant 0$, that is $\square$
$$x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y) \geqslant 0$$
This inequality is symmetric with respect to $x, y, z$. Without loss of generality, assume $x \geqslant y \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,289 |
2. Prove that for all $n \geqslant 2$, the inequality $\frac{x_{1}^{2}}{x_{1}^{2}+x_{2} x_{3}}+\frac{x_{2}^{2}}{x_{2}^{2}+x_{3} x_{4}}+\cdots+\frac{x_{n-1}^{2}}{x_{n-1}^{2}+x_{n} x_{1}}+$ $\frac{x_{n}^{2}}{x_{n}^{2}+x_{1} x_{2}} \leqslant n-1$ holds. (27th IMO Shortlist Problem) | 2. Let $y_{i}=\frac{x_{i}^{2}}{x_{i+1} x_{i+2}}, i=1,2, \cdots, n$, and $x_{n+i}=x_{i}$. Notice that $y_{1} y_{2} \cdots y_{n}=1$, thus,
$$\begin{array}{c}
\frac{x_{i}^{2}}{x_{i}^{2}+x_{i+1} x_{i+2}}=\frac{x_{i}^{2}+x_{i+1} x_{i+2}-x_{i+1} x_{i+2}}{x_{i}^{2}+x_{i+1} x_{i+2}}=1-\frac{x_{i+1} x_{i+2}}{x_{i}^{2}+x_{i+1} x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,290 |
3. Given that $a, b, c$ are positive numbers, and $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1$, prove: $a+b+c \geqslant \frac{3}{2}$. | 3. Let $x=\frac{a}{1+a}, y=\frac{b}{1+b}, z=\frac{c}{1+c}$, then $x+y+z=1, a=\frac{x}{1-x}=\frac{x}{y+z}$, similarly $b=\frac{y}{z+x}, c=\frac{z}{x+y}$-thus from $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac{3}{2}$ we get $a+b+c \geqslant \frac{3}{2}$. | a+b+c \geqslant \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 733,291 |
4. Given that $x, y, z$ are positive numbers, and $x+y+z=1$, prove that: $\left(\frac{1}{x}-x\right)\left(\frac{1}{y}-y\right)\left(\frac{1}{z}-z\right) \geqslant$
$$\left(\frac{8}{3}\right)^{3}$$ | 4. Let $\frac{1}{x}=a, \frac{1}{y}=b-\frac{1}{z}=c$, substituting into the given conditions we get
$$\frac{1}{a}+\frac{\mathrm{k}}{b}+\frac{1}{c}=1$$
From (1) and the AM-GM inequality, we have
$$a b c \geqslant 27$$
Equation (1) is equivalent to
$$a b c=a b+b c+c a$$
Therefore,
$$\begin{array}{l}
\left(\frac{1}{x}-x... | \left(\frac{8}{3}\right)^{3} | Inequalities | proof | Yes | Yes | inequalities | false | 733,292 |
5. Let $\alpha_{0}, \alpha_{1}, \alpha_{2}, \cdots \alpha_{n} \in\left(0, \frac{\pi}{2}\right)$, such that $\sum_{i=0}^{n} \tan \left(\alpha_{i}-\frac{\pi}{4}\right) \geqslant n-1$, prove that $\tan \alpha_{i} \geqslant n^{n+1}$ (1998 27th US Mathematical Olympiad Problem) | 5. From the condition, we have $\sum_{i=0}^{n} \tan \left(\alpha_{i}-\frac{\pi}{4}\right)=\sum_{i=0}^{n} \frac{\tan \alpha_{i}-1}{\tan \alpha_{i}+1} \geqslant n-1$, from which we can derive
$$\sum_{i=0}^{n} \frac{1}{\tan \alpha_{i}+1} \leqslant 1$$
Let $y_{i}=\frac{1}{1+\tan \alpha_{i}}$, then $\tan \alpha_{i}=\frac{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,293 |
6. Let real numbers $a, b, c$ be such that the sum of any two is greater than the third, then $\frac{2}{3}(a+b+c)\left(a^{2}+b^{2}+\right.$ $\left.c^{2}\right) \geqslant a^{3}+b^{3}+c^{3}+a b c$. | 6. The inequality to be proved is equivalent to
$$2(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \geqslant 3\left(a^{3}+b^{3}+c^{3}+a b c\right)$$
Let $2 x=-a+b+c, 2 y=a-b+c, 2 z=a+b-c$, then $a=y+z, b=z+x, c=x+y, (1) \Leftrightarrow$
Thus, $a+b+c=2(x+y+z), a^{2}+b^{2}+c^{2}=2\left(x^{2}+y^{2}+z^{2}+x y+y z+z x\right)$, $a^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,294 |
7. Let $a \geqslant b \geqslant c \geqslant 0$, and $a+b+c=3$, prove: $a b^{2}+b c^{2}+c a^{2} \leqslant \frac{27}{8}$, and determine the condition for equality. (2002 Hong Kong Mathematical Olympiad Problem) | 7. Let $x=\frac{c}{3}, y=\frac{b-c}{3}, z=\frac{a-b}{3}$, then $x+y+z=\frac{a}{3}, x+y=\frac{b}{3}, x=\frac{c}{3}$, so $3 x+2 y+z=1$, and
$$\begin{aligned}
a & =\frac{3(x+y+z)}{3 x+2 y+z} \\
b & =\frac{3(x+y)}{3 x+2 y+z} \\
c & =\frac{3 x}{3 x+2 y+z}
\end{aligned}$$
Substituting $a, b, c$ into the original inequality,... | a=b=\frac{3}{2}, c=0 | Inequalities | proof | Yes | Yes | inequalities | false | 733,295 |
8. If positive numbers $a, b, c$ satisfy $\frac{a}{b+c}=\frac{b}{a+c}-\frac{c}{a+b}$, prove that $\frac{b}{a+c} \geqslant \frac{\sqrt{17}-1}{4}$. | 8. Given the condition $\frac{b}{a+c}=\frac{a}{b+c}+\frac{c}{a+b}$, let $a+b=x, b+c=y, c+a=z$, then $a=\frac{x+z-y}{2}, b=\frac{x+y-z}{2}, c=\frac{y+z-x}{2}$.
Thus, the original condition becomes
$$\begin{array}{c}
\frac{x+y-z}{2 z}=\frac{y+z-x}{2 x}+\frac{x+z-y}{2 y} \\
\frac{x+y}{z}=\frac{y+z}{x}+\frac{x+z}{y}-1 \geq... | \frac{\sqrt{17}-1}{4} | Algebra | proof | Yes | Yes | inequalities | false | 733,296 |
9. Given that $a, b, c, d$ are non-negative real numbers, and $a b c d=1$, prove: $\frac{1+a b}{1+a}+\frac{1+b c}{1+b}+\frac{1+c d}{1+c}+$ $\frac{1+d a}{1+d} \geqslant 4 .(2002$ Turkish Mathematical Olympiad Problem) | 9. Let $a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{t}{z}, d=\frac{x}{t}$, then $\frac{1+a b}{1+a}+\frac{1+b c}{1+b}+\frac{1+c d}{1+c}+\frac{1+d a}{1+d} \geqslant$ 4 can be transformed into $\frac{x+z}{x+y}+\frac{y+t}{y+z}+\frac{z+x}{z+t}+\frac{t+y}{t+x} \geqslant 4$.
From the inequality $\frac{1}{u}+\frac{1}{v} \geqslant \f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,297 |
Example 4 Find all real numbers $k$ such that $a^{3}+b^{3}+c^{3}+d^{3}+1 \geqslant k(a+b+c+d)$, for all $a, b, c, d \in[-1,+\infty)$: (2004 China Western Mathematical Olympiad) | When $a=b=c=d=-1$, we have $3 \geqslant k(-4)$, so $k \geqslant 3$. When $a=b=c=d=\frac{1}{2}$, we have $4 \times \frac{1}{8}+1 \geqslant k\left(4 \times \frac{1}{2}\right)$, so $k \leqslant \frac{3}{4}$. Therefore, $k=\frac{3}{4}$. Below is the proof that
$$a^{3}+b^{3}+c^{3}+d^{3}+1 \geqslant \frac{3}{4}(a+b+c+d)$$
ho... | \frac{3}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,299 |
11. Prove that for any positive real numbers $a, b, c$, we have $1<\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}} \leqslant \frac{3 \sqrt{2}}{2}$. (2004 China Western Mathematical Olympiad) | 11. First, prove the left inequality. Let $x=\frac{b^{2}}{c^{2}}, y=\frac{c^{2}}{a^{2}}, z=\frac{a^{2}}{b^{2}}$, then $x, y, z \in \mathbf{R}^{+}, x y z=1$, so we only need to prove $\square$
$$\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}>F$$
Assume without loss of generality that $x \leqslant y \leq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,301 |
12. If $a, b, c$ are all positive numbers, prove: $a b c \geqslant(-a+b+c)(a-b+c)(a+b-c)$. | 12. Let $a \geqslant b \geqslant c>0$, and set $b=c+t_{1}, a=c+t_{1}+t_{2}, t_{1} \geqslant 0, t_{2} \geqslant 0$, then $a b c-(-a+b+c)=(a-b+c)(a+b-c)=\left(c+t_{1}+t_{2}\right)\left(c+t_{1}\right) c-\left(c-t_{2}\right)$ $\left(c+t_{2}\right)\left(c+2 t_{1}+t_{2}\right)=\left(t_{1}^{2}+t_{1} t_{2}+t_{2}^{2}\right) c+2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,302 |
13. Let $x, y, z \geqslant 0$, and satisfy $x y+y z+z x=1$, prove that: $x\left(1-y^{2}\right)\left(1-z^{2}\right)+y(1-$ $\left.z^{2}\right)\left(1-x^{2}\right)+z\left(1-x^{2}\right)\left(1-y^{2}\right) \leqslant \frac{4 \sqrt{3}}{9}$. (1994 Hong Kong Mathematical Olympiad Problem) | 13. Let $x=\tan \frac{\alpha}{2}, y=\tan \frac{\beta}{2}, z=\tan \frac{\gamma}{2}, 0 \leqslant \alpha, \beta ; \gamma<\pi$, then from $x y+y z+z x=1$ we get $\frac{x+y}{1-x y}=\frac{F}{z}$, i.e., $\tan \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=\cot \frac{\gamma}{2}$, so $\alpha+\beta+\gamma=\pi$, and
$$\begin{array... | \frac{4 \sqrt{3}}{9} | Inequalities | proof | Yes | Yes | inequalities | false | 733,303 |
14. Let $a, b, c, d$ all be positive numbers, prove the inequality: $\frac{a}{b+2 c+3 d}+\frac{b}{c+2 d+3 a}+$ $\frac{c}{d+2 a+3 b}+\frac{d}{a+2 b+3 c} \geqslant \frac{2}{3}$. (34th IMO Shortlist) | 14. Make the linear substitution $\left\{\begin{array}{l}x=b+2 c+3 d \\ y=c+2 d+3 a \\ z=d+2 a+3 b \\ w=a+2 b+3 c\end{array}\right.$, treating $a, b, c, d$ as variables, and solve to get
$$\left\{\begin{array}{l}
a=-\frac{5}{24} x+\frac{7}{24} y+\frac{1}{24} z+\frac{1}{24} w \\
b=\frac{1}{24} x-\frac{5}{24} y+\frac{7}{... | \frac{2}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 733,304 |
16. If $n$ is a positive integer, and $x_{k}>0, k=1,2, \cdots, n+1$. If $\sum_{k=1}^{n+1} \frac{1}{1+x_{k}} \geqslant n$, then $\sum_{k=1}^{n+1} \frac{1}{x_{k}} \geqslant n^{n+1} \cdot$ (J. Berkes Inequality) | 16. Let $x_{k}=\tan \alpha_{k}, k=1,2, \cdots, n+1$. Then we have $\frac{1}{1+x_{k}}=\cos ^{2} \alpha_{k}, \frac{1}{x_{k}}=\frac{1}{\sin ^{2} \alpha_{k}}-1$, thus the original condition can be expressed as
$$\sum_{k=1}^{n+1} \cos ^{2} \alpha_{k} \geqslant n$$
which implies $\square$
$$\sum_{k=1}^{n+1} \sin ^{2} \alpha... | \prod_{k=1}^{n+1} \frac{1}{x_{k}} \geqslant n^{n+1} | Inequalities | proof | Yes | Yes | inequalities | false | 733,306 |
17. Let $a, b, c \in \mathbf{R}^{+}$, when $a^{2}+b^{2}+c^{2}+a b c=4$, prove: $a+b+c \leqslant 3$. | 17. In $\triangle A B C$, $\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2 \cos A \cos B \cos C=1$. In fact,
$$\begin{array}{l}
\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2 \cos A \cos B \cos C=\frac{1+\cos 2 A}{2}+\frac{1+\cos 2 B}{2}+ \\
\cos ^{2} C+2 \cos A \cos B \cos C= \\
1+\cos (A+B) \cos (A-B)+\cos ^{2} C+2 \cos A \cos B \cos C... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,307 |
19. Let $x, y, z$ be positive numbers, and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$, prove that: $\sqrt{x+y z}+\sqrt{y+z x}+$ $\sqrt{z+x y} \geqslant \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z} \cdot(2002$ Asia Pacific Mathematical Olympiad Problem) | 19. Let $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$, then $a+b+c=1$.
The original inequality is equivalent to
$$\sqrt{\frac{1}{a}+\frac{1}{b c}}+\sqrt{\frac{1}{b}+\frac{1}{c a}}+\sqrt{\frac{1}{c}+\frac{1}{a b}} \geqslant \sqrt{\frac{1}{a b c}}+\sqrt{\frac{1}{a}} \pm \sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}}$$
which is
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,309 |
20. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive numbers satisfying $\sum_{k=1}^{n} \frac{1}{x_{k}+1998}=\frac{1}{1998}$, prove: $\frac{\sqrt[n]{x_{1} x_{2} \cdots x_{n}}}{n-1} \geqslant$ 1998. (1998 Vietnam Mathematical Olympiad Problem) | 20. The condition $\sum_{k=1}^{n} \frac{1}{x_{k}+1998}=\frac{1}{1998}$ can be transformed into $\sum_{k=1}^{n} \frac{1998}{x_{k}+1998}=\sum_{k=1}^{n} \frac{1}{\frac{x_{k}}{1998}+1}=1$. Let $\frac{x_{k}}{1998}=t_{k}$, then $\sum_{k=1}^{n} \frac{1}{t_{k}+1}=1$. Let $y_{k}=\frac{1}{t_{k}+1}$, then $\sum_{k=1}^{n} y_{k}=1$... | \frac{\sqrt[n]{x_{1} x_{2} \cdots x_{n}}}{n-1} \geqslant 1998 | Inequalities | proof | Yes | Yes | inequalities | false | 733,310 |
21. Let $a, b, c$ be positive numbers, satisfying $a b + b c + c a = a b c$, prove the inequality: $\frac{a^{4}+b^{4}}{a b\left(a^{3}+b^{3}\right)}+$ $\frac{b^{4}+c^{4}}{b c\left(b^{3}+c^{3}\right)}+\frac{c^{4}+a^{4}}{c a\left(c^{3}+a^{3}\right)} \geqslant 1$. (2006 Polish Mathematical Olympiad) | 21. Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$. From the given $a b+b c+c a=a b c$, we get $x+y+z=1$.
$$\frac{x^{4}+y^{4}}{x^{3}+y^{3}}-\frac{x+y}{2}=\frac{(x-y)^{2}\left(x^{2}+x y+z^{2}\right)}{2\left(x^{3}+y^{3}\right)} \geqslant 0$$
Thus,
$$\frac{x^{4}+y^{4}}{x^{3}+y^{3}} \geqslant \frac{x+y}{2}$$
Similarly... | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 733,312 |
24. Prove that in the open interval $(0,1)$, there must exist 4 pairs of distinct positive numbers $(a, b)(a \neq b)$, satisfying $\sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}>\frac{a}{2 b}+\frac{b}{2 a}-a b-\frac{1}{8 a b} .(1994$ Czechoslovakia Mathematics | 24. Let $a=\cos \alpha, b=\cos \beta$, where
$$\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$$
Thus $\square$
$$a b+\sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}=\cos \alpha \cos \beta+\sin \alpha \sin \beta=\cos (\alpha-\beta)$$
Squaring both sides of the above equation, we get $a^{2} b^{2}+2 a b \sqrt{\left(1-a^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,315 |
25. Let the set $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}, 0<a_{i}<a_{i+1}(i=1,2,3)$. Is it true that in the set $A$, there must exist two elements $x, y$ such that the inequality $(2+\sqrt{3})\lfloor x-y \rfloor (x+1)(y+1)+xy$ holds? If so, provide a proof; if not, explain why. (2005 Hebei Province Mathematics Comp... | 25. There exist such two numbers. Without loss of generality, let $x>y$, then the original inequality transforms into
$$\begin{aligned}
\frac{x-y}{(x+1)(y+1)+x y} & 0$ we know, $\alpha_{i} \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$.
Divide $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ into three equal parts. By the pi... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,316 |
26. Let $a_{1}, a_{2}, \cdots, a_{n} \in[-2,2]$, and $\sum_{k=1}^{n} a_{k}=0$, prove: $\left|a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}\right| \leqslant$
2n. (1996 USA National Training Team Problem) | 26. Let $a_{k}=2 \cos \theta_{k} \cdot(k=1,2, \cdots, n)$, then by the triple angle formula $\cos 3 \theta=4 \cos ^{3} \theta- 3 \cos \theta$ we can get $a_{k}^{3}=3 a_{k}+2 \cos 3 \theta_{k}$.
Therefore, $\left|a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}\right|=\left|3 \sum_{k=1}^{n} a_{k}+2 \sum_{k=1}^{n} \cos 3 \theta_{k}\... | 2n | Inequalities | proof | Yes | Yes | inequalities | false | 733,317 |
27. Given $a, b, c > 0$, prove: $\sqrt{\left(a^{2} b+b^{2} c+c^{2} a\right)\left(a b^{2}+b c^{2}+c a^{2}\right)} \geqslant a b c + \sqrt[3]{\left(a^{3}+a b c\right)\left(b^{3}+a b c\right)\left(c^{3}+a b c\right)} \cdot (2001$ Korean Mathematical Olympiad Problem) | 27. Let $x^{2}=\left(a^{2} b+b^{2} c+c^{2} a\right)\left(a b^{2}+b c^{2}+c a^{2}\right), y=a b c$. By the AM-GM inequality, we have $x^{2} \geqslant 9 y^{2}$, so, $x \geqslant 3 y$. Since $\left(a^{2} b+b^{2} c+c^{2} a\right)\left(a b^{2}+b c^{2}+c a^{2}\right)=\left(a^{2}+b c\right)$ $\left(b^{2}+a c\right)\left(c^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,318 |
28. Given $a, b, c > 0$, and $abc = 1$, prove: $\frac{a}{b^{2}(c+1)} + \frac{b}{c^{2}(a+1)} + \frac{c}{a^{2}(b+1)} \geqslant \frac{3}{2}$. (2005 Romanian Mathematical Olympiad Problem) | 28. Let $a=\frac{x}{z}, b=\frac{y}{x}, c=\frac{z}{y}$, then $\frac{a}{b^{2}(c+1)}+\frac{b}{c^{2}(a+1)}+\frac{c}{a^{2}(b+1)} \geqslant \frac{3}{2} \Leftrightarrow$ $\frac{x^{3}}{y z(y+z)}+\frac{y^{3}}{z x(z+x)}+\frac{z^{3}}{x y(x+y)} \geqslant \frac{3}{2} \Leftrightarrow \frac{x^{4}}{x y z(y+z)}+\frac{y^{4}}{x y z(z+x)}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,319 |
29. Let $a, b, c$ be distinct real numbers. Prove: $\left(\frac{2 a-b}{a-b}\right)^{2}+\left(\frac{2 b-c}{b-c}\right)^{2}+\left(\frac{2 c-a}{c-a}\right)^{2} \geqslant 5$. | 29. Let \( x = \frac{a}{a-b}, y = \frac{b}{b-c}, z = \frac{c}{c-a} \)
Then
\[
(x-1)(y-1)(z-1) = \frac{b}{a-b} \cdot \frac{c}{b-c} \cdot \frac{a}{c-a} = \frac{a}{a-b} \cdot \frac{b}{b-c} \cdot \frac{c}{c-a} = x y z
\]
Expanding and simplifying the above equation, we get
\[
x + y + z = x y + y z + z x + 1
\]
Therefore... | 5 | Inequalities | proof | Yes | Yes | inequalities | false | 733,320 |
30. Given $a, b, c>0$, and $a^{2}+b^{2}+c^{2}=1$, prove: $\frac{a}{a^{2}+b c}+\frac{b}{b^{3}+c a}+$ $\frac{c}{c^{3}+a b}>3 .(2005$ Yakut Mathematical Olympiad problem) | $$\begin{array}{l}
\text { 30. Let } x=a^{2}, y=b^{2}, z=c^{2}, p=x y z, \frac{a}{a^{3}+b c}+\frac{b}{b^{3}+c a}+\frac{c}{c^{3}+a b}>3 \Leftrightarrow \frac{x}{x^{2}+p}+ \\
\frac{y}{y^{2}+p}+\frac{z}{z^{2}+p}>3 \Leftrightarrow x\left(y^{2}+p\right)\left(z^{2}+p\right)+y\left(z^{2}+p\right) \cdot\left(x^{2}+p\right)+z\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,321 |
12. Let $x_{1}, x_{2}, \cdots, x_{n}$ be real numbers $(n \geqslant 3)$, and let $p=\sum_{i=1}^{n} x_{i}, q=\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}$. Prove:
(1) $\frac{n-1}{n} p^{2}-2 q \geqslant 0$;
(2) $\left|x_{i}-\frac{p}{n}\right| \leqslant \frac{n-1}{n} \sqrt{p^{2}-\frac{2 n}{n-1} q}(i=1,2, \cdots, n)$. (C... | 12. (1) From the given condition $\sum_{i=1}^{n} x_{i}^{2}=p^{2}-2 q$, we have
$$\begin{array}{l}
\sum_{i<k}\left(x_{i}-x_{k}\right)^{2}=(n-1) x_{i}^{2}-2 q= \\
(n-1)\left(p^{2}-2 q\right)-2 q=(n-1) p^{2}-2 n q \geqslant 0
\end{array}$$
Thus, $\frac{n-1}{n} p^{2}-2 q \geqslant 0$.
(2) Now, apply this relation to the r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,322 |
33- Given that $a, b, c$ are all positive numbers, prove: $\frac{a^{3}}{(a+b)^{3}}+\frac{b^{3}}{(b+c)^{3}}+\frac{c^{3}}{(c+a)^{3}} \geqslant \frac{3}{8}$. | 33. By the power mean inequality $\sqrt[3]{\frac{x^{3}+y^{3}+z^{3}}{3}} \geqslant \sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}$, to prove $\frac{a^{3}}{(a+b)^{3}}+$ $\frac{b^{3}}{(b+c)^{3}}+\frac{c^{3}}{(c+a)^{3}} \geqslant \frac{3}{8}$, it is sufficient to prove
$$\frac{a^{2}}{(a+b)^{2}}+\frac{b^{2}}{(b+c)^{2}}+\frac{c^{2}}{(c+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,325 |
35. Given that $a, b, c, d$ are all positive numbers, and satisfy $\frac{1}{1+a^{4}}+\frac{1}{1+b^{4}}+\frac{1}{1+c^{4}}+\frac{1}{1+d^{4}}=1$, prove: $a b c d \geqslant 3$. (2002 Latvia Mathematical Olympiad Problem) | 35. Let \(a^{2}=\tan A, b^{2}=\tan B, c^{2}=\tan C, d^{2}=\tan D\), where \(A, B, C, D\) are all acute angles. Given the condition, it can be transformed to \(\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+\cos ^{2} D=1\). Applying the AM-GM inequality, we get
$$\sin ^{2} A=1-\cos ^{2} A=\cos ^{2} B+\cos ^{2} C+\cos ^{2} D \geqsl... | a b c d \geqslant 3 | Inequalities | proof | Yes | Yes | inequalities | false | 733,327 |
36. Given that $a, b, c$ are positive numbers, and $a+b+c=1$, prove: $\frac{a}{a+b c}+\frac{b}{b+c a}+\frac{\sqrt{a b c}}{c+a b} \leqslant 1+$
$$\frac{3 \sqrt{3}}{4}$$ | 36. $\frac{a}{a+b c}+\frac{b}{b+c a}+\frac{\sqrt{a b c}}{c+a b} \leqslant 1+\frac{3 \sqrt{3}}{4}$ is equivalent to
$$\frac{1}{1+\frac{b c}{a}}+\frac{1}{1+\frac{c a}{b}}+\frac{\sqrt{\frac{a b}{c}}}{1+\frac{a b}{c}} \leqslant 1+\frac{3 \sqrt{3}}{4}$$
Let $x=\sqrt{\frac{b c}{a}}, y=\sqrt{\frac{c a}{b}}, z=\sqrt{\frac{a b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,328 |
39. Let $a, b, c$ be positive numbers, prove the inequality: $2 \sqrt{a b+b c+c a} \leqslant \sqrt{3} \sqrt[3]{(b+c)(c+a)(a+b)}$ (1992 Poland-Austria Mathematical Olympiad) | 39. Since the degrees at both ends are equal, we might as well set \(a b+b c+c a=1\), then let \(a=\tan \alpha, b=\tan \beta, c=\tan \gamma\), where \(\alpha+\beta+\gamma=90^{\circ}\). \(\alpha, \beta, \gamma\) are all acute angles. Then the problem is reduced to proving
$$(b+c)(c+a)(a+b) \geqslant \frac{8 \sqrt{3}}{9}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,331 |
40. Let $x, y, z \in \mathbf{R}^{+}$, and $x+y+z=1$, prove that: $\frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+x z}}+\frac{x z}{\sqrt{x z+x y}} \leqslant \frac{\sqrt{2}}{2} .(2006$ National Training Team Exam Question) | 40. Let $a=\sqrt{x}, b=\sqrt{y}, c=\sqrt{z}$, then $a^{2}+b^{2}+c^{2}=1$, and by the inequality $\sqrt{a^{2}+b^{2}} \geqslant \frac{\sqrt{2}}{2}(a+b)$
$$\begin{aligned}
\frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+x z}}+\frac{x z}{\sqrt{x z+x y}}= & \frac{a^{2} b}{\sqrt{a^{2}+c^{2}}}+\frac{b^{2} c}{\sqrt{b^{2}+a^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,332 |
13. Given $a, b, c \in\left(-\frac{3}{4},+\infty\right)$, and $a+b+c=1$, prove: $\frac{a}{a^{2}+1}+\frac{b}{b^{2}+1}+$ $\frac{c}{c^{2}+1} \leqslant \frac{9}{10}$. 1 (1996 Polish Mathematical Olympiad Problem) | 13. Considering that the equality holds when $a, b, c=\frac{1}{3}$, at this time if $x=\frac{1}{3}, f(x)=\frac{x}{1+x^{2}}=$ $\frac{36 x+3}{50}$, it is only necessary to prove $\frac{x}{1+x^{2}} \leqslant \frac{36 x+3}{50},-\frac{3}{4}<x<1$.
This is equivalent to proving $(3 x-1)^{2}(4 x+3) \geqslant 0$, which is obvi... | \frac{a}{a^{2}+1}+\frac{b}{b^{2}+1}+\frac{c}{c^{2}+1} \leqslant \frac{9}{10} | Inequalities | proof | Yes | Yes | inequalities | false | 733,333 |
41. Let $x_{1}, x_{2}, \cdots, x_{n}>1$, prove: $\frac{x_{1} x_{2}}{x_{3}-1}+\frac{x_{2} x_{3}}{x_{4}-1}+\frac{x_{3} x_{4}}{x_{5}-1}+\cdots+\frac{x_{n-1} x_{n}}{x_{1}-1}+$ $\frac{x_{n} x_{1}}{x_{2}-1} \geqslant 4 n$. (2009 Indonesia Mathematical Olympiad Problem) | 41. Substitution: Let $y_{i}=x_{i}-1(i=1,2, \cdots, n)$, and denote $x_{n+1}=x_{1}, x_{n+2}=x_{2}$, then
$$\begin{aligned}
\sum_{i=1}^{n} \frac{x_{i} x_{i+1}}{x_{i+2}-1}= & \sum_{i=1}^{n} \frac{\left(-y_{i}+1\right)\left(y_{i+1}+1\right)}{y_{i+2}}=\sum_{i=1}^{n}\left(\frac{y_{i} y_{i+1}}{y_{i+2}}+\frac{1}{y_{i+2}}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,334 |
42. Given that $x, y, z$ are positive real numbers, and $x^{2}+y^{2}+z^{2}=1$, prove that $\frac{x}{1+x^{2}}+\frac{y}{1+y^{2}}+\frac{z}{1+z^{2}} \leqslant \frac{3 \sqrt{3}}{4}$. (1998 Bosnia and Herzegovina Mathematical Olympiad Problem) | 42. Let $x=\tan \alpha, y=\tan \beta, z=\tan \gamma (\alpha, \beta, \gamma$ are acute angles $)$, then $\tan ^{2} \alpha+$ $\tan ^{2} \beta+\tan ^{2} \gamma=1$, by the Cauchy-Schwarz inequality we have 3 $\left(\tan ^{2} \alpha+\tan ^{2} \beta+\tan ^{2} \gamma\right) \geqslant(\tan \alpha+$ $\tan \beta+\tan \gamma)^{2}... | \frac{3 \sqrt{3}}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 733,335 |
43. Let $a, b, c$ be positive numbers, and $abc=1$, prove the inequality: $\left(\frac{a}{1+ab}\right)^{2}+\left(\frac{b}{1+bc}\right)^{2}+$ $\left(\frac{c}{1+ca}\right)^{2} \geqslant \frac{3}{4} . (2009$ Spanish Mathematical Olympiad problem) | 43. Let $a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{x}{z}$, then
$$\left(\frac{a}{1+a b}\right)^{2}+\left(\frac{b}{1+b c}\right)^{2}+\left(\frac{c}{1+c a}\right)^{2} \geqslant \frac{3}{4}$$
is equivalent to
$$\left(\frac{x}{y+z}\right)^{2}+\left(\frac{y}{z+x}\right)^{2}+\left(\frac{z}{x+y}\right)^{2} \geqslant \frac{3}{4}$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,336 |
44. Let $a, b, c$ be real numbers, prove the inequality: $(a+b)^{4}+(b+c)^{4}+(c+a)^{4} \geqslant \frac{4}{7}(a+b+c)^{4}$. (1996 Vietnam National Training Team Problem) | 44. Let $a+b=2z, a+b=2z, b+c=2x, c+a=2y$, the inequality becomes $\sum(y+z-$
$$\begin{aligned}
x)^{4} \leqslant & 28 \sum x^{4} . \\
\sum(y+z-x)^{4}= & \sum\left(\sum x^{2}+2yz-2xy-2zx\right)^{2}= \\
& 3\left(\sum x^{2}\right)^{2}+4\left(\sum x^{2}\right)\left(\sum(yz-xy-zx)\right)+ \\
& 4 \sum(yz-xy-zx)^{2}+16 \sum x^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,337 |
45. Let $0 \leqslant a, b, c \leqslant 1$, prove: $\sqrt{a(1-b)(1-c)}+\sqrt{b(1-a)(1-c)}+$ $\sqrt{c(1-b)(1-a)} \leqslant 1+\sqrt{a b c}$. (1981 Leningrad Mathematical Olympiad Problem) | 45. Proof Let $a=\sin ^{2} \alpha, b=\sin ^{2} \beta, c=\sin ^{2} \gamma, 0 \leqslant \alpha, \beta, \gamma \leqslant \frac{\pi}{2}$, then
$$\begin{array}{l}
\sqrt{a(1-b)(1-c)}+\sqrt{b(1-a)(1-c)}+\sqrt{c(1-b)(1-a)}-\sqrt{a b c}= \\
\cos \gamma(\sin \alpha \cos \beta+\cos \alpha \sin \beta)+\sin \gamma(\cos \alpha \cos ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,338 |
Example 2 Let $P(x)$ be a polynomial with real coefficients $P(x)=a x^{3}+b x^{2}+c x+d$. Prove: If for any $|x|<1$, we have $|P(x)| \leqslant 1$, then $|a|+|b|+|c|+|d| \leqslant 7$. (37th IMO) | Prove that if $P(x)$ is a continuous function and $|x|<1$, then $|P(x)| \leqslant 1$. Hence, if $|x| \leqslant 1$, then $|P(x)| \leqslant 1$. By setting $x=\lambda$ and $\frac{\lambda}{2}$ (where $\lambda = \pm 1$), we get
$$\begin{array}{c}
|\lambda a+b+\lambda c+d| \leqslant 1 \\
\left|\frac{\lambda}{8} a+\frac{1}{4}... | 7 | Algebra | proof | Yes | Yes | inequalities | false | 733,340 |
Example 4 Given $x_{i} \in \mathbf{R}(i=1,2, \cdots, n)$, satisfying $\sum_{i=1}^{n}\left|x_{i}\right|=1, \sum_{i=1}^{n} x_{i}=0$, prove: $\left.\vdash \sum_{i=1}^{n} \cdot \frac{x_{i}}{i} \right\rvert\, \leqslant \frac{1}{2}-\frac{1}{2 n} \cdot$ (1989 National High School Mathematics League Test, Second Round) | Proof: Let $x_{1}, x_{2}, \cdots, x_{n}$ be real numbers where the positive ones are $x_{k_{1}}, x_{k_{2}}, \cdots, x_{k_{l}}$, and the non-positive ones are $x_{k_{l+1}}, x_{k_{l+2}}, \cdots, x_{k_{n}}$. From the given conditions, we have
$$\begin{array}{l}
\sum_{i=1}^{l} x_{k_{i}}=\frac{1}{2} \\
\sum_{i=l+1}^{n} x_{k... | \left|\sum_{i=1}^{n} \frac{x_{i}}{i}\right| \leqslant \frac{1}{2}-\frac{1}{2 n} | Inequalities | proof | Yes | Yes | inequalities | false | 733,342 |
Example 5 Given the functions $F(x)=a x^{2}+b x+c$, and $G(x)=c x^{2}+b x+a$, where $|F(0)| \leqslant 1,|F(1)| \leqslant 1,|F(-1)| \leqslant 1$.
Prove: For $|x| \leqslant 1$, we have $|F(x)| \leqslant \frac{5}{4},|G(x)| \leqslant 2$. (26th IMO Preliminary Selection | Proof
(1) $F(0)=c, F(-1)=a-b+c, F(1)=a+b+c$,
Therefore,
$$\begin{aligned}
F(x)= & \frac{x(x-1)}{2} F(-1)-\left(x^{2}-1\right) F(0)+\frac{x(x+1)}{2} F(1) \\
2|F(x)|= & |x(x-1)| \cdot|F(-1)|+2\left|x^{2}-1\right| \\
& |F(0)|+|x(x-1)| \cdot|F(1)| \leqslant \\
& |x(x-1)|+2\left|x^{2}-1\right|+|x(x-1)|
\end{aligned}$$
For... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,343 |
14. Given $a, b, c \geqslant 1$, prove: $\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1} \leqslant \sqrt{(a b+1) c}$. | $$\begin{array}{l}
\text { 14. Let } x=\sqrt{a-1}, y=\sqrt{b-1}, z=\sqrt{c-1}, \text { then } \sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1} \leqslant \\
\sqrt{(a b+1) c} \Leftrightarrow x+y+z \leqslant \sqrt{\left[\left(1+x^{2}\right)\left(1+y^{2}\right)+1\right]\left(1+z^{2}\right)} \Leftrightarrow(x+y+z)^{2} \leqslant \\
{\left[\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,344 |
Example 6 Given $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ are all greater than 1, and $\left|a_{k+1}-a_{k}\right|<1, k=1$, $2, \cdots, n-1$. Prove: $\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}+\frac{a_{n}}{a_{1}}<2 n-1$. (21st All-Russian Mathematical Olympiad) | Prove that we have $a_{n}-a_{1}=\sum_{j=1}^{n-1}\left(a_{j+1}-a_{j}\right)$, the original inequality is equivalent to $\sum_{j=1}^{n-1} \frac{a_{j}-a_{j+1}}{a_{j+1}}+$ $\frac{a_{n}-a_{1}}{a_{1}}<n-1$ or $\sum_{j=1}^{n-1}\left(a_{j}-a_{j+1}\right)\left(\frac{1}{a_{j+1}}-\frac{1}{a_{1}}\right)<n-1$.
However,
$$\begin{ali... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,345 |
Example 8 Let $a_{1}, a_{2}, \cdots, a_{n}$ be real numbers, $s$ be a non-negative number, satisfying $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, a_{1}+$ $a_{2}+\cdots+a_{n}=0$, $\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right|=s$, prove that: $a_{n}-a_{1} \geqslant \frac{2 s}{n}$. (1996 Aus... | Prove that when $s=0$, the inequality obviously holds. Therefore, we may assume $s>0$. From $\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right|=s$, we know that at least one of $a_{1}, a_{2}, \cdots, a_{n}$ is not zero. Also, since $a_{1}+a_{2}+\cdots+a_{n}=0$, the non-zero terms among $a_{1}, a_{2}, \cdot... | \delta \geqslant \frac{2 s}{n} | Inequalities | proof | Yes | Yes | inequalities | false | 733,347 |
Example 11 Given real numbers $a_{1}, a_{2}, \cdots, a_{n}$. For each $i(1 \leqslant i \leqslant n)$, define: $d_{i}=\max \left\{a_{j} \mid\right.$ $1 \leqslant j \leqslant i\}-\min \left\{a_{j} \mid i \leqslant j \leqslant n\right\}$, and let $d=\max \left\{d_{i} \mid 1 \leqslant i \leqslant n\right\}$.
(1) Prove: For... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Inequalities | proof | Yes | Yes | inequalities | false | 733,350 |
3. Prove that for any $a_{1}, a_{2}, \cdots, a_{n} \in [0 ; 2], n \geqslant 2$, we have $\sum_{i, j=1}^{n}\left|a_{i}-a_{j}\right| \leqslant n^{2}$. Determine for which $a_{1}, a_{2}, \cdots, a_{n}$ the equality holds. (1982 Polish Mathematical Olympiad Problem) | 3. Proof: Without loss of generality, let $a_{i} \geqslant a_{2} \geqslant, \cdots \geqslant a_{n}$, then
$$\begin{array}{l}
S=\sum_{i=1}^{n} |a_{i}-a_{j}|=2 \sum_{1 \leqslant i<j \leqslant n} (a_{i}-a_{j})=2 \sum_{i=1}^{n}(n-2i+1)a_{i} \\
\left\{\begin{array}{l}
a_{i} \geqslant 0, \quad 1 \leqslant i \leqslant \frac{n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,353 |
5. Let real numbers $\theta_{1}, \theta_{2}, \cdots, \theta_{n}$ satisfy $\sin \theta_{1}+\sin \theta_{2}+\cdots+\sin \theta_{n}=0$, prove that: $\left|\sin \theta_{1}+2 \sin \theta_{2}+\cdots+n \sin \theta_{n}\right| \leqslant\left[\frac{n^{2}}{4}\right]$. (28th IMO Shortlist Problem) | 5. Let $x_{k}=\sin \theta_{k}, k=1,2, \cdots, n$.
When $n=2 m$,
$$\begin{aligned}
{\left[\frac{n^{2}}{4}\right]-\sum_{k=1}^{n} k x_{k}=} & m^{2}-\sum_{k=1}^{n} k x_{k}=\sum_{k=1}^{m}(m+k)-\sum_{k=1}^{m} k-\sum_{k=1}^{2 m} k x_{k}= \\
& \sum_{k=1}^{m}(m+k)\left(1-x_{m+k}\right)-\sum_{k=1}^{m}\left(1+x_{k}\right) \geqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,356 |
6. Let $\alpha, \beta$ be real numbers, and $\cos \alpha \neq \cos \beta, k$ be a positive integer greater than 1, prove: $\left|\frac{\cos k \beta \cos \alpha-\cos k \alpha \cos \beta}{\cos \beta-\cos \alpha}\right|<k^{2}-1$. (17th Putnam Mathematical Competition Problem) | 6. Let $x=\frac{1}{2}(\alpha-\beta), y=\frac{1}{2}(\alpha+\beta)$, then $\cos \beta-\cos \alpha=2 \sin x \sin y$,
$$\begin{aligned}
\cos k \beta \cos \alpha-\cos k \alpha \cos \beta= & \frac{1}{2}[\cos (k \beta+\alpha)+\cos (k \beta-\alpha)]- \\
& \frac{1}{2}[\cos (k \alpha+\beta)+\cos (k \alpha-\beta)]= \\
& \frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,357 |
7. Given $a_{1}, a_{2}, \cdots, a_{n}$ are pairwise distinct real numbers, find the minimum value of the function defined by $f(x)=\sum_{i=1}^{n}\left|x-a_{i}\right|$. Here, $x$ is a real number. (1969 Polish Mathematical Olympiad) | 7. First, note that when $a < b$,
$$x - a + \left| x - b \right| = \left\{\begin{array}{l}
a + b - 2x, \quad x \leqslant a \\
-a + b, \quad a \leqslant x \leqslant b \\
2x - a - b, \quad x \geqslant b
\end{array}\right.$$
Therefore, at each point in the interval $[a, b]$, the sum $|x - a| + |x - b|$ reaches its minimu... | -a_{1} - a_{2} - \cdots - a_{m} + a_{m+1} + \cdots + a_{n} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,358 |
8. Let $n$ be a given positive integer, and the sum $\sum_{1 \leqslant i<j \leqslant n}\left|x_{i}-x_{j}\right|=\left|x_{1}-x_{2}\right|+\left|x_{1}-x_{3}\right|+\cdots+$ $\left|x_{1}-x_{n}\right|+\left|x_{2}-x_{3}\right|+\left|x_{2}-x_{4}\right|+\cdots+\left|x_{2}-x_{n}\right|+\left|x_{n-2}-x_{n}\right|+\left|x_{n-2}-... | 8. Let $0 \leqslant x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n} \leqslant 1$. Let $S=\sum_{i \leqslant i<j \leqslant n}\left|x_{i}-x_{j}\right|=\sum_{1 \leqslant i<j \leqslant n}\left(x_{j}-\right.$ $\left.x_{i}\right)$.
This sum has $\mathrm{C}_{n}^{2}$ terms. Each $x_{k}$ appears in $n-1$ terms $(1 \leqsla... | \left[\frac{n^{2}}{4}\right] | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 733,359 |
9. Prove that if for any $x_{1}, x_{2}, \cdots, x_{n} \in\{-1,1\}$, the number $M$ and the array
$$\begin{array}{c}
a_{11}, a_{12}, \cdots, a_{1 n} \\
a_{21}, a_{22}, \cdots, a_{2 n} \\
\vdots \\
a_{n 1}, a_{n 2}, \cdots, a_{n n}
\end{array}$$
have $\sum_{j=1}^{n}\left|a_{j 1} x_{1}+a_{j 2} x_{2}+\cdots+a_{j n} x_{n}\... | 9. For any $x_{1}, x_{2}, \cdots, x_{n} \in\{-1,1\}$, we have
$$-\sum_{j=1}^{n}\left|a_{j 1} x_{1}+a_{j 2} x_{2}+\cdots+a_{j n} x_{n}\right| \leqslant M$$
Therefore,
$$\frac{1}{2^{n}} \sum_{\left(x_{1}, x_{2}, \cdots, x_{n}\right)}\left(\sum_{j=1}^{n}\left|a_{j 1} x_{1}+a_{j 2} x_{2}+\cdots+a_{j n} x_{n}\right|\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,360 |
10. Let $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}, g(x)=c_{n+1} x^{n+1}+c_{n} x^{n}+\cdots+$ $c_{1} x+c_{0}$ be two polynomials with real coefficients, and there exists a real number $r$ such that $g(x)=(x-r) f(x)$. Let $A=$ $\max \left\{\left|a_{n}\right|,\left|a_{n-1}\right|, \cdots,\left|a_{0}\right|\ri... | 10. Since
$$c_{n+1} x^{n+1}+c_{n} x^{n}+\cdots+c_{1} x+c_{0}=(x-r)\left(a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\right)$$
Therefore,
$$c_{n+1}=a_{n}, c_{n}=a_{n-1}-r a_{n}, \cdots, c_{1}=a_{0}-r a_{1}, c_{0}=-r a_{0}$$
Thus, $a_{n}=c_{n+1}, a_{n-1}=c_{n}+r c_{n+1}, a_{n-2}=c_{n-1}+r c_{n}+r^{2} c_{n+1}, \cdot... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,361 |
11. Let $0<\alpha, \beta, \gamma<\frac{\pi}{2}$ satisfy $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+2 \cos \alpha \cos \beta \cos \gamma=1$, prove that: $\left|\frac{\sin \alpha-\sin \beta}{\sin \alpha+\sin \beta}+\frac{\sin \beta-\sin \gamma}{\sin \beta+\sin \gamma}+\frac{\sin \gamma-\sin \alpha}{\sin \gamma+\s... | 11. From the given conditions, we have $(\cos \gamma+\cos \alpha \cos \beta)^{2}=\left(1-\cos ^{2} \alpha\right)\left(1-\cos ^{2} \beta\right)=$ $\sin ^{2} \alpha \sin ^{2} \beta$
Since $0<\alpha, \beta, \gamma<\frac{\pi}{2}$, we have $\cos \gamma+\cos \alpha \cos \beta=\sin \alpha \sin \beta$, thus
$$\cos \gamma=-\co... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,362 |
14. Let $a_{1}, a_{2}, \cdots, a_{n}$ be real numbers, prove: $\sum_{i j}\left|a_{i}+a_{j}\right| \geqslant n \sum_{i=1}^{n}\left|a_{i}\right|$. | 14. $\left.\left|x_{1}, x_{2}, \cdots, x_{n}\right|=\mid a_{1}, a_{2}, \cdots, a_{r}\right\} \cup\left|-b_{1},-b_{2}, \cdots,-b_{s}\right|$
where $r+s=n, b_{i} \geqslant 0, c_{i}>0$ (divide $x_{1}, x_{2}, \cdots, x_{n}$ into non-negative and negative parts, and assume without loss of generality that $r \geqslant s$, o... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,365 |
16. Let $a, b, c \in \mathbf{R}^{+}$, prove: $a^{3}+b^{3}+c^{3}-3 a b c \geqslant \max \left\{(a-b)^{2} a,(a-b)^{2} b\right.$, $\left.(b-c)^{2} b,(b-c)^{2} c,(c-a)^{2} c,(c-a)^{2} a\right\} .(1999$ MOP Problem) | 16. By symmetry, without loss of generality, let $a \geqslant b \geqslant c$, then $(a-c)^{2} a$ is the largest.
$$a^{3}+b^{3}+c^{3}-3 a b c-(a-c)^{2} a=2 a^{2} c-a\left(3 b c+c^{2}\right)+b^{3}+c^{3}$$
Let $f(x)=2 c x^{2}-\left(3 b c+c^{2}\right) x+b^{3}+c^{3}$, then its axis of symmetry is $x=\frac{3 b+c}{4} \leqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,366 |
16. Let $\left|a_{n}\right|(n \geqslant 1)$ be a sequence, and satisfy $\left|a_{n+1}-a_{n}\right| \leqslant 1$, the sequence $\left|b_{n}\right|(n \geqslant 1)$ is defined as follows $b_{n}=a_{1}+a_{2}+\cdots+a_{n}$, prove: $\left|b_{n+1}-b_{n}\right| \leqslant \frac{1}{2}(2008$ Romanian Math | 16. Since for any $i(1 \leqslant i \leqslant n)$, we have $\left|a_{i}-a_{n+1}\right| \leqslant\left|a_{i}-a_{i+1}\right|+\mid a_{i+1}-a_{i+2}$ $|+\cdots+| a_{n}-a_{n+1} \mid \leqslant n+1-i$, therefore
$\left|b_{n}-b_{n+1}\right|=\left|\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}-\frac{a_{1}+a_{2}+\cdots+a_{n+1}}{n+1}\right|=$
... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,368 |
Example 1 Let real numbers $a_{1}, a_{2}, \cdots, a_{100}$ satisfy $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{100} \geqslant 0$, $a_{1}+a_{2} \leqslant 100, a_{3}+a_{4}+\cdots+a_{100} \leqslant 100$, determine the maximum value of $a_{1}^{2}+a_{2}^{2}+\cdots+$ $a_{100}^{2}$, and find the sequence $a_{1}, a_{2... | Solve: Since $a_{1}+a_{2}+a_{3}+a_{4}+\cdots+a_{100} \leqslant 200$, then
$$\begin{aligned}
a_{1}^{2}+ & a_{2}^{2}+\cdots+a_{100}^{2} \leqslant\left(100-a_{2}\right)^{2}+a_{2}^{2}+a_{3}^{2}+\cdots+a_{100}^{2}= \\
& 100^{2}-200 a_{2}+2 a_{2}^{2}+a_{3}^{2}+\cdots+a_{100}^{2} \leqslant \\
& 100^{2}-\left(a_{1}+a_{2}+a_{3}... | 10000 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,369 |
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