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class | __index_level_0__ int64 0 742k |
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Example 3 Let positive real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$. Prove that:
$$x^{2} y z+y^{2} x z+z^{2} x y \leqslant \frac{1}{3}$$ | Prove that because $\sqrt[3]{x y z} \leqslant \frac{x+y+z}{3} \leqslant \sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}=\frac{1}{\sqrt{3}}$, so
$$\begin{array}{c}
x y z \leqslant \frac{1}{3 \sqrt{3}}, x+y+z \leqslant \sqrt{3}, \\
x^{2} y z+y^{2} x z+z^{2} x y \leqslant \frac{1}{3}
\end{array}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,173 |
Example 4 Let $a, b, c, d \in \mathbf{R}^{+}$, prove that:
$$\sqrt[3]{\frac{a b c+b c d+c d a+a d b}{4}} \leqslant \sqrt{\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}}$$ | Prove that by applying $G_{2} \leqslant A_{2}$ twice, we get
$$\begin{aligned}
& \frac{a b c+b c d+c d a+a d b}{4} \\
= & \frac{1}{2}\left(a b \cdot \frac{c+d}{2}+c d \cdot \frac{a+b}{2}\right) \\
\leqslant & \frac{1}{2}\left[\left(\frac{a+b}{2}\right)^{2} \cdot \frac{c+d}{2}+\left(\frac{c+d}{2}\right)^{2} \cdot \frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,174 |
Example 5 Let $x_{i} \geqslant 1(i=1,2, \cdots, n)$, prove:
$$\frac{\prod\left(x_{i}-1\right)}{\left(\sum\left(x_{i}-1\right)\right)^{n}} \leqslant \frac{\prod x_{i}}{\left(\sum x_{i}\right)^{n}}$$ | Prove that the original inequality is equivalent to
$$\left(\frac{\prod\left(x_{i}-1\right)}{\prod x_{i}}\right)^{\frac{1}{n}} \leqslant \frac{\sum\left(x_{i}-1\right)}{\sum x_{i}}$$
This inequality can be derived from the following facts.
By the AM-GM inequality, we have
$$\begin{aligned}
\left(\frac{\prod\left(x_{i}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,175 |
Example 6 Let $x_{i} \in\left[0, \frac{\pi}{2}\right], i=1,2, \cdots, 10$, satisfy $\sin ^{2} x_{1}+\sin ^{2} x_{2}+\cdots+$ $\sin ^{2} x_{10}=1$. Prove:
$$3\left(\sin x_{1}+\cdots+\sin x_{10}\right) \leqslant \cos x_{1}+\cdots+\cos x_{10} \text {. }$$ | Prove that since $\sin ^{2} x_{1}+\sin ^{2} x_{2}+\cdots+\sin ^{2} x_{10}=1$,
$$\cos x_{i}=\sqrt{\sum_{j \neq i} \sin ^{2} x_{j}}$$
then for $1 \leqslant i \leqslant 10$, we have
$$\cos x_{i}=\sqrt{\sum_{j \neq i} \sin ^{2} x_{j}} \geqslant \frac{\sum_{j \neq i} \sin x_{j}}{3}$$
Thus,
$$\sum_{i=1}^{10} \cos x_{i} \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,176 |
Example 7 Let $a_{i} \in \mathbf{R}^{+}, i=1,2, \cdots, n$, and $\sum_{i=1}^{n} a_{i}=1$, find
$$M=\sum_{i=1}^{n} \frac{a_{i}}{1+\sum_{j \neq i, j=1}^{n} a_{j}}$$
the minimum value. | Solve:
\[
\begin{aligned}
M+n & =\left(\frac{a_{1}}{2-a_{1}}+1\right)+\left(\frac{a_{2}}{2-a_{2}}+1\right)+\cdots+\left(\frac{a_{n}}{2-a_{n}}+1\right) \\
& =\frac{2}{2-a_{1}}+\frac{2}{2-a_{2}}+\cdots+\frac{2}{2-a_{n}}\left(\text { by } H_{n} \leqslant A_{n}\right) \\
& \geqslant \frac{n^{2}}{\frac{1}{2}\left(2-a_{1}\r... | \frac{n}{2 n-1} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,177 |
Example 8 Given $a, b, c \in \mathbf{R}^{+}$, and satisfying $\frac{a^{2}}{1+a^{2}}+\frac{b^{2}}{1+b^{2}}+\frac{c^{2}}{1+c^{2}}=1$, prove:
$$a b c \leqslant \frac{\sqrt{2}}{4} .$$ | Let $x=\frac{a^{2}}{1+a^{2}}, y=\frac{b^{2}}{1+b^{2}}, z=\frac{c^{2}}{1+c^{2}}$, then
$$\begin{array}{c}
00$, then we call
$$M_{r}=\left(\frac{\sum_{i=1}^{n} a_{i}^{r}}{n}\right)^{\frac{1}{r}}$$
the $r$-th power mean of $a_{1}, a_{2}, \cdots, a_{n}$.
For $M_{r}$, we have the power mean inequality, that is:
For $\alpha... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,178 |
Example 9 Given a positive integer $k$, when $x^{k}+y^{k}+z^{k}=1$, find the minimum value of $x^{k+1}+y^{k+1}+z^{k+1}$. | From the assumption and the power mean inequality, we have
$$\left(\frac{x^{k+1}+y^{k+1}+z^{k+1}}{3}\right)^{\frac{1}{k+1}} \geqslant\left(\frac{x^{k}+y^{k}+z^{k}}{3}\right)^{\frac{1}{k}}=\left(\frac{1}{3}\right)^{\frac{1}{k}},$$
Therefore,
$$x^{k+1}+y^{k+1}+z^{k+1} \geqslant 3\left(\frac{1}{3}\right)^{\frac{k+1}{k}}=... | 3^{-\frac{1}{k}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,179 |
19 Let $x_{1}, x_{2}, \cdots, x_{n}>0$, and $x_{1} x_{2} \cdots x_{n}=1$. Prove: $\frac{1}{x_{1}\left(1+x_{1}\right)}+\frac{1}{x_{2}\left(1+x_{2}\right)}$ $+\cdots+\frac{1}{x_{n}\left(1+x_{n}\right)} \geqslant \frac{n}{2}$. | 9. Proof: It is clear that the original inequality is equivalent to $\frac{1+x_{1}+x_{1}^{2}}{x_{1}\left(1+x_{1}\right)}+\frac{1+x_{2}+x_{2}^{2}}{x_{2}\left(1+x_{2}\right)}+\cdots+$ $\frac{1+x_{n}+x_{n}^{2}}{x_{n}\left(1+x_{n}\right)} \geqslant \frac{3 n}{2}$. Note that $4\left(1+x_{i}+x_{i}^{2}\right) \geqslant 3\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,181 |
Example 1 Let $a, b, c, d$ be real numbers, not all zero, find
$$f=\frac{a b+2 b c+c d}{a^{2}+b^{2}+c^{2}+d^{2}}$$
the maximum value. | Assume the maximum value of $f$ is $M$, then
$$a b+2 b c+c d \leqslant M\left(a^{2}+b^{2}+c^{2}+d^{2}\right),$$
Therefore, to establish an inequality of the above form and find a set of values for $a, b, c, d$ that make the inequality an equality.
Let $\alpha, \beta, \gamma>0$, then
$$\frac{\alpha}{2} a^{2}+\frac{b^{... | \frac{\sqrt{2}+1}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,183 |
Example 2 Find the largest positive number $\lambda$ such that for any real numbers $x$, $y$, $z$ satisfying $x^{2}+y^{2}+z^{2}=1$, the inequality holds:
$$|\lambda x y+y z| \leqslant \frac{\sqrt{5}}{2} .$$ | Since
$$\begin{aligned}
1 & =x^{2}+y^{2}+z^{2}=x^{2}+\frac{\lambda^{2}}{1+\lambda^{2}} y^{2}+\frac{1}{1+\lambda^{2}} y^{2}+z^{2} \\
& \geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(\lambda|x y|+|y z|) \geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(|\lambda x y+y z|),
\end{aligned}$$
and when $y=\frac{\sqrt{2}}{2}, x=\frac{\sqrt{2... | 2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,184 |
Example 4 Let $a$ be a real number, find the maximum value of the function $f(x)=|\sin x(a+\cos x)|(x \in \mathbf{R})$. | Let $\alpha$ be a parameter such that
$$\begin{aligned}
f^{2}(x) & =\frac{1}{\alpha^{2}} \sin ^{2} x(a \alpha+\alpha \cos x)^{2} \leqslant \frac{1}{\alpha^{2}} \sin ^{2} x\left(\alpha^{2}+\cos ^{2} x\right)\left(a^{2}+\alpha^{2}\right) \\
& \leqslant \frac{1}{\alpha^{2}}\left(\frac{\sin ^{2} x+\alpha^{2}+\cos ^{2} x}{2... | \frac{\sqrt{a^{4}+8 a^{2}}-a^{2}+4}{8} \cdot \sqrt{\frac{\sqrt{a^{4}+8 a^{2}}+a^{2}+2}{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,186 |
Example 5 Let $x, y, z \in \mathbf{R}^{+}$, and $x^{4}+y^{4}+z^{4}=1$, find
$$f(x, y, z)=\frac{x^{3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}+\frac{z^{3}}{1-z^{8}}$$
the minimum value. | Solve by transforming the original expression into
$$f(x, y, z)=\frac{x^{4}}{x\left(1-x^{8}\right)}+\frac{y^{4}}{y\left(1-y^{8}\right)}+\frac{z^{4}}{z\left(1-z^{8}\right)}$$
For $w \in(0,1)$, let $\phi(w)=w\left(1-w^{8}\right)$, and first find the maximum value of $\phi(w)$. Choose a parameter $a$, and use $G_{9} \leq... | \frac{9}{8} \sqrt[4]{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,187 |
Example 6 Find the smallest positive integer $k$, such that for all $a$ satisfying $0 \leqslant a \leqslant 1$ and all positive integers $n$, the inequality
$$a^{k}(1-a)^{n} \leqslant \frac{1}{(n+1)^{3}}$$
holds. | Solve by first eliminating the parameter $a$, then find the minimum value of $k$.
By the AM-GM inequality, we have
Therefore $\square$
$$\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{k+n}=\frac{k}{k+n}$$
$$a^{k}(1-a)^{n} \leqslant \frac{k^{k} n^{n}}{(n+k)^{n+k}}... | 4 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,188 |
Given $0<a<1,0<b<1$, and $a b=\frac{1}{36}$. Find the minimum value of $u=\frac{1}{1-a}+\frac{1}{1-b}$. | 1. By the AM-GM inequality $u \geqslant \frac{2}{\sqrt{(1-a)(1-b)}} \geqslant \frac{2}{\frac{1-a+1-b}{2}}=$ $\frac{4}{2-(a+b)} \geqslant \frac{4}{2-2 \sqrt{a b}}=\frac{4}{2-\frac{1}{3}}=\frac{12}{5}$. When $a=b=\frac{1}{6}$, the equality holds, hence the minimum value of $u$ is $\frac{12}{5}$. | \frac{12}{5} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,189 |
2. Let $a, b, c \in \mathbf{R}^{+}$, and $a+b+c=3$. Prove:
$$\sum \frac{1}{a \sqrt{2\left(a^{2}+b c\right)}} \geqslant \sum \frac{1}{a+b c}$$
where, " $\sum$ " denotes the cyclic sum. | 2. Without loss of generality, let $a \geqslant b \geqslant c$. Then $\frac{(c-a)(c-b)}{3(c+a b)} \geqslant 0$ and $\frac{(a-b)(a-c)}{3(a+b c)}+ \frac{(b-a)(b-c)}{3(b+a c)}=\frac{c(a-b)^{2}}{3}\left[\frac{1+a+b-c}{(a+b c)(b+a c)}\right] \geqslant 0$. Therefore, $\sum \frac{(a-b)(a-c)}{3(a+b c)} \geqslant 0$. And $\sum ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,190 |
3 Let $a_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n, n \geqslant 2)$, prove:
$$\frac{a_{1}}{a_{2}+a_{3}}+\frac{a_{2}}{a_{3}+a_{4}}+\cdots+\frac{a_{n-1}}{a_{n}+a_{1}}+\frac{a_{n}}{a_{1}+a_{2}}>\frac{n}{4}$$ | 3. Let $a_{n+1}=a_{1}, a_{n+2}=a_{2}$. Suppose $a_{i_{1}}=\max _{1 \leqslant i \leqslant n} a_{i}$, the left side of the original equation is $s$, and the term in $s$ that contains $a_{i_{1}}$ in the numerator is $\frac{a_{i_{1}}}{a_{i_{1}+1}+a_{i_{1}+2}}$. Let $a_{i_{2}}=\max \left\{a_{i_{1}+1}, a_{i_{1}+2}\right\}$. ... | \frac{n}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 738,191 |
10 Let $a, b, c>0$ and $a^{2}+b^{2}+c^{2}+(a+b+c)^{2} \leqslant 4$. Prove:
$$\frac{a b+1}{(a+b)^{2}}+\frac{b c+1}{(b+c)^{2}}+\frac{c a+1}{(c+a)^{2}} \geqslant 3$$ | 10. Proof: From $a^{2}+b^{2}+c^{2}+(a+b+c)^{2} \leqslant 4$ we can deduce that $a^{2}+b^{2}+c^{2}+a b+b c+c a \leqslant 2$, hence $\frac{2(a b+1)}{(a+b)^{2}} \geqslant \frac{2 a b+a^{2}+b^{2}+c^{2}+a b+b c+c a}{(a+b)^{2}}=$ $\frac{(a+b)^{2}+(c+a)(c+b)}{(a+b)^{2}}$ which means $\frac{a b+1}{(a+b)^{2}} \geqslant \frac{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,192 |
Let $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 2)$ be positive real numbers, and satisfy $a_{1}+a_{2}+\cdots+a_{n}<1$. Prove:
$$\frac{a_{1} a_{2} \cdots a_{n}\left[1-\left(a_{1}+\cdots+a_{n}\right)\right]}{\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right)} \leqslant ... | 4. Let $a_{n+1}=1-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$, then $a_{n+1} \geqslant 0, a_{1}+\cdots+a_{n+1}=1$. The original inequality is equivalent to $n^{n+1} a_{1} a_{2} \cdots a_{n+1} \leqslant\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right)\left(1-a_{n+1}\right)$. For $i=1,2, \cdots, n+1$, by the... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,193 |
6. Let $x, y \in \mathbf{R}^{+}, x \neq y$. Define
$$Q=\sqrt{\frac{x^{2}+y^{2}}{2}}, A=\frac{x+y}{2}, G=\sqrt{x y}, H=\frac{2 x y}{x+y} .$$
Prove: $G-H<Q-A<A-G$. | $$\begin{array}{l}
\text { 6. } \frac{x+y}{2}-\sqrt{x y}>\sqrt{\frac{x^{2}+y^{2}}{2}}-\frac{x+y}{2} \Leftrightarrow x+y>\sqrt{\frac{x^{2}+y^{2}}{2}}+\sqrt{x y} \Leftrightarrow \\
(x+y)^{2}>\frac{x^{2}+y^{2}}{2}+x y+\sqrt{2 x y\left(x^{2}+y^{2}\right)} \Leftrightarrow \frac{1}{2}(x+y)^{2}>\sqrt{2 x y\left(x^{2}+y^{2}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,195 |
8 Let $x_{1}, x_{2}, \cdots, x_{n}$ be distinct positive rational numbers. Prove:
$$\left(\frac{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}{x_{1}+x_{2}+\cdots+x_{n}}\right)^{x_{1}+x_{2}+\cdots+x_{n}}>x_{1}^{x_{1}} x_{2}^{x_{2}} \cdots x_{n}^{x_{n}}$$ | 8. First, let $x_{i} \in \mathbf{N}$. Since $x_{i}$ are distinct, the left side $=$ $[\frac{\sum_{i=1}^{n} \overbrace{x_{i}+\cdots+x_{i}}^{x_{i} \text { times }}}{x_{1}+x_{2}+\cdots+x_{n}}]^{x_{1}+x_{2}+\cdots+x_{n}}>x_{1}^{x_{1}} \cdot x_{2}^{x_{2}} \cdots \cdots x_{n}{ }^{x_{n}}$. Next, let $x_{i}$ be positive ration... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,197 |
10 Prove: $\left(1-\frac{1}{365}\right)\left(1-\frac{2}{365}\right) \cdots\left(1-\frac{25}{365}\right)<\frac{1}{2}$. | 10. By the inequality of arithmetic means, we get the left side $\leqslant\left\{\frac{1}{25}\left[\left(1-\frac{1}{365}\right)+\left(1-\frac{2}{365}\right)+\cdots+\right.\right.$ $\left.\left.\left(1-\frac{25}{365}\right)\right]\right\}^{25}=\left(1-\frac{13}{365}\right)^{25}$. In the binomial expansion of $\left(1-\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,199 |
II Let $a, d \geqslant 0, b, c>0$ and $b+c \geqslant a+d$. Find the minimum value of $\frac{b}{c+d}+\frac{c}{a+b}$. | 11. From the given, we have $b+c \geqslant \frac{1}{2}(a+b+c+d)$. Without loss of generality, assume $a+b=c+d$, then
$$\frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}+c\left(\frac{1}{a+b}-\frac{1}{c+d}\right) \geqslant \frac{\frac{1}{2}(a+b+c+d)}{c+d}+(c+$$
d) $\left(\frac{1}{a+b}-\frac{1}{c+d}\right)=\frac{a+b}{2(c+d)}+\f... | \sqrt{2}-\frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,200 |
12 For any positive numbers $a_{1}, a_{2}, \cdots, a_{n}, n \geqslant 2$, find the minimum value of $\sum_{i=1}^{n} \frac{a_{i}}{S-a_{i}}$, where $S=$ $\sum_{i=1}^{n} a_{i}$. | 12. Let $b_{i}=S-a_{i}$, then $\sum_{i=1}^{n} b_{i}=(n-1) S$. By the AM-GM inequality, we get $\sum_{i=1}^{n} \frac{a_{i}}{S-a_{i}}$
$$=\sum_{i=1}^{n}\left(\frac{a_{i}}{b_{i}}+1\right)-n=S \sum_{i=1}^{n} \frac{1}{b_{i}}-n \geqslant \frac{n S}{\sqrt[n]{b_{1} \cdots b_{n}}}-n \geqslant \frac{n^{2} S}{b_{1}+\cdots+b_{n}}-... | \frac{n}{n-1} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,201 |
13 Find the maximum value of the product $x^{2} y^{2} z^{2} u$ under the conditions $x, y, z, u \geqslant 0$ and $2 x+x y+z+y z u=1$. | 13. By the AM-GM inequality, we have $\sqrt[4]{2 x^{2} y^{2} z^{2} u} \leqslant \frac{2 x+x y+z+z y u}{4}=\frac{1}{4}$, which implies $x^{2} y^{2} z^{2} u \leqslant \frac{1}{512}$. Moreover, equality holds when $2 x=x y=z=y z u=\frac{1}{4}$, i.e., $x=\frac{1}{8}, y=2, z=\frac{1}{4}$, $u=\frac{1}{2}$. Therefore, the max... | \frac{1}{512} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,202 |
Example 1 Let $a, b, c$ be positive real numbers, prove that:
$$\frac{a^{2}+b c}{b+c}+\frac{b^{2}+c a}{c+a}+\frac{c^{2}+a b}{a+b} \geqslant a+b+c \text {. }$$ | $$\begin{array}{l}
=\frac{a^{2}+b c}{b+c}-a+\frac{b^{2}+c a}{c+a}-b+\frac{c^{2}+a b}{a+b}-c \\
=\frac{a^{2}+b c-a b-a c}{b+c}+\frac{b^{2}+c a-b c-b a}{c+a}+\frac{c^{2}+a b-c a-c b}{a+b} \\
=\frac{(a-b)(a-c)}{b+c}+\frac{(b-c)(b-a)}{c+a}+\frac{(c-a)(c-b)}{a+b} \\
=\frac{\left(a^{2}-b^{2}\right)\left(a^{2}-c^{2}\right)+\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,203 |
Example 2 Real numbers $x, y, z$ satisfy $xy + yz + zx = -1$, prove:
$$x^{2} + 5y^{2} + 8z^{2} \geqslant 4$$ | Prove that
$$\begin{aligned}
& x^{2}+5 y^{2}+8 z^{2}-4 \\
= & x^{2}+5 y^{2}+8 z^{2}+4(x y+y z+z x) \\
= & (x+2 y+2 z)^{2}+(y-2 z)^{2} \geqslant 0,
\end{aligned}$$
Therefore,
$$x^{2}+5 y^{2}+8 z^{2} \geqslant 4 \text {. }$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,204 |
Example 11 Positive real numbers $x, y, z$ satisfy $x y z \geqslant 1$, prove
$$\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 .$$ | Prove that the original inequality can be transformed into
$$\frac{x^{2}+y^{2}+z^{2}}{x^{5}+y^{2}+z^{2}}+\frac{x^{2}+y^{2}+z^{2}}{y^{5}+z^{2}+x^{2}}+\frac{x^{2}+y^{2}+z^{2}}{z^{5}+x^{2}+y^{2}} \leqslant 3$$
By the Cauchy-Schwarz inequality and the condition \(xyz \geqslant 1\), we have
$$\left(x^{5}+y^{2}+z^{2}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,205 |
4 Let $x, y, z \in \mathbf{R}^{+}$, prove:
$$\frac{x+y+z}{3} \cdot \sqrt[3]{x y z} \leqslant\left(\frac{x+y}{2} \cdot \frac{y+z}{2} \cdot \frac{z+x}{2}\right)^{\frac{2}{3}} .$$ | 4. Let $x+y=2a, y+z=2b, z+x=2c$. Then $a, b, c$ can form a triangle, with area $S$ and circumradius $R$. It is easy to see that the original inequality is equivalent to $a+b+c \leqslant 3 \sqrt{3} R$.
By $2 R(\sin A+\sin B+\sin C)=a+b+c \leqslant 2 R \cdot 3 \cdot \sin \frac{A+B+C}{3}=3 \sqrt{3} R$,
thus the original ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,206 |
5 If $x, y \in \mathbf{R}^{+}$, prove:
$$\left(\frac{2 x+y}{3} \cdot \frac{x+2 y}{3}\right)^{2} \geqslant \sqrt{x y} \cdot\left(\frac{x+y}{2}\right)^{3} .$$ | 5. Let $x+y=a, xy=b$, then $a^{2} \geqslant 4 b>0$. Hence
$$\begin{aligned}
\text { LHS } & =\left(\frac{2 a^{2}+b}{9}\right)^{2}=\left(\frac{\frac{1}{4} a^{2}+\frac{1}{4} a^{2}+\cdots+\frac{1}{4} a^{2}+b}{9}\right)^{2} \\
& \geqslant \sqrt[9]{\frac{1}{4^{16}} \cdot a^{32} \cdot b^{2}} \geqslant \sqrt[9]{\frac{1}{2^{27... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,207 |
6 Let real numbers $a, b$ satisfy $ab>0$, prove that: $\sqrt[3]{\frac{a^{2} b^{2}(a+b)^{2}}{4}} \leqslant \frac{a^{2}+10 a b+b^{2}}{12}$, and determine the condition for equality. Generally, for any real numbers $a, b$, prove:
$$\sqrt[3]{\frac{a^{2} b^{2}(a+b)^{2}}{4}} \leqslant \frac{a^{2}+a b+b^{2}}{3}$$ | 6. (1) Let $a b=x>0, a+b=y$, then $y^{2} \geqslant 4 x$. Therefore, the right side of the inequality $=\frac{y^{2}+8 x}{12}=$ $\frac{y^{2}}{12}+\frac{x}{3}+\frac{x}{3} \geqslant 3 \cdot \sqrt[3]{\frac{x^{2} y^{2}}{12 \cdot 3^{2}}}=\sqrt[3]{\frac{x^{2} y^{2}}{4}}=$ left side, hence the inequality holds, and the equality... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,208 |
7 Let $a, b, c \in \mathbf{R}^{+}, abc=1$, prove:
$$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a} \leqslant \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c} .$$ | 7. Let $x=a+b+c \geqslant 3 \sqrt[3]{a b c}=3, y=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a b+b c+c a \geqslant 3 \sqrt[3]{a^{2} b^{2} c^{2}}=3$, then the original inequality is equivalent to
$$\begin{array}{l}
\frac{(1+b+c)(1+c+a)+(1+c+a)(1+a+b)+(1+a+b)(1+b+c)}{(1+a+b)(1+b+c)(1+c+a)} \\
\leqslant \frac{(2+b)(2+c)+(2+c)(2+a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,209 |
8 Given that $a, b, c, d, e$ are positive numbers, and $a b c d e=1$, prove:
$$\begin{array}{c}
\frac{a+a b c}{1+a b+a b c d}+\frac{b+b c d}{1+b c+b c d e}+\frac{c+c d e}{1+c d+c d e a}+ \\
\frac{d+d e a}{1+d e+d e a b}+\frac{e+e a b}{1+e a+e a b c} \geqslant \frac{10}{3}
\end{array}$$ | 8. Let $a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{u}{z}, d=\frac{v}{u}, e=\frac{x}{v}, x, y, z, u, v \in \mathbf{R}^{+}$, then the original inequality is equivalent to $\frac{u+y}{x+z+v}+\frac{z+v}{x+y+u}+\frac{x+u}{y+z+v}+\frac{y+v}{x+z+u}+$ $\frac{x+z}{y+u+v} \geqslant \frac{10}{3}$
Adding 5 to both sides, and then mult... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,210 |
9 Let $a, b, c$ be positive real numbers, prove that:
$$a^{2}+b^{2}+c^{2} \geqslant \frac{c\left(a^{2}+b^{2}\right)}{a+b}+\frac{b\left(c^{2}+a^{2}\right)}{c+a}+\frac{a\left(b^{2}+c^{2}\right)}{b+c} .$$ | 9. Let $a \geqslant b \geqslant c$, and set $\frac{a}{c}=x, \frac{b}{c}=y$, then $x \geqslant y \geqslant 1$.
The original inequality transforms to $x^{2}+y^{2}+1 \geqslant \frac{x^{2}+y^{2}}{x+y}+\frac{y\left(1+x^{2}\right)}{1+x}+\frac{x\left(1+y^{2}\right)}{1+y}$.
Clearing the denominators and rearranging, we get $\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,211 |
10 Let $x, y, z \in \mathbf{R}^{+}$, and satisfy $x y z + x + z = y$, find the maximum value of $p = \frac{2}{x^{2}+1} - \frac{2}{y^{2}+1} + \frac{3}{z^{2}+1}$. | \begin{array}{l}\text { 10. Given, } x+(-y)+z=x \cdot(-y) \cdot z \text {. } \\ \text { Let } x=\tan \alpha, y=-\tan \beta, z=\tan \gamma,(\alpha+\beta+\gamma=k \pi) . \\ \text { Therefore, } p=2 \cos ^{2} \alpha-2 \cos ^{2} \beta+3 \cos ^{2} \gamma=2 \cos ^{2} \alpha-2 \cos ^{2} \beta+3 \cos ^{2}(\alpha+\beta)= \\ -2 ... | \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,212 |
11 Prove: In the open interval $(0,1)$, there must exist four pairs of distinct positive numbers $(a, b)(a \neq b)$, satisfying:
$$\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}$$ | 11. Let $a=\cos \alpha, b=\cos \beta, \alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, then
$$a b+\sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}=\cos (\alpha-\beta)$$
Square both sides, we have $\sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}=\frac{1}{2 a b} \cdot\left[\cos ^{2}(\alpha-\beta)-1\right]+\frac{a}{2 b}+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,213 |
12 Let $s$ be the set of all triangles satisfying the following condition:
$$5\left(\frac{1}{A P}+\frac{1}{B Q}+\frac{1}{C R}\right)-\frac{3}{\min \{A P, B Q, C R\}}=\frac{6}{r},$$
where $r$ is the inradius of $\triangle A B C$, and $P, Q, R$ are the points where the incircle touches sides $A B, B C, C A$, respectivel... | 12. Let $a=\max \{a, b, c\}$, then $A P=\min \{A P, B Q, C R\}$, from the problem we have
$$\begin{array}{l}
\frac{4}{-a+b+c}+\frac{10}{-b+a+c}+\frac{10}{-c+a+b}=\frac{6}{r} \\
\quad \text { Let }-a+b+c=2 x,-b+a+c=2 y,-c+a+b=2 z, x, y, z>0 \text {. }
\end{array}$$
Then the above equation is equivalent to: $\frac{2}{x}... | proof | Geometry | proof | Yes | Yes | inequalities | false | 738,214 |
13 Let $a, b, c \in \mathbf{R}^{+}$, and satisfy $a b c=1$, prove:
$$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1$$ | 13. Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, and $x, y, z \in \mathbf{R}^{+}$, then the original inequality is equivalent to $\left(\frac{x}{y}-1+\frac{z}{y}\right)\left(\frac{y}{z}-1+\frac{x}{z}\right)\left(\frac{z}{x}-1+\frac{y}{x}\right) \leqslant 1$, i.e., $(x-y+z)(y-z+x)(z-x+y) \leqslant x y z$. $x-y+z, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,215 |
Example 12 If $x, y \in \mathbf{R}, y \geqslant 0$, and $y(y+1) \leqslant(x+1)^{2}$, prove: $y(y-1) \leqslant x^{2}$.
---
The translation maintains the original text's line breaks and format. | Prove that if $0 \leqslant y \leqslant 1$, then $y(y-1) \leqslant 0 \leqslant x^{2}$.
If $y>1$, by the problem statement,
$$\begin{array}{c}
y(y+1) \leqslant(x+1)^{2} \\
y \leqslant \sqrt{(x+1)^{2}+\frac{1}{4}}-\frac{1}{2}
\end{array}$$
To prove $y(y-1) \leqslant x^{2}$, it is sufficient to prove
$$\begin{array}{cc}
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,216 |
14 Let $a, b, c \in \mathbf{R}^{+}$, prove that:
$$\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1 .$$ | 14. Let $x=\frac{a}{\sqrt{a^{2}+8 b c}}, y=\frac{b}{\sqrt{b^{2}+8 a c}}, z=\frac{c}{\sqrt{c^{2}+8 a b}}, x, y, z \in \mathbf{R}^{+}$, then $\left(\frac{1}{x^{2}}-1\right)\left(\frac{1}{y^{2}}-1\right)\left(\frac{1}{z^{2}}-1\right)=512$.
Assume $x+y+z < 1$, then
\begin{aligned}
& \frac{\left[(x+y+z)^{2}-x^{2}\right] \c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,217 |
Example 1: Prove that for any real numbers $x, y, z$, the following three inequalities cannot all hold simultaneously:
$$|x|<|y-z|,|y|<|z-x|,|z|<|x-y| \text {. }$$ | Prove by contradiction, assuming all three inequalities hold, then
we have
$$\left\{\begin{array}{l}
x^{2}<(y-z)^{2} \\
y^{2}<(z-x)^{2} \\
z^{2}<(x-y)^{2}
\end{array}\right.$$
$$\left\{\begin{array}{l}
(x-y+z)(x+y-z)<0, \\
(y-z+x)(y+z-x)<0, \\
(z-x+y)(z+x-y)<0 .
\end{array}\right.$$
Multiplying the above three inequa... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,218 |
Example 2 If $a, b, c, d$ are non-negative integers, and $(a+b)^{2}+3a+2b=(c+d)^{2}+3c+$
$2d$. Prove:
$$a=c, b=d$$ | Prove $a+b=c+d$ first. Use proof by contradiction.
If $a+b \neq c+d$, without loss of generality, assume $a+b>c+d$, then $a+b \geqslant c+d+1$. Therefore,
$$\begin{aligned}
(a+b)^{2}+3 a+2 b & =(a+b)^{2}+2(a+b)+a \\
& \geqslant(c+d+1)^{2}+2(c+d+1)+a \\
& =(c+d)^{2}+4(c+d)+3+a \\
& >(c+d)^{2}+3 c+2 d .
\end{aligned}$$
... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,219 |
Example 3 Given 12 real numbers $a_{1}, a_{2}, \cdots, a_{12}$ satisfy:
$$\left\{\begin{array}{l}
a_{2}\left(a_{1}-a_{2}+a_{3}\right)<0, \\
a_{3}\left(a_{2}-a_{3}+a_{4}\right)<0, \\
\cdots \cdots . \\
a_{11}\left(a_{10}-a_{11}+a_{12}\right)<0 .
\end{array}\right.$$
Prove: At least 3 positive numbers and 3 negative num... | Proof by contradiction, suppose that among $a_{1}, a_{2}, \cdots, a_{12}$ there are at most two negative numbers, then there exists $1 \leqslant k \leqslant 9$ such that $a_{k}, a_{k+1}, a_{k+2}, a_{k+3}$ are all non-negative real numbers.
From the problem, we have $\quad\left\{\begin{array}{l}a_{k+1}\left(a_{k}-a_{k+... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,220 |
64 real numbers $a, b, c, d$ satisfy:
(1) $a \geqslant b \geqslant c \geqslant d$;
(2) $a+b+c+d=9$;
(3) $a^{2}+b^{2}+c^{2}+d^{2}=21$.
Prove: $a b-c d \geqslant 2$. | Proof: If $a+b<5$, then $4<c+d \leqslant a+b<5$, thus
$$\begin{aligned}
(a b+c d)+(a c+b d)+(a d+b c) & =\frac{(a+b+c+d)^{2}-\left(a^{2}+b^{2}+c^{2}+d^{2}\right)}{2} \\
& =30
\end{aligned}$$
And $a b+c d \geqslant a c+b d \geqslant a d+b c(\Leftrightarrow(a-d)(b-c) \geqslant 0,(a-b)(c-d) \geqslant 0)$, so
$$a b+c d \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,223 |
Example 7 Let $x, y, z \in \mathbf{R}^{+}$, prove that:
$$\sqrt{x+\sqrt[3]{y+\sqrt[4]{z}}} \geqslant \sqrt[32]{x y z}$$ | Prove by contradiction.
If there exist positive real numbers $x_{0}$, $y_{0}$, $z_{0}$, such that $\sqrt{x_{0}+\sqrt[3]{y_{0}+\sqrt[4]{z_{0}}}}<\sqrt[32]{x_{0} y_{0} z_{0}}$, then we have $\square$
$$\begin{array}{l}
\left\{\begin{array}{l}
\sqrt{x_{0}}<\sqrt[32]{x_{0} y_{0} z_{0}}, \\
\sqrt[6]{y_{0}}<\sqrt[32]{x_{0} y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,224 |
Example 8 Suppose for any real number $x$ we have $\cos (a \sin x)>\sin (b \cos x)$, prove that:
$$a^{2}+b^{2}<\frac{\pi^{2}}{4}$$ | Proof by contradiction. Suppose $a^{2}+b^{2} \geqslant \frac{\pi^{2}}{4}$, express $a \sin x+b \cos x$ in the form $\sqrt{a^{2}+b^{2}} \sin (x+$
$\varphi)$. Here, $\cos \varphi=\frac{a}{\sqrt{a^{2}+b^{2}}}, \sin \varphi=\frac{b}{\sqrt{a^{2}+b^{2}}}$.
Since $\sqrt{a^{2}+b^{2}} \geqslant \frac{\pi}{2}$, there exists a re... | a^{2}+b^{2}<\frac{\pi^{2}}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 738,225 |
Example 13 Let $a, b, c \in \mathbf{R}^{+}$, prove:
$$a+b+c-3 \sqrt[3]{a b c} \geqslant a+b-2 \sqrt{a b} .$$ | $$\begin{array}{l}
a+b+c-3 \sqrt[3]{a b c} \geqslant a+b-2 \sqrt{a b} \\
\Leftrightarrow \quad c+2 \sqrt{a b} \geqslant 3 \sqrt[3]{a b c}. \\
\text { Since } \\
c+2 \sqrt{a b}=c+\sqrt{a b}+\sqrt{a b} \\
\geqslant 3 \sqrt[3]{c \sqrt{a b} \sqrt{a b}} \\
=3 \sqrt[3]{a b c},
\end{array}$$
thus $\quad a+b+c-3 \sqrt[3]{a b ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,227 |
Example 10 Let $p$ be the product of two consecutive integers greater than 2, prove that there are no integers $x_{1}, x_{2}, \cdots, x_{p}$ that satisfy the equation
$$\sum_{i=1}^{p} x_{i}^{2}-\frac{4}{4 p+1}\left(\sum_{i=1}^{p} x_{i}\right)^{2}=1 .$$ | Proof by contradiction. Let $p=k(k+1), k \geqslant 3$, then $p \geqslant 12,4 p+1 \geqslant 4 p$.
Assume there are integers $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{p}$ satisfying the equation:
$$\begin{aligned}
4 p+1 & =(4 p+1) \sum_{i=1}^{p} x_{i}^{2}-4\left(\sum_{i=1}^{p} x_{i}\right)^{2} \\
& =4\left[p ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,228 |
Example 12 For a positive integer $n(n \geqslant 2)$, assume that $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ has all real coefficients, all complex roots of $f(x)$ have negative real parts, and $f(x)$ has a pair of equal real roots. Prove: there must exist $k, 1 \leqslant k \leqslant n-1$, satisfying:
$$a_... | Prove that when $n=2$, $f(x)=a_{2}(x+a)^{2}$, where $a$ is a positive real number, so
$$f(x)=a_{2}\left(x^{2}+2 a x+a^{2}\right)$$
Thus, $a_{1}=2 a a_{2}, a_{0}=a^{2} a_{2}, a_{1}^{2}-4 a_{0} a_{2}=0$, the conclusion holds.
Assume $n>2$ is a positive integer, and let $a_{n}>0$ (if not, multiply each coefficient by $(-... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,230 |
1 Given $a, b, c \in \mathbf{R}$, and $a+b+c>0, ab+bc+ca>0, abc>0$. Prove:
$$a>0, b>0, c>0 .$$ | 1. Otherwise, since $a b c>0$, without loss of generality, let $a0$. From $a+b+c>0$ we get $c>|a+b|$, and $a b>c \cdot|a+b|$, thus $a b>|a+b|^{2}$, which is a contradiction! | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,231 |
3. Let real numbers $a, b, c, d, p, q$ satisfy:
$$a b + c d = 2 p q, \quad a c \geqslant p^{2} > 0,$$
Prove: $b d \leqslant q^{2}$. | 3. If $b d>q^{2}$, then $4 a b c d=4(a c)(b d)>4 p^{2} q^{2}=(a b+c d)^{2}=a^{2} b^{2}+$ $2 a b c d+c^{2} d^{2}$, so $(a b-c d)^{2}<0$. Contradiction! | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,233 |
4 Let $a, b, c$ be positive real numbers, and $a+b+c \geqslant a b c$, prove:
$$a^{2}+b^{2}+c^{2} \geqslant a b c$$ | 4. If $a^{2}+b^{2}+c^{2}<a b c$, then $a<b c, b<c a, c<a b$. Hence $a+b+c<a b+b c+c a \leqslant a^{2}+b^{2}+c^{2}<a b c$, contradiction! | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,234 |
5 Let $f(x), g(x)$ be real-valued functions on $[0,1]$. Prove: there exist $x_{0}, y_{0} \in[0,1]$, such that
$$\left|x_{0} y_{0}-f\left(x_{0}\right)-g\left(y_{0}\right)\right| \geqslant \frac{1}{4}$$ | 5. By contradiction, if not, then for all $x, y \in[0,1]$, we have: $\mid x y-f(x)-$ $g(y) \left\lvert\,<\frac{1}{4}\right.$, taking $(x, y)=(0,0) 、(0,1) 、(1,0) 、(1,1)$, we get: $\mid f(0)+$ $g(0)\left|<\frac{1}{4},\right| f(0)+g(1)\left|<\frac{1}{4},\right| f(1)+g(0)\left|<\frac{1}{4},\right| 1-f(1)-$ $g(1) \left\lver... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,235 |
6 Prove: For any real numbers $a, b$, there exist $x$ and $y$ in $[0,1]$, such that
$$|x y-a x-b y| \geqslant \frac{1}{3}$$
and ask whether the above proposition still holds if $\frac{1}{3}$ is changed to $\frac{1}{2}$ or 0.33334. | 6. If the negation of the proposition is not true, then there exist real numbers $a, b$ such that for any $x, y$ in $[0,1]$, we have $|x y-a x-b y|<\frac{1}{3}$.
Taking $(x, y)=(1,0);(0,1);(1,1)$, we get $|a|<\frac{1}{3},|b|<\frac{1}{3}$, $|1-a-b|<\frac{1}{3}$, then $1=|a+b+1-a-b| \leqslant|a|+|b|+|1-a-b|<1$, contradi... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,236 |
7 Let $m, n \in \mathbf{Z}^{+}, a_{1}, a_{2}, \cdots, a_{m}$ be distinct elements of the set $\{1,2, \cdots, n\}$, and whenever $a_{i}+a_{j} \leqslant n, 1 \leqslant i \leqslant j \leqslant m$, there exists some $k, 1 \leqslant k \leqslant m$, such that $a_{i}+a_{j}=a_{k}$. Prove: $\frac{1}{m}\left(a_{1}+a_{2}+\cdots+a... | 7. Let's assume $a_{1}>a_{2}>\cdots>a_{m}$, and we will prove the following: For any positive integer $i$ satisfying $1 \leqslant i \leqslant m$, we have
$$a_{i}+a_{m+1-i} \geqslant n+1$$
If (1) holds, then $2\left(a_{1}+a_{2}+\cdots+a_{m}\right)=\left(a_{1}+a_{m}\right)+\left(a_{2}+a_{m-1}\right)+\cdots+$ $\left(a_{m... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 738,237 |
Example 14 Given $n \in \mathbf{N}_{+}$, prove:
$$\frac{1}{n+1}\left(1+\frac{1}{3}+\cdots+\frac{1}{2 n-1}\right) \geqslant \frac{1}{n}\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right) .$$ | To prove (1), we only need to prove
$$n\left(1+\frac{1}{3}+\cdots+\frac{1}{2 n-1}\right) \geqslant(n+1)\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right) .$$
The left side of (2) is
$$\frac{n}{2}+\frac{n}{2}+n\left(\frac{1}{3}+\cdots+\frac{1}{2 n-1}\right)$$
The right side of (2) is
$$\begin{aligned}
& n\left(\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,238 |
8 For any $n \in \mathbf{Z}^{+}$ and real number sequence $a_{1}, a_{2}, \cdots, a_{n}$, prove that there exists $k \in \mathbf{Z}^{+}$, satisfying:
$$\left|\sum_{i=1}^{k} a_{i}-\sum_{i=k+1}^{n} a_{i}\right| \leqslant \max _{1 \leqslant i \leqslant n}\left|a_{i}\right|$$ | 8. By contradiction, if for any $k \in\{1,2, \cdots, n\}$, we have $\left|S_{k}\right|=\left|\sum_{i=1}^{k} a_{i}-\sum_{i=k+1}^{n} a_{i}\right|>$ $\max _{1 \leqslant i \leqslant n}\left|a_{i}\right|$. Let $A=\max _{1 \leqslant i \leqslant n}\left|a_{i}\right|$.
Supplementary definition $S_{0}=-S_{n}$, then one of $S_{... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,239 |
10 Given a non-increasing sequence of positive numbers $a_{1} \geqslant a_{2} \geqslant a_{3} \geqslant \cdots \geqslant a_{n} \geqslant \cdots$, where $a_{1}=\frac{1}{2 k}(k \in \mathbf{N}$, $k \geqslant 2)$, and $a_{1}+a_{2}+\cdots+a_{n}+\cdots=1$. Prove that: from the sequence, $k$ numbers can be selected such that ... | 10. By contradiction. If there do not exist such $k$ numbers, then for $a_{1}, a_{2}, \cdots, a_{k}$, we have $a_{k} \leqslant \frac{1}{2} a_{1}$; for $a_{k}, a_{k+1}, \cdots, a_{2 k-1}$, we have $a_{2 k-1} \leqslant \frac{1}{2} a_{k} \leqslant \frac{1}{2^{2}} a_{1} ; \cdots$; for $a_{(n-1)(k-1)+1}$, $a_{(n-1)(k-1)+2},... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 738,240 |
11 Prove or disprove the proposition: If $x, y$ are real numbers and $y \geqslant 0, y(y+1) \leqslant(x+1)^{2}$, then $y(y-1) \leqslant x^{2}$. | 11. Suppose $y(y-1)>x^{2}$, then by $y \geqslant 0$ we know $y>1$. Further, we have $y>\frac{1}{2}+$ $\sqrt{\frac{1}{4}+x^{2}}$. From the assumption $y(y+1) \leqslant(x+1)^{2}$ and $y>1$, we know $y \leqslant-\frac{1}{2}+$ $\sqrt{\frac{1}{4}+(x+1)^{2}}$, thus we get $\frac{1}{2}+\sqrt{\frac{1}{4}+x^{2}}<-\frac{1}{2}+\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,241 |
13 Let real numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 2)$ and $A$ satisfy: $A+\sum_{i=1}^{n} a_{i}^{2}<\frac{1}{n-1}\left(\sum_{i=1}^{n} a_{i}\right)^{2}$. Prove that for $1 \leqslant i<j \leqslant n$, we have $A<2 a_{i} a_{j}$. | 13. By contradiction, suppose there exist $1 \leqslant i<j \leqslant n$ such that $A \geqslant 2 a_{i} a_{j}$.
Without loss of generality, let $i=1, j=2$, then we have
$$A+\sum_{i=1}^{n} a_{i}^{2} \geqslant 2 a_{1} a_{2}+\sum_{i=1}^{n} a_{i}^{2}=\left(a_{1}+a_{2}\right)^{2}+a_{3}^{2}+\cdots+a_{n}^{2},$$
By the Cauchy... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,243 |
14. Let $\left\{a_{k}\right\}$ be an infinite sequence of non-negative real numbers, $k=1,2, \cdots$, satisfying: $a_{k}-2 a_{k+1}+a_{k+2} \geqslant 0$ and $\sum_{j=1}^{k} a_{j} \leqslant 1, k=1,2, \cdots$. Prove that: $0 \leqslant a_{k}-a_{k+1} \leqslant \frac{2}{k^{2}}, k=1,2, \cdots$. | 14. First prove $a_{k}-a_{k+1} \geqslant 0$, which can be done using proof by contradiction.
Assume there exists some $a_{k}k)$ that tends to infinity as $n$ tends to infinity, leading to a contradiction! Therefore, we have $a_{k}-$
$$\begin{array}{l}
a_{k+1} \geqslant 0, k=1,2, \cdots \text {. } \\
\text { Let } b_{k... | 0 \leqslant a_{k}-a_{k+1} \leqslant \frac{2}{k^{2}} | Inequalities | proof | Yes | Yes | inequalities | false | 738,244 |
15 If the roots of the equation $x^{4}+a x^{3}+b x+c=0$ are all real, prove: $a b \leqslant 0$.
| 15. Suppose $ab>0$, without loss of generality, assume $a>0$, then $b>0$. We discuss in three cases:
(1) If $c>0, x^{4}+a x^{3}+b x+c=0$ has all negative roots, which contradicts the coefficient of $x^{2}$ being 0.
(2) If $c-x_{1}-a=x_{2}+x_{3}+x_{4}$. Since the coefficient of $x^{2}$ is 0, we should have $x_{1}\left(x... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,245 |
9 Let $x_{k}, y_{k} \in \mathbf{R}, j_{k}=x_{k}+\mathrm{i} y_{k}(k=1,2, \cdots, n, \mathrm{i}=\sqrt{-1}) . r$ is the absolute value of the real part of $\pm \sqrt{j_{1}^{2}+j_{2}^{2}+\cdots+j_{n}^{2}}$. Prove:
$$r \leqslant\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|$$ | 9. Let $a+\mathrm{i} b$ be any square root of $j_{1}^{2}+j_{2}^{2}+\cdots+j_{n}^{2}$, then $r=|a|$, and $(a+\mathrm{i} b)^{2}=$
$$\sum_{k=1}^{n} j_{k}^{2}=\sum_{k=1}^{n}\left(x_{k}+\mathrm{i} y_{k}\right)^{2}$$
Thus, $a^{2}-b^{2}=\sum_{k=1}^{n} x_{k}^{2}-\sum_{k=1}^{n} y_{k}^{2}, a b=\sum_{k=1}^{n} x_{k} y_{k}$.
Assum... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,246 |
Example 1 Given $a^{2}+b^{2}+c^{2}+d^{2}=1$, prove:
$$\begin{array}{c}
(a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4} \\
\quad+(b+d)^{4}+(c+d)^{4} \leqslant 6
\end{array}$$ | Consider the sum: $(a-b)^{4}+(a-c)^{4}+(a-d)^{4}+(b-c)^{4}+(b-d)^{4}+$ $(c-d)^{4}$. It is not hard to see that it forms an identity with the left side of (1), that is:
$$\begin{array}{c}
(a+b)^{4}+(a-b)^{4}+(a+c)^{4}+(a-c)^{4}+(a+d)^{4}+(a-d)^{4} \\
+(b+c)^{4}+(b-c)^{4}+(b+d)^{4}+(b-d)^{4}+(c+d)^{4}+(c-d)^{4} \\
=6\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,247 |
Example 2 Let $\triangle A_{1} A_{2} A_{3}$ and $\triangle B_{1} B_{2} B_{3}$ have side lengths $a_{1}, a_{2}, a_{3}$ and $b_{1}, b_{2}, b_{3}$, and areas $S_{1}, S_{2}$, respectively. Also, let
$$H=a_{1}^{2}\left(-b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)+a_{2}^{2}\left(b_{1}^{2}-b_{2}^{2}+b_{3}^{2}\right)+a_{3}^{2}\left(b... | Prove that from Heron's formula, we have
$$\begin{array}{l}
16 S_{1}^{2}=2 a_{1}^{2} a_{2}^{2}+2 a_{2}^{2} a_{3}^{2}+2 a_{3}^{2} a_{1}^{2}-a_{1}^{4}-a_{2}^{4}-a_{3}^{4} \\
16 S_{2}^{2}=2 b_{1}^{2} b_{2}^{2}+2 b_{2}^{2} b_{3}^{2}+2 b_{3}^{2} b_{1}^{2}-b_{1}^{4}-b_{2}^{4}-b_{3}^{4} .
\end{array}$$
Let \( D_{1}=\sqrt{\la... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,248 |
Example 15 Given $a, b, c \in \mathbf{R}^{+}, a b c=1$. Prove:
$$(a+b)(b+c)(c+a) \geqslant 4(a+b+c-1) .$$ | Assume without loss of generality that $a \geqslant 1$, then the original inequality is equivalent to
$$a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)+6 \geqslant 4(a+b+c)$$
which is $\square$
$$\begin{aligned}
& \left(a^{2}-1\right)(b+c)+b^{2}(c+a)+c^{2}(a+b)+6 \\
\geqslant & 4 a+3(b+c)
\end{aligned}$$
Since $(a+1)(b+c) \geqslant... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,249 |
Example 3 Given $a, b, c \in(-2,1)$, prove:
$$a b c>a+b+c-2$$ | Proof: Let $f(x)=(b c-1) x-b-c+2$, then we have
$$\begin{aligned}
f(-2) & =-2 b c-b-c+4 \\
& =-2\left(b+\frac{1}{2}\right)\left(c+\frac{1}{2}\right)+\frac{9}{2} .
\end{aligned}$$
Since $b, c \in(-2,1)$, it follows that $b+\frac{1}{2}, c+\frac{1}{2} \in\left(-\frac{3}{2}, \frac{3}{2}\right)$, hence
$$\left(b+\frac{1}{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,250 |
Example 4 Let $x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3} \in \mathbf{R}$, and satisfy $x_{1}^{2}+x_{2}^{2}+x_{3}^{2} \leqslant 1$, prove:
$$\begin{aligned}
& \left(x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}-1\right)^{2} \\
\geqslant & \left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-1\right)\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}-1\right)
\end... | Prove that when $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1$, the original inequality obviously holds.
When $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}<1$, construct a quadratic function
$$\begin{aligned}
f(t) & =\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-1\right) t^{2}-2\left(x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}-1\right) t+\left(y_{1}^{2}+y_{2}^{2}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,251 |
Example 5 Let the three sides of $\triangle ABC$ be $a, b, c$ satisfying: $a+b+c=1$. Prove:
$$5\left(a^{2}+b^{2}+c^{2}\right)+18 a b c \geqslant \frac{7}{3} .$$ | Prove that from $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a)$
$$=1-2(a b+b c+c a)$$
it follows that the original inequality is equivalent to
$$\begin{array}{c}
\frac{5}{9}(a b+b c+c a)-a b c \leqslant \frac{4}{27} \\
\text { Let } f(x)=(x-a)(x-b)(x-c)=x^{3}-x^{2}+(a b+b c+c a) x-a b c
\end{array}$$
Then
$$f\left(\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,252 |
Example 6 Given the inequality
$$\sqrt{2}(2 a+3) \cos \left(\theta-\frac{\pi}{4}\right)+\frac{6}{\sin \theta+\cos \theta}-2 \sin 2 \theta<3 a+6$$
for $\theta \in\left[0, \frac{\pi}{2}\right]$ to always hold, find the range of values for $a$. | Let $\sin \theta + \cos \theta = x$, then $x \in [1, \sqrt{2}]$, and
$$\begin{aligned}
\sin 2 \theta & = 2 \sin \theta \cos \theta = x^2 - 1 \\
\cos \left(\theta - \frac{\pi}{4}\right) & = \frac{\sqrt{2}}{2} \cos \theta + \frac{\sqrt{2}}{2} \sin \theta = \frac{\sqrt{2}}{2} x
\end{aligned}$$
Thus, the original inequali... | a > 3 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,253 |
Example 8 Prove: For any real number $x$, we have
$$\left|\sqrt{x^{2}+x+1}-\sqrt{x^{2}-x+1}\right|<1 .$$ | Prove that
$$\begin{aligned}
& \left|\sqrt{x^{2}+x+1}-\sqrt{x^{2}-x+1}\right| \\
= & \left|\sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}-\sqrt{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\right|
\end{aligned}$$
The above expression can be seen as the difference in dis... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,255 |
Example 10 Let $x, y, z, \alpha, \beta, \gamma$ be positive numbers, with any two of $\alpha, \beta, \gamma$ summing to more than the third and all belonging to the interval $[0, \pi)$, prove:
$$\sqrt{x^{2}+y^{2}-2 x y \cos \alpha}+\sqrt{y^{2}+z^{2}-2 y z \cos \beta} \geqslant \sqrt{z^{2}+x^{2}-2 z x \cos \gamma} .$$ | Prove that because $\alpha<\beta+\gamma<\pi, \beta<\gamma+\alpha<\pi, \gamma<\alpha+\beta<\pi, \gamma<\alpha+\beta<\pi$, a triangular face $P-ABC$ can be constructed from a point $P$ in space such that:
$$\begin{array}{c}
\angle A P B=\alpha, \\
\angle B P C=\beta, \\
\angle C P A=\gamma ; \\
P A=x, P B=y, \\
P C=z (\t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,257 |
Example 11 Given $|u| \leqslant \sqrt{2}, v$ is a positive real number, prove:
$$S=(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2} \geqslant 8 .$$ | Prove that the key is to see that the expression for $S$ is precisely the square of the distance between point $A\left(u, \sqrt{2-u^{2}}\right)$ and point $B\left(u, \frac{9}{v}\right)$ in Cartesian coordinates.
Obviously, point $A$ lies on the circle $x^{2}+y^{2}=2$, and point $B$ lies on the hyperbola $x y=9$.
There... | 8 | Inequalities | proof | Yes | Yes | inequalities | false | 738,258 |
Example 12 If $a_{1}+a_{2}+\cdots+a_{n}=1$, prove:
$$\begin{array}{c}
\frac{a_{1}^{4}}{a_{1}^{3}+a_{1}^{2} a_{2}+a_{1} a_{2}^{2}+a_{2}^{3}}+\frac{a_{2}^{4}}{a_{2}^{3}+a_{2}^{2} a_{3}+a_{2} a_{3}^{2}+a_{3}^{3}}+\cdots \\
+\frac{a_{n}^{4}}{a_{n}^{3}+a_{n}^{2} a_{1}+a_{1}^{2} a_{n}+a_{1}^{3}} \geqslant \frac{1}{4}
\end{ar... | Let the left side of the original inequality be $A$.
Construct the dual form
$$B=\frac{a_{2}^{4}}{a_{1}^{3}+a_{1}^{2} a_{2}+a_{1} a_{2}^{2}+a_{2}^{3}}+\cdots+\frac{a_{1}^{4}}{a_{n}^{3}+a_{n}^{2} a_{1}+a_{1}^{2} a_{n}+a_{1}^{3}},$$
Then
$$\begin{aligned}
A-B= & \frac{\left(a_{1}^{2}+a_{2}^{2}\right)\left(a_{1}+a_{2}\ri... | \frac{1}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 738,259 |
Example 16 Let $x, y, z$ be 3 non-zero real numbers, find the maximum value of $\frac{xy+2yz}{x^2+y^2+z^2}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve: Because
$$\begin{array}{l}
x^{2}+\frac{1}{5} y^{2} \geqslant \frac{2}{\sqrt{5}} x y \\
\frac{4}{5} y^{2}+z^{2} \geqslant \frac{4}{\sqrt{5}} y z
\end{array}$$
Therefore, we have
$$\begin{array}{c}
x^{2}+y^{2}+z^{2} \geqslant \frac{2}{\sqrt{5}}(x y+2 y z) \\
\frac{x y+2 y z}{x^{2}+y^{2}+z^{2}} \leqslant \frac{\sq... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,260 |
Example 13 Let $x_{n}=\sqrt{2+\sqrt[3]{3+\cdots+\sqrt[n]{n}}}$, prove that:
$$x_{n+1}-x_{n}<\frac{1}{n!}, n=2,3, \cdots$$ | Prove that when $n=2$, $x_{3}-x_{2}=\sqrt{2+\sqrt[3]{3}}-\sqrt{2}2$, $\frac{n+1}{n^{n-1}}<\frac{2 n}{n^{2}}<1$ is obvious.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
(Note: The note is not part of the translation, it is... | null | Inequalities | proof | Yes | Yes | inequalities | false | 738,261 |
Example 14 Real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy: $a_{1}+a_{2}+\cdots+a_{n}=0$, prove that: $\max _{1 \leqslant k \leqslant n}\left(a_{k}^{2}\right) \leqslant \frac{n}{3} \sum_{i=1}^{n-1}\left(a_{i}-a_{i+1}\right)^{2}$. | To prove that it is sufficient to show the inequality holds for any $1 \leqslant k \leqslant n$.
Let $d_{k}=a_{k}-a_{k+1}, k=1,2, \cdots, n-1$, then
$$\begin{array}{c}
a_{k}=a_{k} \\
a_{k+1}=a_{k}-d_{k}, a_{k+2}=a_{k}-d_{k}-d_{k+1}, \cdots, a_{n}=a_{k}-d_{k}-d_{k+1}-\cdots-d_{n-1} \\
a_{k-1}=a_{k}+d_{k-1}, a_{k-2}=a_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,262 |
Example 15 Given two sets of numbers $x_{1}, x_{2}, \cdots, x_{n}$ and $y_{1}, y_{2}, \cdots, y_{n}$, it is known that
$$\begin{array}{c}
x_{1}>x_{2}>\cdots>x_{n}>0, y_{1}>y_{2}>\cdots>y_{n}>0 \\
x_{1}>y_{1}, x_{1}+x_{2}>y_{1}+y_{2}, \cdots \\
x_{1}+x_{2}+\cdots+x_{n}>y_{1}+y_{2}+\cdots+y_{n}
\end{array}$$
Prove: For ... | Prove that because $a_{1}>a_{2}>\cdots>a_{n}>0$, there exist positive numbers $b_{1}, b_{2}, \cdots b_{n-1}, b_{n}$ such that:
$$\begin{array}{c}
a_{n}=b_{1} \\
a_{n-1}=b_{1}+b_{2} \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
a_{2}=b_{1}+b_{2}+\cdots+b_{n-1} \\
a_{1}=b_{1}+b_{2}+\cdots+b_{n}
\end{array... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,263 |
Example 16 For a given positive integer $n$ greater than 1, do there exist $2n$ distinct positive integers $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n}$, simultaneously satisfying the following conditions:
(1) $a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n}$;
(2) $n-1>\sum_{i=1}^{n} \frac{a_{i}-b_{i}}{... | The answer is affirmative.
Take
$$\begin{array}{c}
a_{1}=N+1, a_{2}=N+2, \cdots, a_{n-1}=N+(n-1) \\
b_{1}=1, b_{2}=2, \cdots, b_{n-1}=n-1
\end{array}$$
Then, by (1) we have $b_{n}-a_{n}=N(n-1)$. Let $a_{n}=N^{2}, b_{n}=N^{2}+N(n-1)$.
Therefore,
$$\begin{aligned}
\sum_{i=1}^{n} \frac{a_{i}-b_{i}}{a_{i}+b_{i}} & =n-1-\l... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 738,264 |
Example 17 Find all positive integers $n$ greater than 1, such that for any positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$, the inequality
$$\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} \geqslant n\left(x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{n} x_{1}\right)$$
holds. | When $n=2$, the inequality is $\left(x_{1}+x_{2}\right)^{2} \geqslant 2\left(x_{1} x_{2}+x_{2} x_{1}\right)$, which is equivalent to $\left(x_{1}-x_{2}\right)^{2} \geqslant 0$. Therefore, $n=2$ satisfies the condition.
When $n=3$, the inequality $\left(x_{1}+x_{2}+x_{3}\right)^{2} \geqslant 3\left(x_{1} x_{2}+x_{2} x_... | 2, 3, 4 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,265 |
Example 18 Given a positive integer $n \geqslant 2$, real numbers $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}>0, b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}>0$, and it is known that: $a_{1} a_{2} \cdots a_{n}=b_{1} b_{2} \cdots b_{n}, \sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right) \le... | Not necessarily.
Let
$$\begin{array}{c}
a_{1}=a_{2}=\cdots=a_{n-1}=h, a_{n}=\frac{1}{h^{n-1}} \\
b_{1}=k, b_{2}=b_{3}=\cdots=b_{n-1}=1, b_{n}=\frac{1}{k}(k \geqslant 1)
\end{array}$$
(The idea is to take $h$ sufficiently large, so that $\sum_{i=1}^{n} a_{i}$ can be sufficiently large. To prevent $\sum_{1 \leq i0$, it i... | not found | Inequalities | proof | Yes | Yes | inequalities | false | 738,266 |
1 Given $k>a>b>c>0$, prove:
$$k^{2}-(a+b+c) k+(a b+b c+c a)>0$$ | 1. Consider the identity $(k-a)(k-b)(k-c)=k^{3}-(a+b+c) k^{2}+(a b+b c+$ ca) $k-a b c$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,267 |
3 Let $F(x)=|f(x) \cdot g(x)|$, where $f(x)=a x^{2}+b x+c, x \in[-1,1]$; $g(x)=c x^{2}+b x+a, x \in[-1,1]$, and for any parameters $a, b, c$, it always holds that $|f(x)| \leqslant 1$. Find the maximum value of $F(x)$. | 3. $|g(x)|=\left|c x^{2}-c+b x+a+c\right| \leqslant|c| \cdot\left|x^{2}-1\right|+|b x+a+c| \leqslant$ $1+|b x+a+c|$.
Consider the linear function $T(x)=|b x+a+c|, -1 \leqslant x \leqslant 1$.
When $x=-1$, $T(-1)=|a-b+c| \leqslant 1$; when $x=1$, $T(1)=|a+b+c| \leqslant 1$. Therefore, when $|x| \leqslant 1$, $T(x) \leq... | 2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,269 |
4 Let $a, b, c$ be real numbers, prove that:
$$\frac{|a+b+c|}{1+|a+b+c|} \leqslant \frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}+\frac{|c|}{1+|c|} .$$ | 4. Construct the function $f(x)=\frac{x}{1+x}$, it is easy to prove that $f(x)$ is an increasing function on $[0,+\infty)$, thus $f(|a+b+c|) \leqslant f(|a|+|b|+|c|)$, i.e.,
$$\begin{aligned}
\frac{|a+b+c|}{1+|a+b+c|} \leqslant & \frac{|a|+|b|+|c|}{1+|a|+|b|+|c|} \\
= & \frac{|a|}{1+|a|+|b|+|c|}+\frac{|b|}{1+|a|+|b|+|c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,270 |
5 Find all real numbers $a$ such that any positive integer solution of the inequality $x^{2}+y^{2}+z^{2} \leqslant a(x y+y z+z x)$ are the lengths of the sides of some triangle. | 5. Taking $x=2, y=z=1$, we have $a \geqslant \frac{6}{5}$. Therefore, when $a \geqslant \frac{6}{5}$, the original inequality has integer roots $(2,1,1)$, but $(x, y, z)$ cannot form a triangle, so $a<\frac{6}{5}$.
When $a<1$, $x^{2}+y^{2}+z^{2} \leqslant a(x y+y z+z x)<x y+y z+z x$, which is a contradiction! Hence, $... | 1 \leqslant a < \frac{6}{5} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,272 |
6 Let $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ be positive real numbers, prove that:
$$\begin{aligned}
& \left(a_{1} b_{2}+a_{2} b_{1}+a_{2} b_{3}+a_{3} b_{2}+a_{3} b_{1}+a_{1} b_{3}\right)^{2} \\
\geqslant & 4\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}\right)\left(b_{1} b_{2}+b_{2} b_{3}+b_{3} b_{1}\right)
\end{aligne... | 6. Let's assume $\frac{b_{1}}{a_{1}} \geqslant \frac{b_{2}}{a_{2}} \geqslant \frac{b_{3}}{a_{3}}$. Consider the equation: $\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}\right) x^{2}-\left(a_{1} b_{2}+ a_{2} b_{1}+a_{3} b_{2}+a_{2} b_{3}+a_{3} b_{1}+a_{1} b_{3}\right) x+\left(b_{1} b_{2}+b_{2} b_{3}+b_{3} b_{1}\right)=\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,273 |
8 Let $x, y, z$ be any real numbers, prove:
$$\sqrt{x^{2}+x y+y^{2}}+\sqrt{x^{2}+x z+z^{2}} \geqslant \sqrt{y^{2}+y z+z^{2}}.$$ | 8. Establish a Cartesian coordinate system $x O y$, take three points $A(x, 0)$, $B\left(-\frac{y}{2},-\frac{\sqrt{3}}{2} y\right)$, $C\left(-\frac{z}{2}, \frac{\sqrt{3}}{2} z\right)$, then the original inequality is transformed into: $|A B|+|A C| \geqslant|B C|$, which is obvious. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,275 |
9 Let $x, y, z>0$, prove:
$$\begin{aligned}
& 3 \sqrt{x y+y z+z x} \\
\leqslant & \sqrt{x^{2}+x y+y^{2}}+\sqrt{y^{2}+y z+z^{2}}+\sqrt{z^{2}+z x+x^{2}} \\
\leqslant & 2(x+y+z)
\end{aligned}$$ | 9. From the given conditions, we can construct the figure as shown, in $\triangle ABC$ (with an internal point $P$ satisfying the conditions $\angle APB = \angle BPC = \angle CPA = 120^\circ$. Let $PA = x, PB = y, PC = z$, then $AB = \sqrt{x^2 + xy + y^2}$, $BC = \sqrt{y^2 + yz + z^2}$, $CA = \sqrt{z^2 + zx + x^2}$. By... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,276 |
10. Given that $\alpha, \beta, \gamma$ are acute angles, and $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$, prove:
$$\frac{3 \pi}{4}<\alpha+\beta+\gamma<\pi$$ | 10. Given the conditions, construct a rectangular parallelepiped $A B C D-A_{1} B_{1} C_{1} D_{1}$ with length, width, and height $\cos \alpha$, $\cos \beta$, and $\cos \gamma$ respectively. As shown in the figure, $A B=\cos \alpha, B C=\cos \beta, B B_{1}=\cos \gamma$. The length of the diagonal of this rectangular pa... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,277 |
III If $p, q$ are real numbers, and for $0 \leqslant x \leqslant 1$, the inequality holds:
$$\left|\sqrt{1-x^{2}}-p x-q\right| \leqslant \frac{\sqrt{2}-1}{2} .$$
Prove: $p=-1, q=\frac{1+\sqrt{2}}{2}$. | 11. The original inequality is $\sqrt{1-x^{2}}-\frac{\sqrt{2}-1}{2} \leqslant p x+q \leqslant \sqrt{1-x^{2}}+\frac{\sqrt{2}-1}{2}(0 \leqslant$ $x \leqslant 1$ ).
Construct circles $A$ and $B$ with centers at points $A\left(0, \frac{\sqrt{2}-1}{2}\right)$ and $B\left(0,-\frac{\sqrt{2}-1}{2}\right)$, respectively, and a... | p=-1, q=\frac{1+\sqrt{2}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 738,278 |
12 (Minkowski Inequality) Prove: For any $2 n$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$, we have
$$\begin{array}{c}
\sqrt{a_{1}^{2}+b_{1}^{2}}+\sqrt{a_{2}^{2}+b_{2}^{2}}+\cdots+\sqrt{a_{n}^{2}+b_{n}^{2}} \\
\geqslant \sqrt{\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}+\left(b_{1}+b_{... | 12. Regarding the term in the form of $\sqrt{x^{2}+y^{2}}$ as the length of the hypotenuse of a right-angled triangle, we can construct a figure as shown.
Thus, the left side of the inequality $=O A_{1}+A_{1} A_{2}+\cdots+A_{n-1} A_{n}$, the right side of the inequality $=O A_{n}$.
Since the length of the broken line ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,279 |
13 Let $0<a_{i} \leqslant a(i=1,2, \cdots, 6)$. Prove:
(1) $\frac{\sum_{i=1}^{4} a_{i}}{a}-\frac{a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}+a_{4} a_{1}}{a^{2}} \leqslant 2$.
(2) $\frac{\sum_{i=1}^{6} a_{i}}{a}-\frac{a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{6} a_{1}}{a^{2}} \leqslant 3$. | 13. (1) The original inequality is $a_{1}\left(a-a_{2}\right)+a_{2}\left(a-a_{3}\right)+a_{3}\left(a-a_{4}\right)+a_{4}\left(a-a_{1}\right) \leqslant 2 a^{2}$. Construct a square $ABCD$ with side length $a$.
Take $U, V$ such that $SD=a_{1}, CR=a_{2}, BQ=a_{3}, AP=a_{4}$.
Then $SDRM$ and $CRVQ$ do not overlap, and $BPN... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,280 |
15 Let $n$ be an odd number greater than 2. Prove that the following inequality holds for any $a_{1}$, $a_{2}, \cdots, a_{n} \in \mathbf{R}$ if and only if $n=3$ or 5:
$$\begin{array}{c}
\left(a_{1}-a_{2}\right)\left(a_{1}-a_{3}\right) \cdots\left(a_{1}-a_{n}\right)+\left(a_{2}-a_{1}\right)\left(a_{2}-a_{3}\right) \cdo... | $$\begin{array}{l}
\text { 15. Without loss of generality, let } a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} . \\
\quad \text { Then } \quad A_{3}=a_{1}^{2}+a_{2}^{2}+a_{3}^{2}-a_{1} a_{2}-a_{2} a_{3}-a_{3} a_{1} \geqslant 0 . \\
A_{5}=\left(a_{1}-a_{2}\right)\left(a_{1}-a_{3}\right)\left(a_{1}-a_{4}\right)\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,282 |
16. Let $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 2)$ all be greater than -1 and have the same sign, prove:
$$\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)>1+a_{1}+a_{2}+\cdots+a_{n} .$$ | 16. Construct the sequence $x_{n}=\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)-\left(1+a_{1}+a_{2}+\cdots+a_{n}\right)$ $(n \geqslant 2)$, then $x_{n+1}-x_{n}=a_{n+1}\left[\left(1+a_{1}\right) \cdots\left(1+a_{n}\right)-1\right]$.
If $a_{i}>0(i=1,2, \cdots, n+1)$, it is easy to see from the abov... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,283 |
17 Let $a_{i}$ be positive real numbers $(i=1,2, \cdots, n)$, and let:
$$\begin{array}{c}
k b_{k}=a_{1}+a_{2}+\cdots+a_{k}(k=1,2, \cdots, n), \\
C_{n}=\left(a_{1}-b_{1}\right)^{2}+\left(a_{2}-b_{2}\right)^{2}+\cdots+\left(a_{n}-b_{n}\right)^{2}, \\
D_{n}=\left(a_{1}-b_{n}\right)^{2}+\left(a_{2}-b_{n}\right)^{2}+\cdots+... | 17. Construct the sequences $x_{n}=2 C_{n}-D_{n}, y_{n}=D_{n}-C_{n}, n \in \mathbf{N}_{+}$. Then
$$\begin{aligned}
x_{n+1}-x_{n}= & 2\left(C_{n+1}-C_{n}\right)-\left(D_{n+1}-D_{n}\right) \\
= & 2\left(a_{n+1}-b_{n+1}\right)^{2}-\left(a_{n+1}-b_{n+1}\right)^{2}-n\left(b_{n+1}^{2}-b_{n}^{2}\right) \\
& +2\left(b_{n+1}-b_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,284 |
Example 1 Let $a, b, c, d \in \mathbf{R}^{+}$, prove:
$$\begin{aligned}
& \frac{a^{3}+b^{3}+c^{3}}{a+b+c}+\frac{b^{3}+c^{3}+d^{3}}{b+c+d}+\frac{c^{3}+d^{3}+a^{3}}{c+d+a}+\frac{d^{3}+a^{3}+b^{3}}{d+a+b} \\
\geqslant & a^{2}+b^{2}+c^{2}+d^{2}
\end{aligned}$$ | Prove: For $x, y, z \in \mathbf{R}^{+}$, we have
$$\frac{x^{3}+y^{3}+z^{3}}{x+y+z} \geqslant \frac{x^{2}+y^{2}+z^{2}}{3}.$$
In fact, by the Cauchy-Schwarz inequality, we get
$$\begin{aligned}
(x+y+z)\left(x^{3}+y^{3}+z^{3}\right) & \geqslant\left(x^{2}+y^{2}+z^{2}\right)^{2} \\
& \geqslant\left(x^{2}+y^{2}+z^{2}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,285 |
Example $\mathbf{2}$ Let $x, y, z \in \mathbf{R}^{+}$, prove:
$$\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{z+x}}+\sqrt{\frac{z}{x+y}} \geqslant 2 .$$ | To prove $\sqrt{\frac{x}{y+z}} \geqslant \frac{2 x}{x+y+z}$.
In fact, by the mean inequality, we have
$$\sqrt{\frac{x}{y+z}}=\frac{x}{\sqrt{x} \sqrt{y+z}} \geqslant \frac{x}{\frac{x+y+z}{2}}=\frac{2 x}{x+y+z}$$
Similarly, we can get
$$\begin{array}{l}
\sqrt{\frac{y}{z+x}} \geqslant \frac{2 y}{x+y+z} \\
\sqrt{\frac{z}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,286 |
Example 3 Let $0 \leqslant a, b, c \leqslant 1$, prove:
$$\frac{a}{b c+1}+\frac{b}{c a+1}+\frac{c}{a b+1} \leqslant 2$$ | Proof We first prove
$$\frac{a}{b c+1} \leqslant \frac{2 a}{a+b+c}$$
Notice that (1) is equivalent to $a+b+c \leqslant 2 b c+2$, i.e.,
$$(b-1)(c-1)+b c+1 \geqslant a .$$
Since $a, b, c \in[0,1]$, the above inequality clearly holds, so (1) is true.
Similarly, we have
$$\begin{array}{l}
\frac{b}{c a+1} \leqslant \frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,287 |
Example 4 Given $x_{i} \geqslant 1, i=1,2, \cdots, n$, and $x_{1} x_{2} \cdots x_{n}=a^{n}$. Let $a_{n+1}=x_{1}$. Prove: When $n>2$, we have $\sum_{i=1}^{n} x_{i} x_{i+1}-\sum_{i=1}^{n} x_{i} \geqslant \frac{n}{2}\left(a^{2}-1\right)$. | Given that $\left(x_{i}-1\right)\left(x_{i+1}-1\right) \geqslant 0$, hence
$$2 x_{i} x_{i+1}-x_{i}-x_{i+1} \geqslant x_{i} x_{i+1}-1$$
Summing the above inequality from $i=1$ to $n$, we have $2\left(\sum_{i=1}^{n} x_{i} x_{i+1}-\sum_{i=1}^{n} x_{i}\right) \geqslant \sum_{i=1}^{n} x_{i} x_{i+1}-n$. Using the arithmetic... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,288 |
Example 5 Let real numbers $a_{1}, a_{2}, \cdots, a_{n} \in(-1,1]$, prove:
$$\sum_{i=1}^{n} \frac{1}{1+a_{i} a_{i+1}} \geqslant \sum_{i=1}^{n} \frac{1}{1+a_{i}^{2}}\left(\text { with the convention } a_{n+1}=a_{1}\right. \text { ). }$$ | Proof: First, prove that if $x, y \in(-1,1]$, then
$$\frac{2}{1+x y} \geqslant \frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}$$
(1) is equivalent to $2\left(1+x^{2}\right)\left(1+y^{2}\right)-(1+x y)\left(2+x^{2}+y^{2}\right) \geqslant 0$, which simplifies to $(x-y)^{2}-x y(x-y)^{2} \geqslant 0$. Hence, (1) holds.
Therefore,
$$\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,289 |
Example 6 Let $x, y, z \geqslant 0$, and $x^{2}+y^{2}+z^{2}=1$, prove:
$$\frac{x}{1+y z}+\frac{y}{1+z x}+\frac{z}{1+x y} \geqslant 1$$ | To prove we only need to prove the local inequality
$$\frac{x}{1+y z} \geqslant x^{2}$$
(1) is equivalent to
$$x+x y z \leqslant 1$$
and
$$\begin{aligned}
x+x y z & \leqslant x+\frac{1}{2} x\left(y^{2}+z^{2}\right)=\frac{1}{2}\left(3 x-x^{3}\right) \\
& =\frac{1}{2}\left[2-(x-1)^{2}(x+2)\right] \leqslant 1 .
\end{alig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,290 |
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