problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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Example 4 Let $x_{i}>0,(i=1,2,3,4,5)$ and $\sum_{i=1}^{n} \frac{1}{1+x_{i}}$ $=1$, prove that: $\sum_{i=1}^{5} \frac{x_{i}}{4+x_{i}^{2}} \leqslant 1$ (2003 China Western Mathematical Olympiad) demonstrating the simplicity and brutality of the tangent line method. | Prove: By setting $\frac{1}{1+x_{i}}=a_{i}$ and $\sum_{i=1}^{n} a_{i}=1(i=1, 2,3,4,5)$, the inequality is transformed into $\sum_{i=1}^{5} \frac{1 / a_{i}-1}{4+\left(1 / a_{i}-1\right)^{2}} \leqslant 1$
which is equivalent to $\sum_{i=1}^{5} \frac{-a_{i}^{2}+a_{i}}{5 a_{i}^{2}-2 a_{i}+1} \leqslant 1$. Let $f(x)=\frac{-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,054 |
Example 5 Let $a, b, c$ be positive real numbers. Prove that $\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}$ $+\frac{(a+2 b+c)^{2}}{2 b^{2}+(a+c)^{2}}+\frac{(a+b+2 c)^{2}}{2 c^{2}+(b+a)^{2}} \leqslant 8$. (2003 USA Mathematical Olympiad Problem) | To prove this problem, we notice the following fact: replacing $a, b, c$ with $\frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c}$ does not change the inequality, so we can assume $0<a, b, c<1, a+b+c=1$. Then, $\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}=\frac{(a+1)^{2}}{2 a^{2}+(1-a)^{2}}=\frac{(a+1)^{2}}{3 a^{2}-2 a+1}$. ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,055 |
Example 6 Let $0<\alpha, \beta, \gamma<\frac{\pi}{2}$, and $\sin ^{3} \alpha+\sin ^{3} \beta+$ $\sin ^{3} \gamma=1$. Prove that
$\tan ^{2} \alpha+\tan ^{2} \beta+\tan ^{2} \gamma \geqslant \frac{3}{\sqrt[3]{9}-1}$. (2005 China Southeast Mathematical Olympiad Problem Strengthened) | Prove: Let $x=\sin ^{3} \alpha, y=\sin ^{3} \beta, z=\sin ^{3} \gamma$, then the inequality is equivalent to in $0<\alpha, \beta, \gamma<\pi$, $(p-q)^{2}>0,\left(2 p^{3}+4 p^{2} q\right)+\left(3 q^{2}\right.$ $-1)(2 p+q)>0$, so (1) holds, with equality if and only if $p=q$, at which point $x=1 / 3$, hence
$$\frac{\sqrt... | \frac{3}{\sqrt[3]{9}-1} | Inequalities | proof | Yes | Yes | inequalities | false | 736,056 |
Theorem For the $n \times m$ matrix
$$\left(\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1 m} \\
a_{21} & a_{22} & \cdots & a_{2 m} \\
\cdots & \cdots & \cdots & \cdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end{array}\right)$$
where $a_{i j} \geqslant 0(i=1,2, \cdots, n, j=1,2, \cdots, m)$, then
$$\left[\prod_... | Let $A_{j}=\sum_{i=1}^{n} a_{i j}(j=1,2, \cdots, m)$,
$$G_{i}=\prod_{j=1}^{m} a_{i j}(i=1,2, \cdots, n)$$
If some $A_{j}=0$, then by $a_{i j} \geqslant 0(i=1,2, \cdots, n)$, we get $a_{1 j}=a_{2 j}=\cdots=a_{n j}=0$.
At this point, $G_{1}=G_{2}=\cdots=G_{n}=0$,
$$\left[\prod_{j=1}^{m}\left(\sum_{i=1}^{n} a_{i j}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,058 |
Example 2 Let $a, b, c, d \geqslant 0, ab + bc + cd + da = 1$. Prove: $\sum \frac{a^{3}}{b+c+d} \geqslant \frac{1}{3}$.
where, " $\sum$ " denotes the cyclic sum. | Prove the construction of the $4 \times 2$ matrix
$$\left(\begin{array}{ll}
\frac{a^{3}}{b+c+d} & b+c+d \\
\frac{b^{3}}{c+d+a} & c+d+a \\
\frac{c^{3}}{d+a+b} & d+a+b \\
\frac{d^{3}}{a+b+c} & a+b+c
\end{array}\right)$$
Using Carleman's inequality, we get
$$\left[\sum \frac{a^{3}}{b+c+d} \cdot 3(a+b+c+d)\right]^{\frac{1... | \sum \frac{a^{3}}{b+c+d} \geqslant \frac{1}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 736,059 |
Example 2 Let $a, b, c, d>0$, and $d=\max \{a, b, c, d\}$, prove:
$$a(d-b)+b(d-c)+c(d-a)<d^{2} .$$ | Prove that subtracting the left side from the right side of equation (4) and rearranging for \(d\) yields
$$d^{2}-d(a+b+c)+(a b+b c+c a)$$
Recall the identity
$$\begin{aligned}
& (d-a)(d-b)(d-c) \\
= & d^{3}-d^{2}(a+b+c)+d(a b+b c+c a)-a b c
\end{aligned}$$
By comparing (5) and (6), we can see that (4) holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,060 |
Example 11 (An Zhenping Inequality)
Let the side lengths, semi-perimeters, and areas of $\triangle A_{1} A_{2} A_{3}$ and $\triangle B_{1} B_{2} B_{3}$ be $a_{1}, a_{2}, a_{3}, p_{1}, S_{1}$ and $b_{1}, b_{2}, b_{3}, p_{2}, S_{2}$, respectively. Prove:
$$\begin{array}{l}
b_{1}\left(p_{2}-b_{1}\right)\left(p_{1}-a_{2}\r... | Prove by introducing 6 positive numbers in the following way:
$$\left\{\begin{array} { l }
{ x = a _ { 2 } + a _ { 3 } - a _ { 1 } , } \\
{ y = a _ { 3 } + a _ { 1 } - a _ { 2 } , } \\
{ z = a _ { 1 } + a _ { 2 } - a _ { 3 } , } \\
{ p = b _ { 2 } + b _ { 3 } - b _ { 1 } , } \\
{ q = b _ { 3 } + b _ { 1 } - b _ { 2 } ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,061 |
3. For positive real numbers $a, b, c, d$, prove:
$$\frac{c}{a}(8 b+c)+\frac{d}{b}(8 c+d)+\frac{a}{c}(8 d+a)+\frac{b}{d}(8 a+b) \geqslant 9(a+b+c+d),$$
with equality if and only if $a=c$ and $b=d$. | 3. By the Arithmetic Mean-Geometric Mean Inequality, we have
$$\begin{array}{l}
\frac{b c}{a}+\frac{d a}{c} \geqslant 2 \sqrt{b d} \\
\frac{c d}{b}+\frac{a b}{d} \geqslant 2 \sqrt{a c}
\end{array}$$
Therefore, we only need to prove that
$$\frac{c^{2}}{a}+\frac{a^{2}}{c}+\frac{d^{2}}{b}+\frac{b^{2}}{d}+16 \sqrt{a c}+16... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,063 |
6. For non-negative real numbers $x, y, z$, prove:
$$\begin{aligned}
& \frac{(y+z-x)^{2}}{x^{2}+(y+z)^{2}}+\frac{(z+x-y)^{2}}{y^{2}+(z+x)^{2}}+\frac{(y+x-z)^{2}}{z^{2}+(y+x)^{2}} \\
\geqslant & \frac{3\left(x^{2}+y^{2}+z^{2}\right)}{(x+y+z)^{2}+2(y z+z x+x y)} .
\end{aligned}$$ | 6. Using incremental transformation, we know
$$\begin{array}{l}
\frac{(y+z-x)^{2}}{x^{2}+(y+z)^{2}}+\frac{(z+x-y)^{2}}{y^{2}+(z+x)^{2}}+\frac{(y+x-z)^{2}}{z^{2}+(y+x)^{2}} \\
-\frac{3\left(x^{2}+y^{2}+z^{2}\right)}{(z+y+z)^{2}+2(y z+z x+x y)} \\
\equiv \frac{8 F(x, y, z)}{\left(x^{2}+(y+z)^{2}\right)\left(y^{2}+(z+x)^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,066 |
8. For positive real numbers $x, y, z$, prove:
$$\frac{x y}{z(z+x)}+\frac{y z}{x(x+y)}+\frac{z x}{y(y+z)} \geqslant \frac{x}{z+x}+\frac{y}{x+y}+\frac{z}{y+z} .$$ | 8. Let $x=\min \{x, y, z\}$, then
$$\begin{aligned}
& \prod_{\text {sym }} x(y+z)\left(\sum_{\text {cyc }} \frac{x y}{z(z+x)}-\sum_{\text {cyc }} \frac{x}{z+x}\right) \\
= & \sum_{\text {cyc }} y^{2} z^{4}+\sum_{\text {sym }} y^{3} z^{3}-\sum_{\text {cyc }} x y^{2} z^{3}-3 x^{2} y^{2} z^{2} \\
\equiv & F(x, y, z)=F(x, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,068 |
9. For non-negative real numbers $x, y, z$, prove:
$$\begin{array}{l}
\sqrt{2\left(y^{2}+z^{2}\right)}+\sqrt{2\left(z^{2}+x^{2}\right)}+\sqrt{2\left(x^{2}+y^{2}\right)} \\
\geqslant \sqrt[3]{9\left((y+z)^{3}+(z+x)^{3}+(x+y)^{3}\right)}
\end{array}$$ | 9. $\sum_{\mathrm{cyc}} \sqrt{2\left(y^{2}+z^{2}\right)} \geqslant \sum_{\mathrm{cyc}} \frac{7 y^{2}+6 y z+7 z^{2}}{5(y+z)} \geqslant \sqrt[3]{9 \sum_{\mathrm{cyc}}(y+z)^{3}}$.
In fact,
$$\begin{aligned}
& 2\left(y^{2}+z^{2}\right)-\frac{\left(7 y^{2}+6 y z+7 z^{2}\right)^{2}}{25(y+z)^{2}}=\frac{(y-z)^{2}\left(y^{2}+1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,069 |
14. For non-negative real numbers $a, b, c, d$, prove:
$$\sqrt{\frac{a}{a+b+c}}+\sqrt{\frac{b}{b+c+d}}+\sqrt{\frac{c}{c+d+a}}+\sqrt{\frac{d}{d+a+b}} \leqslant \frac{4}{\sqrt{3}} .$$ | $$\begin{array}{l}
\text { 14. } \sum_{\mathrm{cyc}} \sqrt{\frac{a}{a+b+c}} \\
\leqslant \frac{2}{3 \sqrt{3}} \sum_{\text {cyc }} \frac{8 a^{2}+b^{2}+c^{2}+8 d^{2}+24 a b+24 a c+26 a d+4 b c+6 b d+6 c d}{3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+10(a b+b c+a c+b c+b d+c d)} \\
=\frac{4}{\sqrt{3}} .
\end{array}$$ | \frac{4}{\sqrt{3}} | Inequalities | proof | Yes | Yes | inequalities | false | 736,074 |
15. Find the best constant $k$ that satisfies the following inequality:
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k \cdot \frac{y z+z x+x y}{x^{2}+y^{2}+z^{2}} \geqslant 3+k \text {, }$$
where $x, y, z, k$ are positive real numbers. | 15. The best constant $k=3 \sqrt[3]{4}-2 \approx 2.7622$, equality holds when:
$$\begin{aligned}
x & =\frac{1}{3}+\sqrt[3]{2}-\frac{\sqrt[3]{4}}{3}+\frac{2}{3} \sqrt{\sqrt[3]{4}+8 \sqrt[3]{2}-11} \cos \left(\frac{1}{3} \arccos \sqrt{\frac{17-3 \sqrt[3]{4}}{20}}\right) \\
& \approx 1.5949 \\
y & =\frac{2}{3} \sqrt[3]{4}... | 3 \sqrt[3]{4}-2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,075 |
16. Let non-negative real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$. Prove:
$$\sum_{c y c} \frac{1}{1-y z} \leqslant \frac{9}{2}$$ | 16.
$$\begin{array}{c}
\sum_{\text {cyc }} \frac{1}{1-y z} \leqslant \frac{4 \sum x \sum y z}{\prod(y+z)} \leqslant \frac{9}{2} ; \\
\sum_{\text {cyc }} \frac{1}{1-y z} \leqslant \sum_{\text {cyc }} \frac{27 x^{2}}{2(2 x+y)(2 x+z)} \leqslant \frac{9}{2} .
\end{array}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,076 |
17. Let non-negative real numbers $x, y, z$ satisfy $x+y+z=1$. Prove:
$$\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{1}{1+z^{2}} \geqslant \frac{5}{2},$$
with equality when $x=1, y=z=0$. | $\begin{array}{l}\text { 17. } \frac{(x+y+z)^{2}}{(x+y+z)^{2}+x^{2}}-\frac{x+2 y+2 z}{2(x+y+z)} \\ =\frac{x(y+z)}{2(x+y+z)\left((x+y+z)^{2}+x^{2}\right)} \geqslant 0 \Rightarrow \\ \sum_{\text{cyc}} \frac{1}{1+x^{2}}=\sum_{\text{cyc}} \frac{(x+y+z)^{2}}{(x+y+z)^{2}+x^{2}} \geqslant \sum_{\text{cyc}} \frac{x+2 y+2 z}{2(... | \frac{5}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,077 |
18. For positive real numbers $x, y, z$, prove:
$$2 \sum_{\text {cyc }} x \sqrt{\frac{y z+z x+x y}{(x+y)(x+z)}} \geqslant 3 \sum_{\text {cyc }} x \sqrt{\frac{y+z}{4 x+y+z}} .$$ | 18. This is an algebraic inequality related to the Gergonne point and Ceva lines in a triangle.
$$\begin{array}{l}
90(x+y+z) \sum_{\text {cyc }} x \sqrt{(y+z)(4 x+y+z)(4 y+z+x)(4 z+x+y)(y z+z x+x y)} \\
\geqslant 180\left(x^{4}(y+z)+y^{4}(x+z)+z^{4}(x+y)\right)+1079\left(x^{3}\left(y^{2}+z^{2}\right)\right. \\
\left.\q... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,078 |
19. Let positive real numbers $a, b, c$ satisfy $a+b+c=1$. Prove:
(1) $\frac{a}{\sqrt{a+b c}}+\frac{b}{\sqrt{b+c a}}+\frac{c}{\sqrt{c+a b}} \leqslant \frac{3}{2}$;
(2) $\frac{1}{\sqrt{a+b c}}+\frac{1}{\sqrt{b+c a}}+\frac{1}{\sqrt{c+a b}} \geqslant \frac{9}{2}$. | 19.
$$\text { (1) } \begin{aligned}
2 \frac{a}{\sqrt{a+a b}} & =2 \frac{a}{\sqrt{(a+c)(a+b)}} \\
& =2 \sqrt{\frac{a}{a+c} \cdot \frac{a}{a+b}} \leqslant \frac{a}{a+c}+\frac{a}{a+b}
\end{aligned}$$
(2) By the Cauchy-Schwarz inequality, it is easy to get $2 \geqslant \sum_{\mathrm{cyc}} \sqrt{a+b c} \Leftrightarrow 3 \sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,079 |
20. Let positive real numbers $x, y, z$ satisfy $x+y+z=1$. Prove:
$$\sqrt{1-3 y z}+\sqrt{1-3 z x}+\sqrt{1-3 x y} \geqslant \sqrt{6} .$$ | 20. Since
$$\begin{aligned}
& 200(x+y+z)^{2}\left((x+y+z)^{2}-3 y z\right)-3\left(6 x^{2}+19 x(y+z)+7 y^{2}\right. \\
& \left.+2 y z+7 z^{2}\right)^{2} \\
= & (y+z-2 x)^{2}\left(23 x^{2}+52 x(y+z)+2(y+z)^{2}\right) \\
& +3(y-z)^{2}\left(9(y+z-x)^{2}+5 x^{2}+4 x(y+z)+4\left(2 y^{2}+13 y z\right.\right. \\
& \left.\left.... | \sqrt{6}(x+y+z) | Inequalities | proof | Yes | Yes | inequalities | false | 736,080 |
21. For non-negative real numbers $x, y, z$, prove:
$$\sum_{\text {cyc }} \sqrt{2\left(x^{2}+y^{2}\right)} \geqslant \sqrt[3]{9 \sum_{\text {cyc }}(x+y)^{3}} .$$ | $$\begin{array}{l}
\text { 21. }\left\{\begin{array}{l}
\sqrt{2 y^{2}+2 z^{2}} \geqslant y+z, \\
32\left(y^{2}+z^{2}\right)^{3}-(y+z)^{2}\left(5 y^{2}-2 y z+5 z^{2}\right)^{2} \\
=(y-z)^{4}\left(7 y^{2}-2 y z+7 z^{2}\right) \geqslant 0, \\
3 \sqrt{2 \prod_{\mathrm{cyc}}\left(y^{2}+z^{2}\right)}=\sqrt{6 \sum_{\mathrm{cy... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,081 |
22. If $a, b, c$ are any real numbers, satisfying $a+b+c=p$, set $ab+bc+ca=\frac{p^2-q^2}{3}(q \geqslant 0), r=abc$. Prove: $\frac{(p+q)^2(p-2q)}{27} \leqslant r \leqslant \frac{(p-q)^2(p+2q)}{27}$ equality holds if and only if $(a-b)(b-c)(c-a)=0$. | 22. Consider the function $f(x)=(x-a)(x-b)(x-c)=x^{3}-p x^{2}+\frac{p^{2}-q^{2}}{3} x -r$.
We have $f^{\prime}(x)=3 x^{2}-2 p x+\frac{p^{2}-q^{2}}{3}$, with zeros $x_{1}=\frac{p+q}{3}, x_{2}=\frac{p-q}{3}$, and for $x_{2}<x<x_{1}$, $f^{\prime}(x)<0$; for $x<x_{2}$ or $x>x_{1}$, $f^{\prime}(x)>0$.
Furthermore, there ar... | \frac{(p+q)^{2}(p-2 q)}{27} \leqslant r \leqslant \frac{(p-q)^{2}(p+2 q)}{27} | Inequalities | proof | Yes | Yes | inequalities | false | 736,082 |
23. Non-negative real numbers $x, y, z$ satisfy $9(x+y+z)+10 \geqslant 8 x y z$. Prove: $x+y+z$ $\geqslant 2(\sqrt{x y}+\sqrt{y z}+\sqrt{z x})$ | 23. Let $x=a^{2}, y=b^{2}, z=c^{2}$, then the original inequality is equivalent to
$$\begin{aligned}
& 9\left(a^{2}+b^{2}+c^{2}\right)+10 \geqslant 8 a^{2} b^{2} c^{2} \\
\Rightarrow & 6+a^{2}+b^{2}+c^{2} \geqslant 2(a b+b c+c a)(a, b, c \in \mathbf{R})
\end{aligned}$$
Let $a+b+c=p, a b+b c+c a=\frac{p^{2}-q^{2}}{3}, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,083 |
24. Let $x, y, z>0$ and $x+y+z=3$. Prove:
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3 x y z \geqslant x^{2}+y^{2}+z^{2}-3 .$$ | 24.
$$\begin{array}{c}
\frac{\sum_{\mathrm{cyc}} x y}{x y z}-3 x y z \geqslant\left(\sum_{\mathrm{cyc}} x\right)^{2}-2 \sum_{\mathrm{cyc}} x y-3 \\
\sum_{\mathrm{cyc}} x=3, \sum_{\mathrm{cyc}} x y=\frac{9-q^{2}}{3}, x y z=r \\
\frac{9-q^{2}}{3 r}-3 r \geqslant q-3-\frac{2}{3}\left(9-q^{2}\right)
\end{array}$$
That is,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,084 |
25. Let $x, y, z>0$, and satisfy $x y z=1$. Prove: $\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}}$
$$\leqslant \frac{3 \sqrt{2}}{2}$$ | 25. Let $x \leqslant y \leqslant z$, then $x y \leqslant 1$,
and
$$\begin{array}{l}
\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}} \leqslant 2 \cdot \sqrt{\frac{\left(\frac{1}{1+x}+\frac{1}{1+y}\right)}{2}}, \\
\frac{\left(\frac{1}{1+x}+\frac{1}{1+y}\right)}{2} \leqslant \frac{1}{1+\sqrt{x y}} \\
\Leftrightarrow(2 \sqrt{x ... | \frac{3 \sqrt{2}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,085 |
26. Let $a, b, c$ be non-negative. Prove $\sum_{\text {cyc }} \sqrt{a^{2}-a b+b^{2}} \cdot \sqrt{b^{2}-b c+c^{2}} \geqslant a^{2}+$
$$b^{2}+c^{2} \text {. }$$ | 26. $\sum_{\mathrm{cyc}} \sqrt{a^{2}-a b+b^{2}} \cdot \sqrt{b^{2}-b c+c^{2}} \geqslant a^{2}+b^{2}+c^{2}$
Because
$$\sqrt{a^{2}-a b+b^{2}} \geqslant \frac{a^{2}+b^{2}}{a+b}$$
So we only need to prove
$$\sum_{\mathrm{cyc}} \frac{a^{2}+b^{2}}{a+b} \cdot \frac{b^{2}+c^{2}}{b+c} \geqslant \sum_{\mathrm{cyc}} a^{2}$$
Exp... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,086 |
Example 2 Let $x, y, z>1$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. Prove:
$$\sqrt{x+y+z} \geqslant \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$$ | Proof: First, perform algebraic transformation, set $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$, thus
$$\sqrt{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \geqslant \sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}$$
where $a, b, c \in(0,1)$, and $a+b+c=2$. Substituting the condition $a+b+c=2$, we get that equa... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,087 |
Example 3 Let $a, b, c \geqslant 0$ and $t \in(0,3]$. Prove:
$$(3-t)+t(a b c)^{\frac{2}{4}}+\sum_{\text {cyc }} a^{2} \geqslant 2 \sum_{\text {cyc }} a b .$$ | First, let \( x = a^{\frac{2}{3}}, y = b^{\frac{2}{3}}, z = c^{\frac{2}{3}} \), then equation (5) can be rewritten as
$$3 - t + t(x y z)^{\frac{3}{4}} + \sum_{\mathrm{cyc}} x^{3} \geqslant 2 \sum_{\mathrm{cyc}} (x y)^{\frac{3}{2}}.$$
To prove equation (6), it suffices to show that
$$3 - t + t(x y z)^{\frac{3}{t}} \geq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,088 |
Example 3 Let $x, y, z$ be non-negative real numbers, and the sum of any two of them is not zero. Prove:
$$\sum_{\mathrm{cyc}} \frac{2 x^{2}+y z}{y+z} \geqslant \frac{9\left(x^{2}+y^{2}+z^{2}\right)}{2(x+y+z)}$$ | $$\text { Prove } \begin{aligned}
& \sum_{\text {cyc }} \frac{2 x^{2}+y z}{y+z}-\frac{9\left(x^{2}+y^{2}+z^{2}\right)}{2(x+y+z)} \\
= & \sum_{\text {cyc }} \frac{(y-z)^{2}\left(2(x-y-z)^{2}+y z\right)}{2(x+y)(x+z)(x+y+z)} \\
\geqslant & 0,
\end{aligned}$$
so inequality (5) holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,089 |
Example 5 Let positive real numbers $a, b, c$ satisfy $abc=1$. Try to prove:
$$\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1} \leqslant 1$$ | To prove that equation (10) can be rewritten as
$$\frac{1}{a+b+(a b c)^{\frac{1}{3}}}+\frac{1}{b+c+(a b c)^{\frac{1}{3}}}+\frac{1}{c+a+(a b c)^{\frac{1}{3}}} \leqslant \frac{1}{(a b c)^{\frac{1}{3}}}.$$
Let $a=x^{3}, b=y^{3}, c=z^{3}$, then the above equation can be rewritten as
$$\frac{1}{x^{3}+y^{3}+x y z}+\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,090 |
Example 6 (Muirhead's Theorem) Let real numbers $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ satisfy
$$\begin{array}{r}
a_{1} \geqslant a_{2} \geqslant a_{3} \geqslant 0, b_{1} \geqslant b_{2} \geqslant b_{3} \geqslant 0, a_{1} \geqslant b_{1} \\
a_{1}+a_{2} \geqslant b_{1}+b_{2}, a_{1}+a_{2}+a_{3}=b_{1}+b_{2}+b_{3} .
\e... | Prove that when $b_{1} \geqslant a_{2}$,
$$\begin{aligned}
\sum_{\mathrm{sym}} x^{a_{1}} y^{a_{2}} z^{a_{3}} & =\sum_{\mathrm{cyc}} z^{a_{3}}\left(x^{a_{1}} y^{a_{2}}+x^{a_{2}} y^{a_{1}}\right) \\
& \geqslant \sum_{\mathrm{cyc}} z^{a_{3}}\left(x^{a_{1}+a_{2}-b_{1}} y^{b_{1}}+x^{b_{1}} y^{a_{1}+a_{2}-b_{1}}\right) \\
& ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,091 |
Example 4 Let $x, y, z>0$, and $y z+z x+x y=1$. Try to prove:
$$\frac{1+y^{2} z^{2}}{(y+z)^{2}}+\frac{1+z^{2} x^{2}}{(z+x)^{2}}+\frac{1+x^{2} y^{2}}{(x+y)^{2}} \geqslant \frac{5}{2}$$ | Prove that
$$\begin{aligned}
& \sum_{\text {cyc }} \frac{1+y^{2} z^{2}}{(y+z)^{2}}-\frac{5}{2} \\
= & \frac{(y-z)^{2}(z-x)^{2}(x-y)^{2}}{2(y+z)^{2}(z+x)^{2}(x+y)^{2}} \\
& +\sum_{\text {cyc }} \frac{\left(x(y+z)\left(y^{2}+z^{2}-2 x^{2}\right)+(y-z)^{2}\left(x^{2}+y z\right)\right)^{2}}{6(y+z)^{2}(z+x)^{2}(x+y)^{2}} \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,092 |
Example 12 Non-negative real numbers $x, y, z$ satisfy $x y + y z + z x = 1$. Prove:
$$\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \geqslant \frac{5}{2}$$ | To prove the desired inequality (26), it suffices to show that
$$(x y+y z+z x)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)^{2} \geqslant\left(\frac{5}{2}\right)^{2}$$
which is equivalent to
$$\begin{aligned}
& 4 \sum_{\text {sym }} x^{5} y+\sum_{\text {sym }} x^{4} y z+14 \sum_{\text {sym }} x^{3} y^{2} z+38... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,093 |
Example 13 Proof: When the sum of any two of the non-negative real numbers $x, y, z$ is not zero,
$$\sum_{c y c} \frac{x^{3}+x y z}{y+z} \geqslant x^{2}+y^{2}+z^{2}$$ | Prove that by rearranging formula (27), we get
$$\sum_{c y c} \frac{x(x-y)(x-z)}{y+z} \geqslant 0$$
Without loss of generality, assume \( x \geqslant y \geqslant z \), then
$$\begin{aligned}
& \sum_{\mathrm{cyc}} \frac{x(x-y)(x-z)}{y+z} \\
= & \frac{x(x-y)(x-z)}{y+z}+\frac{y(y-z)(y-x)}{z+x}+\frac{z(z-x)(z-y)}{x+y} \\
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,094 |
Example 14 Prove: When the sum of any two of the non-negative real numbers $x, y, z$ is not zero,
$$\sum_{c y c} \frac{2 x^{2}+y z}{y+z} \geqslant \frac{9\left(x^{2}+y^{2}+z^{2}\right)}{2(x+y+z)}$$ | Prove that by Schur's inequality and Muirhead's theorem,
$$\begin{aligned}
& 2(x+y+z) \sum_{\mathrm{cyc}}\left(2 x^{2}+y z\right)(x+y)(x+z)-9\left(x^{2}+y^{2}\right. \\
& \left.+z^{2}\right) \prod_{\text {cyc }}(y+z) \\
= & 4 \sum_{\text {cyc }} x^{5}-\sum_{\text {cyc }} x^{4}(y+z)-3 \sum_{\text {cyc }} x^{3}\left(y^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,095 |
Example 15 Let $x, y, z>0$, and $y z+z x+x y=1$. Prove:
$$\sum_{\text {cyc }} \frac{1+y^{2} z^{2}}{(y+z)^{2}} \geqslant \frac{5}{2}$$ | Prove that by Schur's inequality and Muirhead's theorem,
$$\begin{aligned}
& 2 \sum_{\text {cyc }}(z+x)^{2}(x+y)^{2}\left((y z+z x+x y)^{2}+y^{2} z^{2}\right)-5(y \\
& +z)^{2}(z+x)^{2}(x+y)^{2}(y z+z x+x y) \\
= & 2 \sum_{\text {cyc }} x^{6}\left(y^{2}+z^{2}\right)+4 \sum_{\text {cyc }} x^{6} y z-\sum_{\text {cyc }} x^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,096 |
Example 2 Proof: For positive real numbers $a, b, c$, we have
$$\begin{array}{c}
\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)}
\end{array}$$ | Prove that dividing both sides of equation (2) by $a b c$ and setting $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$, we get
$$\begin{aligned}
& \sqrt{(x+y+z)(x y+y z+z x)} \\
\geqslant & 1+\sqrt[3]{\left(\frac{x}{z}+1\right)\left(\frac{y}{x}+1\right)\left(\frac{z}{y}+1\right)}
\end{aligned}$$
where $x y z=1$. From $x ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,097 |
Example 3 Proof: For positive real numbers $a, b, c$, we have
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}$$ | By the symmetry of equation (5), without loss of generality, let $a+b+c=1$, where $0<a, b, c<1$. Then,
$$\sum_{\text {cyc }} \frac{a}{b+c}=\sum_{\text {cyc }} f(a) \geqslant \frac{3}{2}$$
where $f(x)=\frac{x}{1-x}$. By the convexity of $f(x)$ on $x \in(0,1)$, we have
$$\frac{1}{3} \sum_{\text {cyc }} f(a) \geqslant f\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,098 |
Example 5 Proof: For positive real numbers $a, b, c$, we have
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}$$ | Prove that using the Cauchy-Schwarz inequality, we have
$$((b+c)+(c+a)+(a+b))\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) \geqslant 3^{2}$$
Therefore,
$$\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b} \geqslant \frac{9}{2}$$
Thus, inequality (8) holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,099 |
Example 6 Proof: For positive real numbers $a, b, c$, we have
$$\begin{array}{l}
\quad \sqrt{a^{4}+a^{2} b^{2}+b^{4}}+\sqrt{b^{4}+b^{2} c^{2}+c^{4}}+\sqrt{c^{4}+c^{2} a^{2}+a^{4}} \\
\geqslant a \sqrt{2 a^{2}+b c}+b \sqrt{2 b^{2}+c a}+c \sqrt{2 c^{2}+a b}
\end{array}$$ | $$\begin{aligned}
\sum_{\mathrm{cyc}} \sqrt{a^{4}+a^{2} b^{2}+b^{4}} & =\sum_{\text {cyc }} \sqrt{\left(a^{4}+\frac{a^{2} b^{2}}{2}\right)+\left(b^{4}+\frac{a^{2} b^{2}}{2}\right)} \\
& \geqslant \frac{1}{\sqrt{2}} \sum_{\text {cyc }}\left(\sqrt{a^{4}+\frac{a^{2} b^{2}}{2}}+\sqrt{b^{4}+\frac{a^{2} b^{2}}{2}}\right) \\
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,100 |
Example 8 Let positive real numbers $a, b, c$ satisfy $\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1} \geqslant 1$. Prove:
$$a+b+\dot{c} \geqslant a b+b c+c a .$$ | Proof: By Cauchy-Schwarz inequality, we have
$$(a+b+1)\left(a+b+c^{2}\right) \geqslant(a+b+c)^{2},$$
which implies
$$\frac{1}{a+b+1} \leqslant \frac{c^{2}+a+b}{(a+b+c)^{2}}$$
Therefore,
$$\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1} \leqslant \frac{a^{2}+b^{2}+c^{2}+2(a+b+c)}{(a+b+c)^{2}}$$
Given that \(\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,101 |
Example 9 Let $a_{ij} (i, j=1,2, \cdots, n)$ be positive real numbers. Prove:
$$\begin{aligned}
& \left(a_{11}^{n}+a_{12}^{n}+\cdots+a_{1 n}^{n}\right) \cdot\left(a_{21}^{n}+a_{22}^{n}+\cdots+a_{2 n}^{n}\right) \cdot \cdots \cdot\left(a_{n 1}^{n}+a_{n 2}^{n}\right. \\
& \left.+\cdots+a_{m}^{n}\right) \\
\geqslant & \le... | Prove that according to the homogeneity of equation (11), we can set
$\left(a_{i 1}^{n}+a_{i 2}^{n}+\cdots+a_{i 1}^{n}\right)^{\frac{1}{n}}=1$ or $a_{i 1}^{n}+a_{i 2}^{n}+\cdots+a_{i i}^{n}=1(i=1,2, \cdots$,
n),
Thus, we only need to prove
$$a_{11} a_{21} \cdots a_{n 1}+a_{12} a_{22} \cdots a_{n 2}+\cdots+a_{1 n} a_{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,102 |
1. For $x_{1}, x_{2}, \cdots, x_{n} \geqslant 0$, prove:
$$\begin{array}{c}
(n-1)\left(x_{1}^{n}+x_{2}^{n}+\cdots x_{n}^{n}\right)+n x_{1} x_{2} \cdots x_{n} \\
\geqslant\left(x_{1}+x_{2}+\cdots x_{n}\right)\left(x_{1}^{n-1}+x_{2}^{n-1}+\cdots x_{n}^{n-1}\right) .
\end{array}$$ | 1. When $n=1,2$, it is relatively easy to prove. When $n=3$, it is Schur's inequality. In fact, the proof of Schur-type inequalities can give us sufficient inspiration. This inequality was proposed by Suranyi. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,103 |
Example 6 Proof: For real numbers $x, y, z$,
$$16 \sum_{\text {cyc }} x^{4}-20 \sum_{\text {cyc }} x^{3}(y+z)+9 \sum_{\text {cyc }} y^{2} z^{2}+15 \sum_{\text {cyc }} x^{2} y z \geqslant 0$$ | Prove that for any real numbers $u, v, x, y, z$, consider the following two expressions:
$$\begin{array}{c}
u^{2}\left(\sum_{\text {cyc }} x^{4}-\sum_{\text {cyc }} y^{2} z^{2}\right)+v^{2}\left(\sum_{\text {cyc }} y^{2} z^{2}-\sum_{\text {cyc }} x^{2} y z\right) \\
u v\left(\sum_{\text {cyc }} x^{3}(y+z)-2 \sum_{\text... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,104 |
2. For positive real numbers $x, y, z$, prove:
$$\frac{x y}{z(z+x)}+\frac{y z}{x(x+y)}+\frac{z x}{y(y+z)} \geqslant \frac{x}{z+x}+\frac{y}{x+y}+\frac{z}{y+z} .$$ | 2. By bringing to a common denominator and using the Muirhead's theorem, we get
$$\begin{aligned}
& x y z(y+z)(z+x)(x+y)\left(\sum_{\text {cyc }} \frac{x y}{z(z+x)}-\sum_{\text {cyc }} \frac{x}{z+x}\right) \\
= & \sum_{\text {cyc }} y^{2} z^{4}+\sum_{\text {cyc }} y^{3} z^{3}-\sum_{\text {cyc }} x y^{2} z^{3}-3 x^{2} y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,105 |
3. Let $a, b, c \geqslant 1$, prove:
$$1+\sqrt{\frac{b c+c a+a b}{3}} \geqslant \sqrt[3]{(1+a)(1+b)(1+c)}$$ | 3. This inequality is equivalent to
$$\left(\frac{a b+b c+c a}{3}\right)^{\frac{3}{2}}+3\left(\frac{a b+b c+c a}{3}\right)^{\frac{1}{2}} \geqslant a+b+c+a b c$$
Let $f(a, b, c)=\left(\frac{a b+b c+c a}{3}\right)^{\frac{3}{2}}+3\left(\frac{a b+b c+c a}{3}\right)^{\frac{1}{2}}-(a+b+c)-$
$a b c, t=\sqrt{c^{2}+a b+b c+c a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,106 |
11. Let $a_{1}, a_{2}, \cdots, a_{9}, b_{1}, b_{2}, \cdots, b_{9} \in [1,2]$, and
$$a_{1}^{2}+a_{2}^{2}+\cdots+a_{9}^{2}=b_{1}^{2}+b_{2}^{2}+\cdots+b_{9}^{2}$$
Prove:
$$\sum_{m=1}^{9} \frac{a_{m}^{3}}{b_{m}} \leqslant \frac{5}{3} \sum_{m=1}^{9} a_{m}^{2}$$ | 11. Here is a more advanced form of the proof for this problem.
Let $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n} \in [1001, 2002]$, and
$$a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2} = b_{1}^{2} + b_{2}^{2} + \cdots + b_{n}^{2}$$
Prove: $\frac{a_{1}^{3}}{b_{1}} + \frac{a_{2}^{3}}{b_{2}} + \cdots + \frac{a_... | \sum_{m=1}^{9} \frac{a_{m}^{3}}{b_{m}} \leqslant \frac{5}{3} \sum_{m=1}^{9} a_{m}^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,108 |
12. Let \(x+y+z=1, x, y, z>0\). Prove:
$$\frac{x}{\sqrt{1-x}}+\frac{y}{\sqrt{1-y}}+\frac{z}{\sqrt{1-z}} \geqslant \sqrt{\frac{3}{2}}$$ | 12. Using the Radon's inequality from Section 5.2,
$$\sum_{\mathrm{cyc}} \frac{x}{\sqrt{y+z}}=\sum_{\mathrm{cyc}} \frac{x^{\frac{3}{2}}}{\sqrt{x y+x z}} \geqslant \frac{(x+y+z)^{\frac{3}{2}}}{\sqrt{2 \sum_{\mathrm{cyc}} x y}} \geqslant \sqrt{\frac{3}{2}}$$ | \sqrt{\frac{3}{2}} | Inequalities | proof | Yes | Yes | inequalities | false | 736,109 |
Example: Let $3 a_{1}, a_{2}, \cdots, a_{2002}$ be non-negative integers, satisfying
$$a_{i}+a_{j} \leqslant a_{i+j} \leqslant a_{i}+a_{j}+1,1 \leqslant i, j \leqslant 2002, i+j \leqslant 2002 .$$
Prove: There exists a real number $x$ such that $a_{n}=[n x], n=1,2, \cdots, 2002$. | Let $I_{n}=\left(\frac{a_{n}}{n}, \frac{a_{n}+1}{n}\right), n=1,2, \cdots, 2002$. If there exists a real number $x \in \bigcap_{n=1}^{2002} I_{n}$, then the proposition is proved.
For this, let $L=\max _{1 \leqslant n \leqslant 2002} \frac{a_{n}}{n}$, and $R=\min _{1 \leqslant n \leqslant 2002} \frac{a_{n}+1}{n}$. We ... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 736,111 |
Example 7 For real numbers $a, b, c$, let $|a+b|=m,|a-b|=n$ and $m n \neq 0$. Prove:
$$\max \{|a c+b|,|a+b c|\} \geqslant \frac{m n}{\sqrt{m^{2}+n^{2}}}$$ | Prove that since
$$\begin{aligned}
& \left(m^{2}+n^{2}\right) \max \left\{|a c+b|^{2},|a+b c|^{2}\right\}-m^{2} n^{2} \\
\geqslant & \left(|a+b|^{2}+|a-b|^{2}\right) \frac{|a c+b|^{2}+|a+b c|^{2}}{2}-|a+b|^{2}|a-b|^{2} \\
= & \left(\left(a^{2}+b^{2}\right) c+2 a b\right)^{2} \geqslant 0,
\end{aligned}$$
thus equation ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,113 |
Example 5 Let two positive sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}$ satisfy:
(1) $a_{0}=1 \geqslant a_{1}, a_{n}\left(b_{n-1}+b_{n+1}\right)=a_{n-1} b_{n-1}+a_{n+1} b_{n+1}, n \geqslant 1$;
(2) $\sum_{i=0}^{n} b_{i} \leqslant n^{\frac{3}{2}}, n \geqslant 1$.
Find the general term of $\left\{a_{n}\right\}$... | From condition (1), we have $a_{n}-a_{n+1}=\frac{b_{n-1}}{b_{n+1}}\left(a_{n-1}-a_{n}\right)$, hence
$$a_{n}-a_{n+1}=\frac{b_{0} b_{1}}{b_{n} b_{n+1}}\left(a_{0}-a_{1}\right)$$
If $a_{1}=a_{0}=1$, then $a_{n}=1$. Below, we discuss $a_{1}a_{0}-a_{n}=b_{0} b_{1}\left(a_{0}-a_{1}\right) \sum_{k=0}^{n-1} \frac{1}{b_{k} b_... | a_{n}=1(n \geqslant 0) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,114 |
Example 7 Let the bounded sequence $\left\{a_{n}\right\}_{n \geqslant 1}$ satisfy
$$a_{n}<\sum_{k=n}^{2 n+2006} \frac{a_{k}}{k+1}+\frac{1}{2 n+2007}, n=1,2,3, \cdots$$
Prove:
$$a_{n}<\frac{1}{n}, n=1,2,3, \cdots .$$ | Proof: Let $b_{n}=a_{n}-\frac{1}{n}$, then
$$b_{n}<\sum_{k=n}^{2 n+2006} \frac{b_{k}}{k+1}, n \geqslant 1$$
We will prove that $b_{n}<0$. Since $a_{n}$ is bounded, there exists a constant $M$ such that $b_{n}<M$. When $n$ is sufficiently large (for example, greater than $10^{6}$), we have
$$\begin{aligned}
b_{n} & <\s... | a_{n}<\frac{1}{n}, n=1,2,3, \cdots | Inequalities | proof | Yes | Yes | inequalities | false | 736,115 |
Example 8 Given a real number $a$ and a positive integer $n$. Prove:
(1) There exists a unique real number sequence $x_{0}, x_{1}, \cdots, x_{n}, x_{n+1}$, satisfying
$$\left\{\begin{array}{l}
x_{0}=x_{n+1}=0, \\
\frac{1}{2}\left(x_{i+1}+x_{i-1}\right)=x_{i}+x_{i}^{3}-a^{3}, i=1,2, \cdots, n
\end{array}\right.$$
(2) Th... | Proof of (1) Existence: From $x_{i+1}=2 x_{i}+2 x_{i}^{3}-2 a^{3}-x_{i-1}, i=1,2, \cdots, n$, and $x_{0}=0$, we know that each $x_{i}$ is a real-coefficient polynomial of $x_{1}$ with degree $3^{i-1}$, thus $x_{n+1}$ is a real-coefficient polynomial of $x_{i}$ with degree $3^{n}$. Since $3^{n}$ is an odd number, there ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,116 |
Example 9 Let $n$ be a positive integer, and $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}, c_{2}, c_{3}, \cdots, c_{2 n}$ be $4 n-1$ positive real numbers, such that $c_{i+j}^{2} \geqslant a_{i} b_{j}, 1 \leqslant i, j \leqslant n$. Let $m=\max _{2 \leqslant i \leqslant 2 n} c_{i}$, prove:
$$\begin{aligned... | Let $X=\max _{1 \leqslant i \leqslant n} a_{i}, Y=\max _{1 \leqslant i \leqslant n} b_{i}$, and replace $a_{i}, b_{i}, c_{i}$ with $a_{i}^{\prime}=\frac{a_{i}}{X}, b_{i}^{\prime}=\frac{b_{i}}{Y}, c_{i}^{\prime}=\frac{c_{i}}{\sqrt{X Y}}$ respectively. Therefore, we can assume $X=Y=1$.
Next, we prove
Thus, $\frac{m+c_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,117 |
Example 12 Find the largest constant $M>0$, such that for any positive integer $n$, there exist positive real number sequences $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$, satisfying:
(1) $\sum_{k=1}^{n} b_{k}=1, 2 b_{k} \geqslant b_{k-1}+b_{k+1}, k=2,3, \cdots, n-1$;
(2) $a_{k}^{2} \leqslant 1+\sum... | Prove a lemma first: $\max _{1 \leqslant k \leqslant n} a_{k} < 2$ and $\sum_{k=1}^{n} b_{k}=1$, hence from the above we get
This is
$$b_{k}>\left\{\begin{array}{l}
\frac{k-1}{m-1} b_{m}, 1 \leqslant k \leqslant m \\
\frac{n-k}{n-m} b_{m}, m \leqslant k \leqslant n
\end{array}\right.$$
Therefore
$$\begin{array}{c}
1=... | \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,118 |
4. Given $a_{0}=1, a_{1}=2, a_{n+1}=a_{n}+\frac{a_{n-1}}{1+a_{n-1}^{2}}, n \geqslant 1$. Prove: $52<a_{1371}<$
65. | 4. First prove $a_{n-1}^{2}+2<a_{n}^{2} \leqslant a_{n-1}^{2}+3(n \geqslant 1)$, then use induction to prove $\sqrt{2 n+1} \leqslant a_{n}$ $\leqslant \sqrt{3 n+2}$ | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,122 |
5. Given $a_{1}=1, b_{1}=2, a_{n+1}=\frac{1+a_{n}+a_{n} b_{n}}{b_{n}}, b_{n+1}=\frac{1+b_{n}+a_{n} b_{n}}{a_{n}}$. Prove: $a_{2008}<5$. | 5. Prove that for any positive integer $n$, $\frac{1}{a_{n}+1}-\frac{1}{b_{n}+1}=\frac{1}{6}$. | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,123 |
8. Given $a_{1}=1, a_{n+1}=\frac{a_{n}}{n}+\frac{n}{a}$. Prove that for $n \geqslant 4$, $\sqrt{n}<a_{n}<\sqrt{n+1}$. | 8. First prove that $f(x)=\frac{x}{n}+\frac{n}{x}$ is a decreasing function on $(0, n)$, then use mathematical induction to prove $\sqrt{n} \leqslant a_{n} \leqslant \frac{n}{\sqrt{n-1}}$, and then prove $a_{n}<\sqrt{n+1}$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,125 |
10. Given $a_{1}=\frac{1}{2}, a_{k+1}=-a_{k}+\frac{1}{2-a_{k}}, k \geqslant 1$.
Prove: $\left(\frac{n}{2 \sum_{i=1}^{n} a_{i}}-1\right)^{n} \leqslant\left(\frac{1}{n} \sum_{i=1}^{n} a_{i}\right)^{n} \prod_{i=1}^{n}\left(\frac{1}{a_{i}}-1\right)$. | 10. First, prove $0<a_{n} \leqslant \frac{1}{2}(n \geqslant 1)$ using mathematical induction, then by Jensen's inequality, we get $\left(\frac{n}{\sum_{i=1}^{n} a_{i}}-1\right)^{n} \leqslant \prod_{i=1}^{n}\left(\frac{1}{a_{i}}-1\right)$, and by Cauchy-Schwarz inequality, we get $\sum_{i=1}^{n}\left(1-a_{i}\right)$ $=\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,127 |
11. Given $a_{1}=2, a_{n+1}=\frac{a_{n}}{2}+\frac{1}{a_{n}}$. Prove: $1<a_{n}<\frac{3}{2}+\frac{1}{n}$. | 11. Prove the strengthened proposition $\sqrt{2}<a_{n}<\sqrt{2}+\frac{1}{n}$ using mathematical induction. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,128 |
12. If $1=a_{0} \leqslant a_{1} \leqslant \cdots \leqslant a_{n} \leqslant \cdots, b_{n}=\sum_{k=1}^{n}\left(1-\frac{a_{k-1}}{a_{k}}\right) \frac{1}{\sqrt{a_{k}}}, n \geqslant 1$. Prove: $0 \leqslant b_{n} \leqslant 2$. | 12. $\begin{aligned}\left(1-\frac{a_{k-1}}{a_{k}}\right) \frac{1}{\sqrt{a_{k}}} & =\frac{a_{k-1}}{\sqrt{a_{k}}}\left(\frac{1}{a_{k-1}}-\frac{1}{a_{k}}\right) \\ & =\left(\sqrt{\frac{a_{k-1}}{a_{k}}}+\frac{a_{k-1}}{a_{k}}\right)\left(\frac{1}{\sqrt{a_{k-1}}}-\frac{1}{\sqrt{a_{k}}}\right) \\ & \leqslant 2\left(\frac{1}{\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,129 |
14. Given $a_{1}=\frac{1}{2008}, a_{n}^{2}-2 a_{n}+2 a_{n-1}=0$. Prove:
(1) $0<a_{n} \leqslant \frac{1}{2}(n=1,2, \cdots, m-1)$;
(2) $\sum_{i=1}^{m} \frac{1}{2-a_{i}}<2008$ | 14. (1) $a_{n}^{2}-2 a_{n}+1=1-2 a_{n-1},\left(a_{n}-1\right)^{2}=1-2 a_{n-1} \geqslant 0, a_{n}^{2}=$ $2\left(a_{n}-a_{n-1}\right) \geqslant 0$.
(2) $\frac{1}{a_{n-1}}=\frac{2}{a_{n}\left(2-a_{n}\right)}=\frac{1}{a_{n}}+\frac{1}{2-a_{n}}$, thus the left side $=\frac{1}{2-a_{1}}+\frac{1}{a_{1}}-\frac{1}{a_{m}}<$ $\frac... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,131 |
15. $\left\{a_{n}\right\}$ satisfies $a_{n+1}=\frac{1}{2} a_{n}^{2}-a_{n}+2(n \geqslant 1)$. Prove:
(1) If $a_{1}=4$, then $a_{n+1} \geqslant\left(\frac{3}{2}\right)^{n} \cdot a_{n}$;
(2) If $a_{1}=1$, then when $n \geqslant 5$, $\sum_{k=1}^{n} \frac{1}{a_{k}}<n-1$. | 15. (1) First prove $a_{n+1} \geqslant a_{n}$, then prove that for $n \geqslant 2$, $a_{n+1}>2 a_{n}$, and then use mathematical induction to prove $\frac{1}{2} a_{n}>\left(\frac{3}{2}\right)^{n}+1$. Therefore, $\frac{a_{n+1}}{a_{n}}=\frac{1}{2} a_{n}+\frac{2}{a_{n}}-1>\frac{a_{n}}{2}-1>$ $\left(\frac{3}{2}\right)^{n}$... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,132 |
16. Let $a>2, x_{1}=a, x_{n+1}=\frac{x_{n}^{2}}{2\left(x_{n}-1\right)}$. Prove:
(1) $x_{n}>2, x_{n+1}<x_{n}$;
(2) If $a \leqslant 3$, then $x_{n}<2+\frac{1}{2^{n-1}}$. | 16. (1) Use mathematical induction.
(2) $x_{n+1}-2=\frac{\left(x_{n}-2\right)^{2}}{2\left(x_{n}-1\right)}<\frac{1}{2}\left(x_{n}-2\right)$. | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,133 |
17. $y=\frac{x^{2}-x+n}{x^{2}+x+1}$ ( $n$ is a positive integer) has a minimum value of $a_{n}$, a maximum value of $b_{n}$, and $c_{n}=\frac{n}{4}\left(1+3 a_{n} b_{n}\right)$. Prove:
$$\frac{3}{2}-\frac{1}{n+1}<\sum_{k=1}^{n} \frac{1}{c_{k}}<2-\frac{1}{n}(n \geqslant 2)$$ | 17. First prove $a_{n} b_{n}=\frac{4 n-1}{3}, c_{n}=n^{2}$. | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,134 |
18. Does there exist a number $\alpha(0<\alpha<1)$, such that there is an infinite sequence of positive numbers $\left\{a_{n}\right\}$, satisfying $1+$ $a_{n+1} \leqslant\left(1+\frac{\alpha}{n}\right) a_{n} ?$ | 18. Does not exist. Proof by contradiction. Suppose such a number $\alpha$ exists, then $a_{n}\left(1+\frac{1}{n}\right)>$ $a_{n}\left(1+\frac{\alpha}{n}\right)>1+a_{n+1}$, which means $a_{n}>\frac{n}{n+1}\left(1+a_{n+1}\right)>\frac{n}{n+1}$. By mathematical induction, if there is a $k \leqslant n$ such that $a_{k}>k\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,135 |
19. Given $a_{1}=1, a_{n+1}=a_{n}+2 n, b_{1}=1, b_{n+1}=b_{n}+\frac{b_{n}^{2}}{n}$. Prove:
$$\frac{1}{2} \leqslant \sum_{k=1}^{n} \frac{1}{\sqrt{a_{k+1} b_{k}+k a_{k+1}-b_{k}-k}}<1 .$$ | 19. By Cauchy's inequality, let the original expression be $I_{n}$, then
$I_{n} \leqslant \sqrt{\sum_{k=1}^{n} \frac{1}{a_{k+1}-1} \cdot \sum_{k=1}^{n} \frac{1}{b_{k}+k}}$. And $\sum_{k=1}^{n} \frac{1}{a_{k+1}-1}=\sum_{k=1}^{n} \frac{1}{k(k+1)}<$ 1 , also $\frac{1}{b_{k+1}}=\frac{1}{b_{k}}-\frac{1}{b_{k}+k}$, hence $\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,136 |
20. Given $a_{0}=\frac{\sqrt{2}}{2}, a_{n+1}=\frac{\sqrt{2}}{2} \sqrt{1-\sqrt{1-a_{n}^{2}}}, n \geqslant 0 . b_{0}=1, b_{n+1}=$ $\frac{\sqrt{1+b_{n}^{2}}-1}{b_{n}}, n \geqslant 0$. Prove that for each $n \geqslant 0, 2^{n+2} a_{n}<\pi<2^{n+2} b_{n}$. | 20. By mathematical induction, we have $a_{n}=\sin \frac{\pi}{2^{n+2}}, b_{n}=\tan \frac{\pi}{2^{n+2}}$. | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,137 |
21. Given $a_{1}=1, a_{n+1}=\sqrt{a_{n}^{2}+\frac{1}{a_{n}}}, n \geqslant 1$. Prove: There exists a positive number $\alpha$, such that for any $n \geqslant 1, \frac{1}{2} \leqslant \frac{a_{n}}{n^{\alpha}} \leqslant 2$. (In fact, we can further study $\lim _{n \rightarrow+\infty} \frac{a_{n}}{n^{\alpha}}$ ). | 21. If the inequality to be proved holds, then we have $\frac{1}{4} n^{2 a}+\frac{1}{2 n^{\alpha}} \leqslant a_{n+1}^{2}=a_{n}^{2}+\frac{1}{a_{n}} \leqslant 4 n^{2 a}+$ $\frac{2}{n^{a}}$. Using the inductive hypothesis, we have $\frac{1}{4}(n+1)^{2 a} \leqslant \frac{1}{4} n^{2 a}+\frac{1}{2 n^{a}}, 4 n^{2 a}+\frac{2}{... | \alpha=\frac{1}{3} | Algebra | proof | Yes | Yes | inequalities | false | 736,138 |
Example 3 Let $S_{n}=1+2+3+\cdots+n, n \in \mathbf{N}$, find the maximum value of $f(n)=\frac{S_{n}}{(n+32) S_{n+1}}$. | \begin{aligned} f(n) & =\frac{S_{n}}{(n+32) S_{n+1}}=\frac{n}{(n+32)(n+2)}=\frac{n}{n^{2}+34 n+64} \\ & =\frac{1}{n+34+\frac{64}{n}}=\frac{1}{\left(\sqrt{n}-\frac{8}{\sqrt{n}}\right)^{2}+50} \leqslant \frac{1}{50} .\end{aligned} | \frac{1}{50} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,140 |
23. Given $S_{n}=a_{1}+a_{2}+\cdots+a_{n}, a_{n}+2 S_{n} S_{n-1}=0(n \geqslant 2), a_{1}=\frac{1}{2}$, $b_{n}=2(1-n) a_{n}(n \geqslant 2)$. Prove: $\sum_{i=2}^{n} b_{i}^{2}<1$. | 23. Prove $\frac{1}{S_{n}}-\frac{1}{S_{n-1}}=2, b_{n}=\frac{1}{n}$. | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,141 |
25. Given $r_{1}=2, r_{n}=r_{n-1}^{2}-r_{n-1}+1$. If positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $\sum_{i=1}^{n} \frac{1}{a_{i}}<1$, prove: $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant \sum_{i=1}^{n} \frac{1}{r_{i}}$. | 25. It is known that $r_{n+1}=r_{1} r_{2} \cdots r_{n}+1$ and $\sum_{i=1}^{n} \frac{1}{r_{i}}=1-\frac{1}{r_{1} r_{2} \cdots \cdots r_{n}}$. Suppose the proposition holds for $n \leqslant k$, and let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{k+1}$. Then we have
$$\left\{\begin{array}{l}
\frac{1}{a_{1}} \leqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,143 |
27. Given $f(x)=\frac{1}{\sqrt{x^{2}-4}}(x>2), f^{-1}(x)$ is its inverse function.
(1) If the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=\frac{1}{f^{-1}\left(a_{n}\right)}$, find $\left\{a_{n}\right\}$;
(2) Let $S_{n}=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}, b_{n}=S_{2 n+1}-S_{n}$, does there exist a larg... | 27. (1) First prove $\frac{1}{a_{n+1}^{2}}-\frac{1}{a_{n}^{2}}=4$, then we can get $a_{n}=\frac{1}{\sqrt{4 n-3}}$.
(2) Prove $b_{n+1}-b_{n}<0$, to get $p=3$. | p=3 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,145 |
28. The sequence of positive real numbers $\left\{x_{n}\right\}$ satisfies $x_{n+1}=\sum_{k=1}^{n} x_{k}$. Prove:
$$\sqrt{x_{n+1} \sum_{i=1}^{n}\left(x_{n+1}-x_{i}\right)} \geqslant \sum_{i=1}^{n} \sqrt{x_{i}\left(x_{n+1}-x_{i}\right)} .$$ | 28. Regarding $x_{n+1}$ in the left expression as $\sum_{k=1}^{n} x_{k}$, the Cauchy-Schwarz inequality can be used to prove it. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,146 |
29. If $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n}$ satisfy
(1) $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}$;
(2) $0<a_{1}=a_{2}, a_{i}+a_{i+1}=a_{i+2}(i=1,2, \cdots, n-2)$;
(3) $0<b_{1} \leqslant b_{2}, b_{i}+b_{i+1}=b_{i+2}(i=1,2, \cdots, n-2)$.
Prove: $a_{n-1}+a_{n} \leqslant b_{n-1}+b_{n}$. | 29. First prove $a_{i} \leqslant b_{i}(i \geqslant 1)$ using mathematical induction, let $F_{1}=1, F_{2}=2, F_{i+2}=$ $F_{i+1}+F_{i}$, construct $a_{1}^{\prime}=a_{1}+a_{2}, a_{2}^{\prime}=a_{3}, \cdots, a_{i}^{\prime}=a_{i+1}+F_{i-2} a_{1}(i=3,4, \cdots$, k) $; b_{1}^{\prime}=b_{1}+b_{2}, b_{2}^{\prime}=b_{3}, \cdots,... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,147 |
30. A non-negative sequence $\left\{a_{i}\right\}$ satisfies that for any positive integers $m, n$, we have $a_{m+n} \leqslant a_{m}+a_{n}$. Prove: for any $n \geqslant m$, we have $a_{n} \leqslant m a_{1}+\left(\frac{n}{m}-1\right) a_{m}$. | 30. It is known that $0 \leqslant a_{n} \leqslant n a_{1}$, so the inequality holds when $n=m$. Now assume $n>m, \frac{a_{n}}{n}-\frac{a_{m}}{m}$ $\leqslant \frac{a_{n-m}+a_{m}}{n}-\frac{a_{m}}{m}=\frac{n-m}{n}\left(\frac{a_{n-m}}{n-m}-\frac{a_{m}}{m}\right) \leqslant \cdots \leqslant \frac{s}{n}\left(\frac{a_{s}}{s}-\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,148 |
31. Find all real numbers $a_{0}$ such that the sequence $\left\{a_{n}\right\}$, satisfying $a_{n+1}=2^{n}-3 a_{n}$, is increasing. | 31. $a_{n}=2^{n-1}-3 a_{n-1}=2^{n-1}-3 \cdot 2^{n-2}+9 a_{n-2}=\cdots=2^{n-1}-3 \cdot 2^{n-2}$ $+\cdots+(-1)^{n-1} \cdot 3^{n-1}+(-1)^{n} \cdot 3^{n} a_{0}$, thus $a_{n}=\frac{2^{n}-(-3)^{n}}{5}+(-3)^{n} \cdot$ $a_{0}$. Let $d_{n}=a_{n}-a_{n-1}=\frac{2^{n}-(-3)^{n}}{5}+(-3)^{n} a_{0}-\frac{2^{n-1}-(-3)^{n-1}}{5}-$ $(-3... | a_{0}=\frac{1}{5} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,149 |
32. Let $\left\{a_{n}\right\}$ be an infinite sequence of positive numbers. If for any $i, a_{i} \leqslant m$, and for any positive integers $i, j(i \neq j),\left|a_{i}-a_{j}\right| \geqslant \frac{1}{i+j}$. Prove: $m \geqslant 1$. | 32. For $n \geqslant 4$, let $k_{1}, k_{2}, \cdots, k_{n}$ be a permutation of $1,2, \cdots, n$, and satisfy $0 < a_{k_{1}} < a_{k_{2}} < \cdots < a_{k_{n}} \leqslant m$. Thus, $a_{k_{i}} - a_{k_{i-1}} \geqslant \frac{1}{k_{i} + k_{i-1}} (i=2,3, \cdots, n)$. By the Cauchy-Schwarz inequality, $m \geqslant a_{k_{n}} - a_... | m \geqslant 1 | Inequalities | proof | Yes | Yes | inequalities | false | 736,150 |
33. Let $\sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|=2001, y_{k}=\frac{1}{k} \sum_{i=1}^{k} x_{i}, k=1,2, \cdots, 2001$. Find $\max \sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right|$ | 33. Let $a_{0}=x_{1}, a_{k}=x_{k+1}-x_{k}, k=1,2, \cdots, 2000$, then $x_{1}=a_{0}, x_{k} = \sum_{i=0}^{k-1} a_{i}, k=2,3, \cdots, 2001$. Thus, the condition becomes $\sum_{k=1}^{2000}\left|a_{k}\right|=2001$, and $y_{k} = \frac{1}{k}\left(a_{0}+\left(a_{0}+a_{1}\right)+\cdots+\left(a_{0}+\cdots+a_{k-1}\right)\right) =... | 2000 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,152 |
34. Construct an infinite bounded sequence $x_{0}, x_{1}, \cdots, x_{n}, \cdots$, such that for every pair of non-negative integers $i \neq j, \left|x_{i}-x_{j}\right| \geqslant \frac{1}{|i-j|}$. | 34. $x_{n}$ can be $4(\sqrt{2} n-[\sqrt{2} n])$. It is only necessary to prove that for positive integers $a, b$, if $a\frac{1}{4 b}$. | proof | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,153 |
35. Let the function $f: \mathbf{N}^{*} \rightarrow \mathbf{R}$ satisfy that for any $m, n \in \mathbf{N}^{*}$, we have $\mid f(m)+f(n)-f(m+n) \mid \leqslant m n$. Prove that: for any $n \in \mathbf{N}^{*}$, we have $\left|f(n)-\sum_{k=1}^{n} \frac{f(k)}{k}\right| \leqslant \frac{n(n-1)}{4}$. | 35. Let $a_{n}=f(n)+\frac{n^{2}}{2}$, then for any positive integers $m, n, a_{m}+a_{n} \leqslant a_{m+n}$. Using mathematical induction, we can prove: $a_{n} \geqslant \sum_{k=1}^{n} \frac{a_{k}}{k}$. Construct $b_{n}=f(n)-\frac{n^{2}}{2}$, we know that for any $m, n, b_{m} +b_{n} \geqslant b_{m+n}$. By mathematical i... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,154 |
Example 1 (Continuous Convex Function)
Prove: If $\varphi$ is continuous, then for any non-negative real numbers $q_{1}, q_{2}, \cdots, q_{n}, q_{1}+q_{2}+\cdots+q_{n}=1$,
$$\varphi\left(\sum q_{i} x_{i}\right) \leqslant \sum q_{i} \varphi\left(x_{i}\right)$$
This is equivalent to equation (1). | Prove that if $\varphi(x)$ satisfies equation (1), then we have
$$\begin{aligned}
4 \varphi\left(\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}\right) & \leqslant 2 \varphi\left(\frac{x_{1}+x_{2}}{2}\right)+2 \varphi\left(\frac{x_{3}+x_{4}}{2}\right) \\
& \leqslant \varphi\left(x_{1}\right)+\varphi\left(x_{2}\right)+\varphi\left(x_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,155 |
Example 1 Let $x_{ij} (i=1,2, \cdots, m, j=1,2, \cdots n)$ be positive real numbers, and positive real numbers $\omega_{1}$, $\omega_{2}, \cdots, \omega_{n}$ satisfy $\omega_{1}+\omega_{2}+\cdots+\omega_{n}=1$, prove
$$\prod_{j=1}^{n}\left(\sum_{i=1}^{m} x_{ij}\right)^{w_{j}} \geqslant \sum_{i=1}^{m}\left(\prod_{j=1}^{... | Proof: By the homogeneity of equation (1), we can set $x_{1 j}+x_{2 j}+\cdots+x_{m j}=1$, where $j \in\{1,2, \cdots, n\}$. Then we only need to prove
that is $\square$
$$\begin{array}{c}
\prod_{j=1}^{n} 1^{\omega_{j}} \geqslant \sum_{i=1}^{m} \prod_{j=1}^{n} x_{i j}^{\omega_{j}} \\
1 \geqslant \sum_{i=1}^{m} \prod_{j=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,157 |
Example 2 Proof: For positive numbers $x, y, z$, we have
$$\sum_{\mathrm{cyc}}(y+z) \sqrt{\frac{y z}{(z+x)(x+y)}} \geqslant \sum_{\mathrm{cyc}} x .$$ | $$\begin{aligned}
& \text { Prove } \quad \text { by Hölder's inequality we know } \\
& \left(\sum_{\mathrm{cyc}}(y+z) \sqrt{\frac{y z}{(z+x)(x+y)}}\right)^{2}\left(\sum_{\mathrm{cyc}} y^{2} z^{2}(y+z)(z+x)(z+y)\right) \\
\geqslant & \left(\sum_{\mathrm{cyc}} y z(y+z)\right)^{3},
\end{aligned}$$
Therefore, to prove in... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,158 |
Example 2 (Mitrinović-Djoković Inequality)
In 1970, D. S. Mitrinović and D. Z. Djoković published the following result in *Analytic Inequalities* (translated by Zhao Hanbin, Guangxi People's Publishing House, 1986 edition):
If \( x_{k} > 0 \ (k=1, \cdots, n), x_{1} + x_{2} + \cdots + x_{n} = 1 \), and \( a > 0 \), the... | Prove by the Arithmetic Mean-Geometric Mean Inequality,
$$\left(\frac{1}{n} \sum_{k=1}^{n}\left(x_{k}+\frac{1}{x_{k}}\right)^{a}\right)^{\frac{1}{a}} \geqslant\left(\prod_{k=1}^{n}\left(x_{k}+\frac{1}{x_{k}}\right)\right)^{\frac{1}{n}}$$
Therefore, to prove inequality (28), it suffices to prove:
$$\prod_{k=1}^{n}\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,160 |
Example 1 (Beckenbach Inequality) Let $x_{i}, y_{i}>0$, prove: when $1 \leqslant t \leqslant 2$, we have
$$\frac{\left(M_{t}(x+y)\right)^{t}}{\left(M_{t-1}(x+y)\right)^{t-1}} \leqslant \frac{\left(M_{t}(x)\right)^{t}}{\left(M_{t-1}(x)\right)^{t-1}}+\frac{\left(M_{t}(y)\right)^{t}}{\left(M_{t-1}(y)\right)^{t-1}}$$
When... | We will prove a more general result: If $t \geqslant 1, p \geqslant 1 \geqslant r$, then
$$\frac{\left(M_{p}(x+y)\right)^{t}}{\left(M_{r}(x+y)\right)^{t-1}} \leqslant \frac{\left(M_{p}(x)\right)^{t}}{\left(M_{r}(x)\right)^{t-1}}+\frac{\left(M_{p}(y)\right)^{t}}{\left(M_{r}(y)\right)^{t-1}},$$
When $0 \leqslant t \leqs... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,161 |
Example 1 If $x_{i} \geqslant 0, \alpha<0<\beta$. Prove:
$$\frac{M_{a}(x)}{M_{a}(x+1)} \leqslant \frac{G(x)}{G(x+1)} \leqslant \frac{M_{\beta}(x)}{M_{\beta}(x+1)},$$
equality holds if and only if $x_{1}=x_{2}=\cdots=x_{n}$. | Prove by the Suspended Mean Monotonicity Theorem and Chebyshev's Inequality,
$$\begin{aligned}
& \left(\sum_{i=1}^{n}\left(x_{i}+1\right)^{\beta}\right)^{\frac{1}{\beta}}\left(\prod_{i=1}^{n}\left(\frac{x_{i}}{x_{i}+1}\right)\right)^{\frac{1}{n}} \\
\leqslant & \left(\sum_{i=1}^{n}\left(x_{i}+1\right)^{\beta}\right)^{\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,162 |
1. Use the properties of continuous convex functions to prove the arithmetic mean-geometric mean inequality. | 1. When $n=2$, it is easy to get
$$\frac{x_{1}+x_{2}}{2} \geqslant \sqrt{x_{1} x_{2}}$$
or
$$\ln \frac{x_{1}+x_{2}}{2} \geqslant \frac{\ln x_{1}+\ln x_{2}}{2}$$
In other words, $\ln x$ is a concave function. By using Jensen's inequality, we get: for $x_{i}>0, p_{i}>0$, we have
$$\ln \frac{p_{1} x_{1}+p_{2} x_{2}+\cdo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,163 |
2. If $x_{i}, y_{i} \geqslant 0, p$ and $q$ are positive numbers satisfying $\frac{1}{p}+\frac{1}{q} \geqslant 1$. Prove the second formula (2) (i.e., Hölder's inequality) holds. | 2. Now assume $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}>1$. Since
$$\frac{\sum_{i=1}^{n} a_{i}^{r}}{\left(\sum_{i=1}^{n} a_{i}\right)^{r}}=\sum_{j=1}^{n} \frac{a_{j}^{r}}{\left(\sum_{i=1}^{n} a_{i}\right)^{r}}=\sum_{j=1}^{n}\left(\frac{a_{j}}{\sum_{i=1}^{n} a_{i}}\right)^{r} \geqslant \sum_{j=1}^{n} \frac{a_{j}}{\sum_{i=1}^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,164 |
3. Prove: If $x_{i} \geqslant 0$, then when $\alpha \leqslant 1 \leqslant \beta$, the inequality
$$\frac{M_{a}(x)}{M_{a}(x+1)} \leqslant \frac{M_{\beta}(x)}{M_{\beta}(x+1)} .$$
holds. | 3. Using Minkowski's inequality, the monotonicity of power means, and the increasing property of $f(x)=\frac{x}{x+1}$, we get
$$\frac{M_{a}(x)}{M_{a}(x+1)} \leqslant \frac{M_{\alpha}(x)}{M_{a}(x)+1} \leqslant \frac{M_{\beta}(x)}{M_{\beta}(x)+1} \leqslant \frac{M_{\beta}(x)}{M_{\beta}(x+1)} .$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,165 |
4. Let non-negative real numbers $x, y, z$ satisfy $x+y+z=1$. Prove:
$$\sqrt{9-32 y z}+\sqrt{9-32 z x}+\sqrt{9-32 x y} \geqslant 7 .$$ | 4. By Hölder's inequality, we have
$$\begin{aligned}
& \sum_{\text {cyc }} \frac{\left(3 x^{2}+2\left(y^{2}+z^{2}\right)+7 x(y+z)\right)^{3}}{9(x+y+z)^{2}-32 y z}\left(\sum_{\text {cyc }} \sqrt{9(x+y+z)^{2}-32 y z}\right)^{2} \\
\geqslant & 343(x+y+z)^{6},
\end{aligned}$$
Therefore, it suffices to prove
$$7(x+y+z)^{4}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,166 |
Example 1 Positive real numbers $a, b, c$ satisfy $a+b+c=3$. Prove:
$$\frac{a \sqrt{a}}{a+b}+\frac{b \sqrt{b}}{b+c}+\frac{c \sqrt{c}}{c+a} \geqslant \frac{a b+b c+c a}{2}$$ | Proof: By Hölder's inequality, we have
$$\left(\sum_{\text {cyc }} \frac{a \sqrt{a}}{a+b}\right)^{2} \sum_{\text {cyc }}(a+b)^{2} \geqslant(a+b+c)^{3},$$
so we only need to prove
$$(a+b+c)^{3} \geqslant\left(\frac{a b+b c+c a}{2}\right)^{2} \sum_{\text {cyc }}(a+b)^{2}$$
In fact, we have
$$\begin{aligned}
(a+b+c)^{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,167 |
Example 3 Positive real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}=3$. Prove:
$$\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2-c} \geqslant 3$$ | Prove that removing the denominator and expanding yields
$$\begin{aligned}
& \sum_{\text {๗c }}(2-b)(2-c) \geqslant 3(2-a)(2-b)(2-c) \\
\Rightarrow & \sum_{\text {०c }}(4-2 b-2 c+b c) \\
& \geqslant 3(8-4 a-4 b-4 c+2 a b+2 b c+2 c a-a b c) \\
\Rightarrow & 12-4(a+b+c)+(a b+b c+c a) \\
\geqslant & 24-12(a+b+c)+6(a b+b c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,168 |
Example 3 Use the superiority relationship of Taylor series to prove the arithmetic mean-geometric mean inequality.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Proof First, it is clear that for $N=1,2, \cdots$, and $x, y \geqslant 0$, we have the relation
Thus we get
$$\begin{array}{c}
\mathrm{e}^{x y}>\frac{x^{N} y^{N}}{N!} \\
\mathrm{e}^{y \sum_{i} x_{i}}>\frac{\left(x_{1} x_{2} \cdots x_{n}\right)^{N} y^{n N}}{(N!)^{n}}
\end{array}$$
Therefore, by comparing the coefficie... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,169 |
Example 4 Positive real numbers $a, b, c$ satisfy $a+b+c=3$. Prove:
$$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant a^{2}+b^{2}+c^{2} .$$ | Proof: By the Arithmetic Mean-Geometric Mean Inequality, we have
$$\begin{array}{l}
\quad(a+b+c)^{4}\left(a^{2} b^{2}+a^{2} c^{2}+b^{2} c^{2}\right) \geqslant 81\left(a^{2}+b^{2}+c^{2}\right) a^{2} b^{2} c^{2} \\
\Leftrightarrow \sum_{\text {sym }}\left(a^{6} b^{2}+4 a^{5} b^{3}+4 a^{5} b^{2} c+3 a^{4} b^{4}+12 a^{4} b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,170 |
Example 5 Proof: For positive real numbers $a, b, c$, we have
$$\frac{a^{2}+b c}{b+3 c}+\frac{b^{2}+c a}{c+3 a}+\frac{c^{2}+a b}{a+3 b} \geqslant \frac{a+b+c}{2}$$ | $$\begin{aligned}
& \frac{a^{2}+b c}{b+3 c}+\frac{b^{2}+c a}{c+3 a}+\frac{c^{2}+a b}{a+3 b} \geqslant \frac{a+b+c}{2} \\
\Leftrightarrow & 2 \sum_{\text {cyc }}\left(a^{2}+b c\right)\left(3 a^{2}+9 a b+a c+3 b c\right) \\
\geqslant & \sum_{\text {cyc }} a \sum_{\text {cyc }}\left(3 a^{2} b+9 a^{2} c+\frac{28}{3} a b c\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,171 |
Example 6 Proof: For non-negative real numbers $a, b, c$, we have
$$\frac{a^{3}+b^{3}}{(a+b)^{3}}+\frac{b^{3}+c^{3}}{(b+c)^{3}}+\frac{c^{3}+a^{3}}{(c+a)^{3}}+\frac{27}{4} \cdot \frac{a b c}{a^{3}+b^{3}+c^{3}} \leqslant 3 .$$ | Proof
$$\begin{aligned}
& \sum_{\text {cyc }} \frac{a^{3}+b^{3}}{(a+b)^{3}}+\frac{27 a b c}{4\left(a^{3}+b^{3}+c^{3}\right)} \leqslant 3 \\
\Leftrightarrow & \sum_{\text {cyc }}\left(1-\frac{a^{2}-a b+b^{2}}{(a+b)^{2}}\right) \geqslant \frac{27 a b c}{4\left(a^{3}+b^{3}+c^{3}\right)} \\
\Leftrightarrow & \sum_{\text {c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,172 |
Example 9 Positive real numbers $a, b, c$ satisfy $b c+c a+a b=1$. Prove:
$$\sum_{\text {cyc }} \sqrt{a^{3}+a} \geqslant 2 \sqrt{a+b+c} .$$ | Proof
$$\begin{aligned}
& \sum_{\text {cyc }} \sqrt{a^{3}+a} \geqslant 2 \sqrt{a+b}+c \\
\Leftrightarrow & \sum_{\text {cyc }} \sqrt{a^{3}+a^{2} b+a^{2} c+a b c} \geqslant 2 \sqrt{(a+b+c)(a b+a c+b c)} \\
\Leftrightarrow & \sum_{\text {cyc }}\left(a^{3}+a^{2} b+a^{2} c+a b c\right) \\
& +2 \sum_{\text {cyc }} \sqrt{\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,173 |
Example 11 Proof: For non-negative real numbers $a, b, c$, we have
$$\sum_{\text {cyc }} \sqrt{2\left(a^{2}+b^{2}\right)} \geqslant \sqrt[3]{9 \sum_{\mathrm{cyc}}(a+b)^{3}} .$$ | $$\begin{array}{l}
\sum_{\text {cyc }} \sqrt{2\left(a^{2}+b^{2}\right)}-\sqrt[3]{9 \sum_{\text {cyc }}(a+b)^{3}} \\
=\sum_{\mathrm{cyc}}\left(\sqrt{2\left(a^{2}+b^{2}\right)}-a-b\right)-\left(\sqrt[3]{9 \sum_{\mathrm{cyc}}(a+b)^{3}}-2(a+b+c)\right) \\
=\sum_{\mathrm{cyc}} \frac{(a-b)^{2}}{a+b+\sqrt{2\left(a^{2}+b^{2}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,174 |
1. Real numbers $x, y, z$ satisfy $x y + y z + z x = -1$. Prove: $x^{2} + 5 y^{2} + 8 z^{2} \geqslant 4$. | \begin{array}{l}\text { 1. } 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{array} | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,175 |
Example 14 Proof: For positive real numbers $a, b, c, d$, we have
$$\begin{aligned}
& (a+b)(b+c)(c+d)(d+a)(1+\sqrt[4]{a b c d})^{4} \\
\geqslant & 16 a b c d(1+a)(1+b)(1+c)(1+d)
\end{aligned}$$ | $$\begin{aligned}
& (a+b)(b+c)(c+d)(d+a)(1+\sqrt[4]{a b c d})^{4} \\
\geqslant & 16 a b c d(1+a)(1+b)(1+c)(1+d) \\
\Leftrightarrow & (a+b)(b+c)(c+d)(d+a)-16 a b c d \\
& +4 \sqrt[4]{a b c d}((a+b)(b+c)(c+d)(d+a) \\
& \left.-4 \sqrt[4]{a^{3} b^{3} c^{3} d^{3}}(a+b+c+d)\right) \\
& +2 \sqrt{a b c d}(3(a+b)(b+c)(c+d)(d+a)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,176 |
Example 15 Proof: For non-negative real numbers $a, b, c, d$, we have
$$\begin{aligned}
& (a+b)(b+c)(c+a)(1+a)(1+b)(1+c) \\
\geqslant & a b c(2+a+b)(2+b+c)(2+c+a) .
\end{aligned}$$ | Prove that the original inequality is equivalent to
$$\frac{(a+b)(b+c)(c+a)}{a b c} \geqslant \frac{(a+b+2)(b+c+2)(c+a+2)}{(a+1)(b+1)(c+1)},$$
i.e.,
$$\begin{array}{c}
\frac{(a+b)(b+c)(c+a)}{a b c}-8 \geqslant \frac{(a+b+2)(b+c+2)(c+a+2)}{(a+1)(b+1)(c+1)}-8 \\
\sum_{\text {cyc }}\left(\frac{(a-b)^{2}}{a b}-\frac{(a-b)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,177 |
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