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20. Let $x, y, z$ be positive real numbers, prove that: $\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}>2 \sqrt[3]{x^{3}+y^{3}+z^{3}} \cdot(2008$ China National Training Team Problem) | 20.
$$\begin{array}{l}
\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}>2 \sqrt[3]{x^{3}+y^{3}+z^{3}} \Leftrightarrow\left(\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}\right)^{3}>8\left(x^{3}+y^{3}+z^{3}\right) \Leftrightarrow \\
\left(\frac{x y}{z}\right)^{3}+\left(\frac{y z}{x}\right)^{3}+\left(\frac{z x}{y}\right)^{3}+6 x y z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,066 |
22. Let $a, b, c$ be positive numbers, and $a+b+c=3$, prove: $\frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+$ $\frac{1}{2+c^{2}+a^{2}} \leqslant \frac{3}{4}$ (2009 Iran National Training Team Problem) | 22. Let $a \geqslant b \geqslant c > 0$, and set $f(a, b, c) = \frac{1}{2 + a^2 + b^2} + \frac{1}{2 + b^2 + c^2} + \frac{1}{2 + c^2 + a^2}$. We will prove that
$$f(a, b, c) \leqslant f\left(a, \frac{b+c}{2}, \frac{b+c}{2}\right)$$
That is,
$$\begin{array}{l}
\frac{2}{2 + a^2 + \left(\frac{b+c}{2}\right)^2} + \frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,068 |
23. Let $x, y, z$ be positive real numbers, and $x y z=1$, prove:
(1) $(1+x+y)^{2}+(1+y+z)^{2}+(1+z+x)^{2} \geqslant 27$; | 23. (1) By Cauchy-Schwarz inequality and AM-GM inequality, we have
$$\begin{array}{l}
3\left[(1+x+y)^{2}+(1+y+z)^{2}+(1+z+x)^{2}\right] \geqslant \\
{[(1+x+y)+(1+y+z)+(1+z+x)]^{2}=} \\
{[3+2(x+y+z)]^{2} \geqslant} \\
{\left[3+2 \times 3 \sqrt[3]{(x y z)^{2}}\right]^{2}=81}
\end{array}$$
Therefore,
$$(1+x+y)^{2}+(1+y+z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,069 |
2. Prove: For any numbers $a, b, c$ greater than 1, $2\left(\frac{\log _{b} a}{a+b}+\frac{\log _{c} b}{b+c}+\frac{\log _{a} c}{c+a}\right) \geqslant \frac{9}{a+b+c}$. (1976 Yugoslav Mathematical Olympiad Problem) | 2. Noting that $\log _{b} a, \log _{c} b, \log _{a} c$ are all positive, and $\log _{b} a \cdot \log _{c} b \cdot \log _{a} c=1$, we have by the AM-GM inequality:
$$\begin{array}{c}
\frac{\log _{b} a}{a+b}+\frac{\log _{c} b}{b+c}+\frac{\log _{a} c}{c+a} \geqslant 3 \sqrt[3]{\frac{\log _{b} a}{a+b} \cdot \frac{\log _{c}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,071 |
4. Prove: For non-negative real numbers $a, b, c$, we have $\frac{(a+b+c)^{2}}{3} \geqslant a \sqrt{b c}+b \sqrt{c a}+c \sqrt{a b}$. (25th All-Soviet Union Mathematical Olympiad Problem) | \begin{array}{l}\text { 4. } a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a \text {, add } 2(a b+b c+c a) \text { to both sides, we get } \\ \quad(a+b+c)^{2} \geqslant 3(a b-b c+c a)- \\ \begin{aligned} \frac{(a+b+c)^{2}}{3} \geqslant & a b+b c+c a=\frac{1}{2} a(b+c)+\frac{1}{2} b(c+a)+\frac{1}{2} c(a+b) \geqslant \\ & a \sqr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,073 |
5. If $a, b, c$ are positive real numbers, prove that $a^{3}+b^{3}+c^{3} \geqslant a^{2} b+b^{2} c+c^{2} a$. (1983 British Mathematical Olympiad) | 5. By the AM-GM inequality, we have $a^{2} b \leqslant \frac{a^{3}+a^{3}+b^{3}}{3}, b^{2} c \leqslant \frac{b^{3}+b^{3}+c^{3}}{3}, c^{2} a \leqslant$ $\frac{c^{3}+c^{3}+a^{3}}{3}$, adding these inequalities yields $a^{3}+b^{3}+c^{3} \geqslant a^{2} b+b^{2} c+c^{2} a$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,074 |
6. (1) If $0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$, prove: $\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \cos ^{2} \beta \sin ^{2} \beta} \geqslant 9$.
(1978 National High School Mathematics Competition Problem)
(2) If $0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$, prove:
$$\frac{5}{\cos ^{2} \alpha}+\... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Inequalities | proof | Yes | Yes | inequalities | false | 733,075 |
Example 2 Prove that if two given positive numbers $p \leqslant q$, then for any $\alpha, \beta, \gamma, \delta, \varepsilon \in[p, q]$, we have
$$(\alpha+\beta+\gamma+\delta+\varepsilon)\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}+\frac{1}{\varepsilon}\right) \leqslant 25+6\left(\sqrt{\frac... | To prove that for any positive numbers $u, v$, considering the function $f(x)=(u+x)\left(v+\frac{1}{x}\right), 0<p \leqslant x \leqslant q$, it can be shown that for any $x \in[p, q]$,
$$f(x) \leqslant \max \{f(p), f(q)\}$$
In fact, without loss of generality, assume $p<q$, and let $\lambda=\frac{q-x}{q-p}$, then $0 \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,077 |
3. Let the sequence of non-negative numbers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfy the condition: $a_{m+n} \leqslant a_{n}+a_{m}, m, n \in \mathbf{N}^{*}$, prove that for any $n \geqslant m$ we have $a_{n} \leqslant m a_{1}+\left(\frac{n}{m}-1\right) a_{m}$. (1997 China Mathematical Olympiad Problem) | 3. From the given, we know that $a_{n} \leqslant n a_{1}$. Fix $m$, and use the second mathematical induction on $n$.
When $n=1$, the original inequality is equivalent to $\left(1-\frac{1}{m}\right) a_{m} \leqslant(m-1) a_{1} \Leftrightarrow(m-1) a_{m} \leqslant (m-1) m a_{1}$.
Assume that for $1 \leqslant n \leqslan... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,080 |
4. Prove that for any $\alpha \leqslant 1$ and any real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $1 \geqslant x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}>0$, we have $\left(1+x_{1}+x_{2}+\cdots+x_{n}\right)^{\alpha} \leqslant 1+1^{\alpha-1} x_{1}^{\alpha}+2^{\alpha-1} x_{2}^{\alpha}+\cdots+n^{\alpha-... | 4. When $n=\mathrm{F}$, $\left(1+x_{1}\right)^{\alpha} \leqslant 1+x_{1}^{\alpha}$ (this can be obtained by using the fact that $f(x)=(1+x)^{\alpha}-\left(1+x^{\alpha}\right)$ is a decreasing function on $(0,1)$), the inequality holds. Assume the inequality holds for $n$, we will prove it also holds for $n+1$. We have
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,081 |
5. Given that $a, b$ are positive real numbers, and $\frac{1}{a}+\frac{1}{b}=1$, prove: For every $n \in \mathbf{N}^{*}, (a+b)^{n}-$ $a^{n}-b^{n} \geqslant 2^{n+1}-2^{n}$. (1988 National High School Mathematics Competition Problem) | 5. (1) When $n=1$, the left side $=0=$ right side, the proposition holds.
(2) Assume when $n=k$, the inequality holds, i.e., $(a+b)^{k}-a^{k}-b^{k} \geqslant 2^{2 k}-2^{k+1}$.
Thus, when $n=k+1$, the left side $=(a+b)^{k+1}-a^{k+1}-b^{k+1}=(a+b)\left((a+b)^{k}-\right.$ $a^{k}-b^{k}+a^{k} b+a b^{k^{2}}$
Since $\frac{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,082 |
6. Let $1<x_{1}<2$, for $n=1,2,3, \cdots$, define $x_{n+1}=1+x_{n}-\frac{1}{2} x_{n}^{2}$, prove that for $n \geqslant 3$, we have $1 x_{n}-\sqrt{2} \left\lvert\,<\frac{1}{2 n}\right.$ (1985 Canadian Mathematical Olympiad) | 6. From $x_{n+1}=1+x_{n}-\frac{1}{2} x_{n}^{2}$ and $1<x_{1}<2$, we can get $1<x_{2}<\frac{3}{2}, \frac{3-\left(\frac{1}{2}\right)^{2}}{2}<x_{3}<\frac{3}{2}$, thus $\frac{\Gamma 1}{8}-\sqrt{2}<x_{3}-\sqrt{2}<\frac{3}{2}-\sqrt{2}, \left| x_{3}-\sqrt{2} \right|<\frac{1}{2^{3}}$. That is, the inequality holds for $n=3$. $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,083 |
7. If $x$ is a positive real number, prove that
$$[-n x] \geqslant \frac{[x]}{1}+\frac{[2 x]}{2}+\frac{[3 x]}{3}+\cdots+\frac{[n x]}{n}$$
where $[t]$ denotes the greatest integer not exceeding $t$. (10th USA Mathematical Olympiad Problem). | 7. By mathematical induction. When $n=1,2$, equation (1) obviously holds.
Assume equation (1) holds for $n \leqslant k-1$.
Let $x_{r}=\frac{[x]}{1}+\frac{[2 x]}{2}+\frac{[3 x]}{3}+\cdots+\frac{[i x]}{i}, i=1,2, \cdots, k$. Then we have
$$\begin{array}{l}
k x_{k}=k x_{k-1}+k x_{k-1}=(k-1) x_{k-1}+x_{k-1}+[k x] \\
(k-1)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,084 |
9. Let $x>0, n \in \mathrm{N}$, prove: $\frac{x^{n}\left(x^{n+1}+1\right)}{x^{n}+1} \leqslant\left(\frac{x+1}{2}\right)^{2 n+1}$. | 9. (1) When $n=1$, $\left(\frac{x+1}{2}\right)^{3} \geqslant \frac{x\left(x^{2}+1\right)}{x+1} \Leftrightarrow (x+1)^{4} \geqslant 8 x\left(x^{2}+1\right) \Leftrightarrow (x-1)^{4} \geqslant 0$, the inequality holds;
(2) Assume that when $n=k$, the inequality holds, i.e., $\left(\frac{x+1}{2}\right)^{2 k+1} \geqslant \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,086 |
Example 12 Let $x_{1}, x_{2}, \cdots, x_{n} \geqslant 0$ satisfy $\sum_{i=1}^{n} \frac{1}{1+x_{i}^{2}}=1$, prove that: $\sum_{i=1}^{n} \frac{x_{i}}{n-1+x^{2}} \leq 1$ | Prove that for $y_{i}=\frac{1}{1+x_{i}}(i=1,2, \cdots, n)$, thus $x_{i}=\frac{1}{y_{i}}-1$, and $\sum_{i=1}^{n} y_{i}=1$, we have
$$\begin{array}{l}
\sum_{i=1}^{n} \frac{x_{i}}{n-1+x_{i}^{2}} \sum_{i=1}^{n-1+\left(\frac{1}{y_{i}}-1\right)^{2}}=\sum_{i=1}^{n} \frac{y_{i}-y_{i}^{2}}{n y_{i}^{2}-2 y_{i}+1}= \\
\frac{1}{n}... | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 733,089 |
12. Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{2 n-1} \geqslant 0$, prove that $: a_{1}^{2}-a_{2}^{2}+a_{3}^{2}-a_{4}^{2}+\cdots+a_{2 n-1}^{2} \geqslant$ $\left(a_{1}-a_{2}+a_{3}-a_{4} \cdots+\sigma_{2 n-1}\right)^{2}$. (7th World Cities Invitational Mathematics Competition Problem) | 12. When $n=1$, the proposition is obviously true. When $n=2$, since $a_{1}^{2}-a_{2}^{2}+a_{2}^{z}-\left(a_{1}-a_{2}+\right.$ $\left.a_{3}\right)^{2}=\left(a_{1}-a_{2}\right)\left(a_{1}+a_{2}-a_{1}+a_{2}-2 a_{3}\right)=2\left(a_{1}-a_{2}\right)\left(a_{2}-a_{3}\right) \geqslant 0$, the proposition holds.
Assume that ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,090 |
14. Let $a$ be a positive real number, prove that for any $n \in \mathbf{N}^{*}$, we have $\frac{1+a^{2}+a^{4}+\cdots+a^{2 n}}{a+a^{3}+a^{5}+\cdots+a^{2 n-1}} \geqslant$
$$\frac{n+1}{n}$$ | 14. When $n=1$, by $\frac{1-a^{2}}{a} \geqslant \frac{1}{a}+a \geqslant 2 \sqrt{\frac{1}{a} \cdot a}=2$ we know the inequality holds. Suppose the proposition holds for $n$, i.e., $\frac{1+a^{2}+a^{4}+\cdots+a^{2 n}}{a+a^{3}+a^{5}+\cdots+a^{2 n-1}} \geqslant \frac{n+1}{n}$.
Then $\frac{a+a^{3}+a^{5}+\cdots+a^{2 n-1}}{1+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,092 |
15. (1) Let $x_{1}, x_{2}, \cdots, x_{n}, y_{1}, y_{2}, \cdots, y_{n} \in \mathbf{R}^{+}$, satisfying:
( i ) $0 < x_{1} y_{1} < x_{2} y_{2} < \cdots < x_{n} y_{n}$; ( ii ) $x_{1} + x_{2} + \cdots + x_{k} \geqslant y_{1} + y_{2} + \cdots + y_{k}, k=1,2, \cdots, n$.
Prove: $\frac{1}{x_{1}} + \frac{1}{x_{2}} + \cdots + \... | 15. (1) When $n=1$, $x_{1} \geqslant y_{1}>0, \frac{1}{x_{1}} \leqslant \frac{1}{y_{1}}$.
When $n=2$, $x_{1}+x_{2} \geqslant y_{1}+y_{2}, x_{1}-y_{1} \geqslant y_{2}-x_{2}, \frac{1}{y_{1}}-\frac{1}{x_{1}}=\frac{x_{1}-y_{1}}{x_{1} y_{1}} \geqslant \frac{y_{2}-x_{2}}{x_{2} y_{2}}=$ $\frac{1}{x_{2}}-\frac{1}{y_{2}}$, so ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,093 |
17. Let $n \geqslant 3$, and suppose a sequence of positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfies the condition: for $i=1,2, \cdots, n$, we have $a_{i-1}+a_{i+1}=k_{i} a_{i}$, where $k_{i}$ is a positive integer, $a_{0}=a_{n}, a_{n+1}=a_{r}$, prove that: $2 n \leqslant k_{1}+ k_{2}+\cdots+k_{n} \leqslant... | 17. Since $k_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$, we have $k_{1}+k_{2}+\cdots+k_{n}=\sum_{i=1}^{n}\left(\frac{a_{i}}{a_{i+1}}+\frac{a_{i+1}}{a_{i}}\right) \geqslant n \times 2$ $=2 n$
To prove the inequality on the right, we use induction on $n$: when $n=3$, it is easy to verify that the conclusion holds. Assume it hol... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,095 |
18. Let the function $f: R \rightarrow R$ satisfy that for any $x_{1}, x_{2} \in \mathbf{R}, t \in(0,1)$, we have $f\left(t x_{1}+(1-t\right.$ $\left.x_{2}\right) \leqslant t f\left(x_{1}\right)+(1-t) f\left(x_{2}\right)$
Prove: For all real numbers $a_{1}, a_{2}, \cdots, a_{2004}$, and $a_{1} \geqslant a_{2} \geqslan... | 18. We prove a stronger general conclusion: If real numbers $a_{1}, a_{2}, \cdots, a_{n+1}$, and $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}, a_{n+1}=a_{1}$, then
$$\sum_{k=1}^{n} f\left(a_{k+1}\right) a_{k} \leqslant \sum_{k=1}^{n} f\left(a_{k}\right) a_{k+1}$$
When $n=2$, the proposition is obviously tru... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,096 |
19. Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ distinct positive integers. Prove: $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant$
$$\frac{2 n+1}{3}\left(a_{1}+a_{2}+\cdots+a_{n}\right) .$$ | 19. Let $a_{1}<a_{2}<\cdots<a_{n}$. When $n=1$, the inequality $a_{1}^{2} \geqslant \frac{2+1}{3} a_{1}$ holds; assuming the inequality holds for $n=k$, i.e., $a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2} \geqslant \frac{2 n+1}{3}\left(a_{1}+a_{2}+\cdots+a_{k}\right)$. When $n=$ $k+1$, it suffices to prove that $a_{k+1}^{2} \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,097 |
20. Let $m, n$ be positive integers. Denote $S_{m}(n)=\sum_{k=1}^{n}\left[\sqrt[k^{2}]{k^{m}}\right]$. Prove: $S_{m}(n) \leqslant n+m\left(\sqrt[4]{2^{m}}-\right.$
1). (1998 Poland-Austria Mathematical Olympiad Problem) | 20. When $n \leqslant m$, it is easy to prove that for any $k \in \mathbf{N}^{*}$, we have $k^{4} \leqslant 2^{k^{2}}$. Therefore, $\sqrt[k^{2}]{k^{m}} \leqslant \sqrt[4]{2^{m}}$, at this time, $S_{m}(n) \leqslant \sum_{k=1}^{n} \sqrt[k^{2}]{k^{m}}=n+\sum_{k=1}^{n}\left(\sqrt[k^{2}]{k^{m}}-1\right) \leqslant n+\sum_{k=... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 733,098 |
21. Let the set $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}=\{1,2, \cdots, n\}$, prove: $\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1}{n} \leqslant$ $\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}} \cdot($ 31st IMO Preliminary Problem $)$ | 21. We prove the stronger inequality:
$$\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1}{n} \leqslant \frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}+\frac{a_{n}}{n}-1$$
Using mathematical induction. For $n=2$, we have
$$\frac{1}{2} \leqslant \frac{1}{2}+\frac{2}{2}-1=\frac{1}{2}$$
and
$$\frac{1}{2} \le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,099 |
Example 13 Let $a, b, c, d$ be positive real numbers, and satisfy $a b c d=1$, prove that: $\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+$ $\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geqslant 1 .(2005$ National Training Team for IMO Problem) | Below is the proof of the original problem. Without loss of generality, let $0 < t \leq \frac{1}{4}$.
If $t \geqslant \frac{1}{4}$, then by the lemma, we have
$$\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geqslant \frac{2}{(\sqrt{t}+1)^{2}}+\frac{2}{\left(\frac{1}{\sqrt{t}}+1\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,100 |
23. Given $0<x_{1}<x_{2}<\cdots<x_{2 n+1}$, prove the inequality: $x_{1}-x_{2}+x_{3}-x_{4}+\cdots+$ $x_{2 n-1}-x_{2 n}+x_{2 n+1}<\sqrt[n]{x_{1}^{n}-x_{2}^{n}+x_{3}^{n}-x_{4}^{n}+\cdots+x_{2 n-1}^{n}-x_{2 n}^{n}+x_{2 n+1}^{n}} \cdot$ (1998, Bari) | 23. When $k \geqslant 2$, we prove that
$$
x_{1}-x_{2}+x_{3}-x_{4}+\cdots+x_{2 n-1}-x_{2 n}+x_{2 n+1}<
\sqrt[k]{x_{1}^{k}-x_{2}^{k}+x_{3}^{k}-x_{4}^{k}+\cdots+x_{2 n-1}^{k}-x_{2 n}^{k}+x_{2 n+1}^{k}}
$$
That is, we need to prove
$$
\begin{array}{l}
\left(x_{1}-x_{2}+x_{3}-x_{4}+\cdots+x_{2 n-1}-x_{2 n}+x_{2 n+1}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,102 |
24. Let $a_{1}, a_{2}, \cdots$ be a sequence of real numbers, and for all $i, j=1,2, \cdots$, satisfy $a_{i+j} \leqslant a_{i}+a_{j}$. Prove that for any positive integer $n$, we have $a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n} \geqslant a_{n}$. (1999 Asia Pacific Mathematical Olympiad) | 24. When $n=1$, $a_{1} \geqslant a_{1}$, the inequality obviously holds.
Assume that for $n=1,2, \cdots, k-1$, the inequality holds, i.e.,
$$\left\{\begin{array}{l}
a_{1} \geqslant a_{r} \\
a_{1}+\frac{a_{2}}{2} \geqslant a_{2} \\
a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{k-1}}{k-1} a_{k-1} \geqslant a_{k-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,103 |
26. Given a quadratic trinomial $f(x)=a x^{2}+b x+c$ where all coefficients are positive, and $a+b+c=1$, prove that for any positive numbers $x_{1}, x_{2}, \cdots, x_{n}$, if $x_{1} x_{2} \cdots x_{n}=1$, then $f\left(x_{1}\right) f\left(x_{2}\right) \cdots f\left(x_{n}\right) \geqslant 1$. (24th All-Soviet Union Mathe... | 26. First, prove that for any $x, y>0$, we have $f(x) f(y) \geqslant[f(\sqrt{x y})]^{2}$.
In fact, if we denote $\sqrt{x y}=z$, then we have
$$\begin{array}{l}
f(x) f(y)-\left[f(\sqrt{x y})^{2}\right]^{2}=a^{2}\left(x^{2} y^{2}-z^{4}\right)+b^{2}\left(x y-z^{2}\right)+c^{2}(1-1)+ \\
a b\left(x^{2} y+x y^{2}-2 z^{3}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,105 |
27. Prove that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have $\sum_{k=1}^{n} \sqrt[k]{a_{1} a_{2} \cdots a_{k}} \leqslant e \sum_{k=1}^{n} a_{k}$. | 27. Consider the strengthened proposition
$$\sum_{k=1}^{n} \sqrt[k]{a_{1} a_{2} \cdots a_{k}}+n \sqrt[n]{a_{1} a_{2} \cdots a_{n}} \leqslant \mathrm{e} \sum_{k=1}^{n} a_{k}$$
When $n=1$, the conclusion is obviously true.
Assume that (1) holds for $n-1$, i.e.,
$$\sum_{k=1}^{n-1} \sqrt[k]{a_{1} a_{2} \cdots a_{k}}+(n-1)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,106 |
28. Let $a_{1}<a_{2}<\cdots<a_{n}$ be real numbers, prove: $a_{1} a_{2}^{4}+a_{2} a_{3}^{4}+\cdots+a_{n-1} a_{n}^{4}+$ $a_{n} a_{1}^{4} \geqslant a_{2} a_{1}^{4}+a_{3} a_{2}^{4}+\cdots+a_{n} a_{n-1}^{4}+a_{1} a_{n}^{4}$. (1998 - 1999 Iranian Mathematical Olympiad) | 28. When $n=2$, the above formula is an equality, and the conclusion holds. When $n=3$,
$$\begin{array}{l}
x y^{4}+y z^{4}+z x^{4}-\left(x^{4} y+y^{4} z+z^{4} x\right)=x y\left(y^{3}-x^{3}\right)+z^{4}(y-x)-z\left(y^{4}-x^{4}\right)= \\
(y-x)\left[x y\left(x^{2}+x y+y^{2}\right)+z^{4}-z(y+x)\left(y^{2}+x^{2}\right)\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,107 |
29. Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n} \geqslant a_{n+1}=0$ be a sequence of real numbers, prove: $\sqrt{\sum_{k=1}^{n} a_{k}} \leqslant$ $\sum_{k=1}^{n} \sqrt{k}\left(\sqrt{a_{k}}-\sqrt{a_{k+1}}\right)$. (38th IMO Shortlist Problem) | 29. Restate the conclusion to be proved as follows:
For each non-increasing, non-negative real sequence $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$, the inequality
$$\sqrt{\sum_{k=1}^{n} a_{k}} \leqslant \sum_{k=1}^{n-1} \sqrt{k}\left(\sqrt{a_{k}}-\sqrt{a_{k+1}}\right)+\sqrt{n a_{n}}$$
holds.
Use mathema... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,108 |
30. Let $a_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$. Prove: For all $n \geqslant 2$, we have $a_{n}^{2}>2\left(\frac{a_{2}}{2}+\right.$ $\left.\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n}\right) \cdot(1998$ Moldova Mathematical Olympiad Problem) | 30. Strengthen the proposition proof $a_{n}^{2}>2\left(\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n}\right)+\frac{1}{n}$.
(1) When $n=2$, $a_{n}^{2}=a_{2}^{2}=\frac{9}{4}$, and $2\left(\frac{a_{2}}{2}\right)^{2}+\frac{1}{2}=2$, the inequality holds.
(2) Assume that when $n=k$, the proposition holds, i.e., $a_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,109 |
31. Let $n>2, a_{1}, a_{2}, \cdots, a_{n}$ be $n$ positive numbers, and $a_{1} a_{2} \cdots a_{n}=1$, prove: $\sum_{k=1}^{n} \frac{1}{1+a_{k}}>$
1. (2005 Thailand Mathematical Olympiad Problem) | 31. Let $a, b>0$. Then we have $\frac{1+a b}{1+a}+\frac{1+a b}{1+b}=\frac{1+a+b+2 a b+1+a^{2} b+a b^{2}}{1+a+b+a b}>1$, so, $\frac{1}{1+a}+\frac{1}{1+b}>\frac{1}{1+a b}$.
By mathematical induction, we can get
$$\sum_{k=1}^{n} \frac{1}{1+a_{k}}>\frac{1}{1+a_{1} a_{2} \cdots a_{n}}, n>2$$
Therefore,
$$\sum_{k=1}^{n} \fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,110 |
Lemma Let $a, b>0, ab=t$, then for a fixed $t$, we have $\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}} \geqslant$
$$\left\{\begin{array}{l}
\frac{2}{(\sqrt{t}+1)^{2}} \quad \text { when } t \geqslant \frac{1}{4} \\
-\frac{1-2 t}{(1-t)^{2}} \quad \text { when } 0<t<\frac{1}{4}
\end{array}\right.$$ | Lemma Proof: Let $u=a+b+1$, then $u=(a+b)+1 \geqslant 2 \sqrt{t}+1$. Since
$$\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}=\frac{u^{2}+1-2 t}{(u+t)^{2}}=\frac{(t-1)^{2}}{(u+t)^{2}}-\frac{2 t}{u+t}+1$$
Let $x=\frac{1}{u+t}$ and $f(x)=(t-1)^{2} x^{2}-2 t x+1$, then
$$\begin{aligned}
x & =\frac{1}{u+t} \leqslant \frac{1}{2 \sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,111 |
32. Let $a_{i}$ be positive real numbers $(i=1,2, \cdots, n)$, and let $b_{k}=\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}(k=1,2, \cdots, n), C_{n}=\left(a_{1}-b_{1}\right)^{2}+\left(a_{2}-b_{2}\right)^{2}+\cdots+\left(a_{n}-b_{n}\right)^{2}, D_{n}=\left(a_{1}-b_{n}\right)^{2}+\left(a_{2}-b_{n}\right)^{2}$ $+\cdots+\left(a_{n}-b... | 32. Let $f(x)=\left(x-a_{1}\right)^{2}+\left(x-a_{2}\right)^{2}+\cdots+\left(x-a_{n}\right)^{2}$
Then $\square$
$$f(x)=n\left(x-a_{n}\right)^{2}+f\left(b_{n}\right)$$
Now we prove $C_{n} \leqslant D_{n} \leqslant 2 C_{n}$ by mathematical induction.
When $n=1$, $C_{1} \leqslant D_{1}$, so $C_{1} \leqslant D_{1} \leqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,112 |
33. Let $m, n \in \mathbf{N}^{*}$, and denote $S_{m}(n)=\sum_{k=1}^{n}\left[\sqrt[k^{2}]{k^{m}}\right] \leqslant n+m\left(\sqrt[4]{2^{m}}-1\right)$. Here $[x]$ represents the greatest integer not exceeding $x$. (1999 Polish Mathematical Olympiad Problem) | 33. When $n \leqslant m$, it is easy to prove that for any $k \in \mathbf{N} *$, we have $k^{4} \leqslant 2^{k^{2}}$, thus $\sqrt[k^{2}]{k^{m}} \leqslant \sqrt[4]{2^{m}}$. At this time, $S_{m}(n) \leqslant \sum_{k=1}^{n} \sqrt[k^{2}]{k^{m}}=n+\sum_{k=1}^{n}\left(\sqrt[k^{2}]{k^{m}}-1\right) \leqslant n+\sum_{k=1}^{m}\l... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 733,113 |
34. Given that $m, n$ are positive integers no less than 2, prove that $\max \sqrt[n]{m}, \sqrt[m]{n} \leq \sqrt[3]{3}$. | 34. We first prove that when $m=n$, we have
$$\sqrt[n]{n} \leqslant \sqrt[3]{3}$$
Let $x_{n}=\sqrt[n]{n}$. When $n=1,2,3,4$, the inequality (1) holds. Assume that for $p \geqslant 3$, the inequality holds, i.e., $p^{3} \leqslant 3^{p}$. Then, since $p \geqslant 3$, we get $3^{p \pm t} \geqslant 3 p^{3}=p^{5}+3 p^{2-}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,114 |
35. Prove that for all $n \geqslant 2$, the inequality $\frac{x_{1}^{2}}{x_{1}^{2}+x_{2} x_{3}}+\frac{x_{2}^{2}}{x_{2}^{2}+x_{3} x_{4}}+\cdots+$ $\frac{x_{n-1}^{2}}{x_{n-1}^{2}+x_{n} x_{1}}+\frac{x_{n}^{2}}{x_{n}^{2}+x_{1} x_{2}} \leqslant n-1$ holds. (27th IMO Preliminary Problem) | 35. Proof - Let $y_{i}=x_{i+1} x_{i+2}^{2}(i=1,2, \cdots, n)$ and $x_{n+i}=x_{i}$, note that $y_{1} y_{2} \cdots \bar{y}_{n}=1$, so $\frac{x_{i}^{2}}{x_{i}^{2}+\bar{x}_{i+1} x_{i+2}}=\frac{x_{i}^{2}+x_{i+1} x_{i+2}-x_{i+1} x_{i+2}}{x_{i}^{2}+x_{i+1} x_{i+2}}=1=\frac{x_{i+1} x_{i+2}}{x_{i}^{2}+x_{i+1} x_{i+2}}=1-\frac{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,115 |
36. Given a positive integer $n>1$, prove that: $\frac{2 n}{3 n+1}<\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}<\frac{25}{36}$. | 36. First prove the left inequality
$$f(n+1)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}+\frac{1}{2 n+1}+\frac{1}{2 n+2}=\frac{2 n+2}{3 n+4}$$
So
$$\begin{array}{l}
f(n+1)-f(n)=\left(\frac{1}{2 n+1}-\frac{1}{2 n+2}\right)+\left(\frac{2 n}{3 n+1}-\frac{2 n+2}{3 n+4}\right)= \\
\frac{2}{(2 n+1)(-2 n+2)}-\frac{1}{(3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,116 |
37. The sequence of real numbers $a_{0}, a_{1}, \cdots$, is defined as: $a_{0}=-1$, and for all positive integers $n$, $\sum_{k=0}^{n}$ $\frac{a_{n-k}}{k+1}=0$. Prove: For all positive integers $n$, $a_{n}>0$. (47th IMO Shortlist) | 37. Using mathematical induction. When $n=1$, we get $a_{1}=\frac{1}{2}>0$. Assume that for $n \geqslant 1$, we have $a_{i}>$
$$\begin{aligned}
0 & (i=1,2, \cdots, n) . \\
& \quad \text { From } \sum_{k=0}^{n} \frac{a_{k}}{n-k+1}=0, \sum_{k=0}^{n+1} \frac{a_{k}}{n-k+2}=0, \text { we get } \\
0= & (n+2) \sum_{k=0}^{n+1}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,117 |
38. Let $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}$ be positive numbers, prove: $\frac{1}{x_{1}}+\frac{x_{1}}{x_{2}}+\frac{x_{1} x_{2}}{x_{3}}+\cdots+\frac{x_{1} x_{2} \cdots x_{n}}{x_{n+1}} \geqslant 4(1$ $\left.-x_{1} x_{2} \cdots x_{n} x_{n+1}\right) .(2007$ Belarusian Mathematical Olympiad Problem) | 38 When $\bar{n}=1$, $1 \geqslant 4\left(1-x_{1}=\left(2 x_{1}-1\right)^{2} \geqslant 0\right)$, so the inequality holds.
Assume that the inequality holds for $n=k$, i.e.,
$$\frac{1}{x_{1}}+\frac{x_{1}}{x_{2}}+\frac{x_{1} x_{2}}{x_{3}}+\cdots+\frac{x_{1} x_{2} \cdots x_{k}}{x_{k+1}} \geqslant 4\left(1-x_{1} x_{2} \cdot... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,118 |
41. Let $\alpha_{i}>0, \beta_{i}>0(-1 \leqslant i \leqslant n, n>1)$, and $\sum_{i=1}^{n} \alpha_{i}=\sum_{i=1}^{n} \beta_{i}=\pi$, then $\sum_{i=1}^{n} \frac{\cos \beta_{i}}{\sin \alpha_{i}} \leqslant \sum_{i=1}^{n} \cot \alpha_{i}$. (29th IMO Shortlist Problem) | 41. When $n=2$, $\frac{\cos \beta_{1}}{\sin \alpha_{1}}+\frac{\cos \beta_{2}}{\sin \alpha_{2}}=\frac{\cos \beta_{\vdash}}{\sin \alpha_{1}}-\frac{\cos \beta_{1}}{\sin \alpha_{1}}=0=\cot \alpha_{1}+\cot \alpha_{2}$. When $n=3$, it is to be proved: Given the internal angles of two triangles are $A, B, C$ and $A_{1}, B_{\m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,121 |
Example 14 Let $x, y, z$ be positive real numbers, and $xyz=1$, prove: $\frac{x^{3}}{(1+y)(1+z)}+$ $\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geqslant \frac{3}{4}$. (39th IMO Shortlist) | To prove that the original inequality is equivalent to $x^{3}+x^{4}+y^{3}+y^{4}+z^{3}+z^{4} \geqslant \frac{3}{4}(1+x)(1+y)(1+z)$, and since for any positive numbers $u, v, w$, we have $u^{3}+v^{3}+w^{3} \geqslant 3 u v w$, we will prove the stronger inequality $x^{3}+x^{4}+y^{3}+y^{4}+z^{3}+z^{4} \geqslant \frac{1}{4}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,122 |
Example 1 Let the sequence of positive terms $\left\{a_{n}\right\}$ satisfy for any $n \in \mathbf{N}^{*}$, $\sum_{i=1}^{n} a_{i} \geqslant \sqrt{n}$. Prove that for any $n \in \mathbf{N}^{*}$, $\sum_{i=1}^{n} a_{i}^{2} \geqslant \frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)$. (1994 USA Mathemati... | Prove that if $b_{i}=a_{i}-\left(\sqrt{i}(\sqrt{i-1}), i=1,2, \cdots\right.$, then for any $n \in \mathbf{N}^{*}, \sum_{i=1}^{n} a_{i} \sqrt{n}$ is equivalent to $\sum_{i=1}^{n} b_{i}=0$ for any $n \in \mathbf{N}^{*}$. Clearly,
$$\begin{aligned}
\sum_{i=1}^{n} a_{i}^{2}= & \sum_{i=1}^{n}\left[(\sqrt{i}-\sqrt{i-1})+b_{i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,123 |
Example 2: Prove that for any positive integer $n, \frac{2}{3} n \sqrt{n}<1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}<\frac{4 n+3}{6} \sqrt{n}$. | Prove the strengthened inequality $\frac{2 n+1}{3} \sqrt{n} < 1 + \sqrt{2} + \sqrt{3} + \cdots + \sqrt{n} < \frac{4 n+3}{6} \sqrt{n} - \frac{1}{6}$
On the other hand, using Abel's sum:
$$\begin{array}{l}
1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}= \\
1 \cdot\left(1-\frac{1}{\sqrt{2}}\right)+(1+2)\left(\frac{1}{\sqrt{2}}-\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,124 |
Example 3 Let $a_{1}, a_{2}, \cdots$ be a sequence of positive real numbers, and for all $i, j=1,2, \cdots$, satisfy $a_{i+j} \leqslant a_{i}+a_{j}$. Prove that for positive integers $n$, we have $a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n} \geqslant a_{n}$. | Let $S_{i}=a_{1}+a_{2}+\cdots+a_{i}, i=1,2, \cdots, n$. By convention, $S_{0}=0$, then
$$2 S_{i}=\left(a_{1}+a_{i}\right)+\left(a_{2}+a_{i-1}\right)+\cdots+\left(a_{i}+a_{1}\right) \geqslant i \bar{a}_{i+1}$$
i.e., $S_{i} \geqslant \frac{i}{2} a_{i+1}$.
Therefore,
$$\begin{array}{l}
a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,125 |
For $1 \leqslant i \leqslant n$. Let $a_{i}$ and $b_{i}$ be real numbers, satisfying:
$$\begin{array}{l}
a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n} \geqslant 0 \\
b_{1} \geqslant a_{1} \\
b_{1} b_{2} \geqslant a_{1} a_{2} \\
b_{1} b_{2} b_{3} \geqslant a_{1} a_{2} a_{3} \\
\vdots \\
b_{1} b_{2} \cdots b_{n}... | Prove that given $c_{i}=\frac{b_{i}}{a_{i}}, 1 \leqslant i \leqslant n_{\llcorner}$, and knowing $c_{1} \geqslant 1, c_{1} c_{2} \geqslant 1, \cdots, c_{1} c_{2} \cdots c_{n} \geqslant 1$, we need to show
$$\left(c_{1}-1\right) a_{1}+\left(c_{2}-1\right) a_{2}+\cdots+\left(c_{n}-1\right) a_{n} \geqslant 0$$
i.e., $\sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,126 |
Example 5 Given two positive integers $n \geqslant 2$ and $T \geqslant 2$, find all positive integers $a$, such that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have
$$\sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}}<T^{2} \sum_{k=1}^{n} \frac{1}{a_{k}}$$
where $S_{k}=a_{1}+a_{2}+\cdots+a_{k}$. (1992 Na... | $$\begin{array}{l}
\sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}}=\sum_{k=1}^{n} \frac{1}{S_{k}}\left[\left(k+\frac{a}{2}\right)^{2}-k^{2}\right]= \\
\frac{1}{S_{1}}\left[\left(1+\frac{a}{2}\right)^{2}-1^{2}\right]+\frac{1}{S_{2}}\left[\left(2+\frac{a}{2}\right)^{2}-2^{2}\right]+\frac{1}{S_{3}}\left[\left(3+\frac{a}{... | 1 ; 2,3, \cdots, 2(T-1) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,127 |
1. Given $a_{1}, a_{2}, \cdots, a_{n}$ are pairwise distinct positive integers. Prove that for any positive integer $n$ we have $\sum_{k=1}^{n} \frac{a_{k}}{k^{2}} \geqslant \sum_{k=1}^{n} \frac{1}{k}$. (20th IMO Problem) | 1. Let $S_{k}=a_{1}+a_{2}+\cdots+a_{k} \geqslant 1+2+\cdots+k=\frac{k(k+1)}{2}, b_{k}=\frac{1}{k^{2}}$, using Abel's identity, we have
$$\begin{array}{l}
\sum_{k=1}^{n} \frac{a_{k}}{k^{2}}=\sum_{k=1}^{n} a_{k} b_{k}=S_{n} b_{n}+\sum_{k=1}^{n-1} S_{k}\left(b_{k}-b_{k+1}\right) \geqslant \\
\frac{1}{n^{2}} S_{n}+\sum_{k=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,128 |
2. Given $x_{i} \in \mathbf{R}(i=1,2, \cdots, n)$, satisfying $\sum_{i=1}^{n}\left|x_{i}\right|=1, \sum_{i=1}^{n} x_{i}=0$, prove: | $\left.\sum_{i=1}^{n} \frac{x_{i}}{i} \right\rvert\, \leqslant \frac{1}{2}-\frac{1}{2 n}$. (1989 National High School Mathematics League Second Trial) | 2. Let $S_{i}=x_{1}+x_{2}+\cdots+x_{i}(i=1,2, \cdots, n)$, then from the given conditions we have $S_{n}=0,\left|S_{i}\right| \leqslant \frac{1}{2}(i=1,2, \cdots, n-1)$. Using Abel's identity, we get
$$\sum_{i=1}^{n} \frac{x_{i}}{i}=S_{n} \cdot \frac{1}{n}+\sum_{i=1}^{n-1} S_{i}\left(\frac{1}{i}-\frac{1}{i+1}\right)$$
... | \frac{1}{2}\left(1-\frac{1}{n}\right) | Inequalities | proof | Yes | Yes | inequalities | false | 733,129 |
3. $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n}$ are real numbers. Prove that the necessary and sufficient condition for the inequality $\sum_{i=1}^{n} a_{i} x_{i} \leqslant \sum_{i=1}^{n} b_{i} x_{i}$ to hold for any real numbers satisfying $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$ is $\su... | 3. First, prove the necessity:
Let $x_{1}=x_{2}=\cdots=x_{n}=1$, we get $\sum_{i=1}^{n} a_{i} \leqslant \sum_{i=1}^{n} b_{i}$. Let $x_{1}=x_{2}=\cdots=x_{n}=-1$, we get $-\sum_{i=1}^{n} a_{i} \leqslant-\sum_{i=1}^{n} b_{i}$. Therefore, $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}$.
Let $x_{1}=x_{2}=\cdots=x_{k}=0, x_{k+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,130 |
5. Given $x_{i}, y_{i}(i=1,2, \cdots, n)$ are real numbers, and $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}, y_{1} \geqslant y_{2} \geqslant \cdots \geqslant$ $y_{n}$, and $z_{1}, z_{2}, \cdots, z_{n}$ is any permutation of $y_{1}, y_{2}, \cdots, y_{n}$. Prove: $\sum_{i=1}^{n}\left(x_{i}-y_{i}\right)^{2} \l... | 5. Since $\sum_{i=1}^{n} y_{i}^{2}=\sum_{i=1}^{n} z_{i}^{2}$, the original inequality is equivalent to $\sum_{i=1}^{n} x_{i} y_{i} \geqslant \sum_{i=1}^{n} x_{i} z_{i}$. Let $A_{i}=\sum_{k=1}^{i} y_{k}, B_{i}=\sum_{k=1}^{i} z_{k}$, then it is easy to see that $A_{i} \geqslant B_{i}$. Thus, by Abel's transformation, we ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,132 |
Example 15 Let $a, b, c$ be positive real numbers; and satisfy $abc=1$, try to prove $a^{3}(b+c)+\frac{1}{b^{3}(c+a)}+$ $\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} \quad($ (36th IMO problem) | Prove that for $f(a, b, c)=\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c(a+b)}$, it is clear that when $a=b=c=1$, $f(a, b, c)=\frac{3}{2}$.
Since $a, b, c$ are positive real numbers and satisfy $abc=1$, one of $a, b, c$ must be no greater than 1. Without loss of generality, let $0 < a \leq 1 \leq c$ and set $q=... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 733,133 |
6. Let $\left\{a_{1}, a_{2}, a_{3}, \cdots\right\}$ be an infinite sequence of positive numbers. Prove the inequality $\sum_{n=1}^{N} \alpha_{n}^{2} \leqslant 4 \sum_{n=1}^{N} a_{n}^{2}$ for any positive integer $N$. Here $\alpha_{n}$ is the average of $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$, i.e., $\alpha_{n}=$ $\frac{a_... | 6. If we set $\frac{1}{c} \sum_{n=1}^{N} \alpha_{n}^{2} \leqslant \sum_{n=1}^{N} \alpha_{n} a_{n}$, then we have $\sum_{n=1}^{N} \alpha_{n} a_{n} \leqslant c \sum_{n=1}^{N} a_{n}^{2}$, and the problem can be transformed into handling by the Abel method:
$$\begin{array}{l}
\sum_{n=1}^{N} \alpha_{n} a_{n}=\sum_{n=1}^{N} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,134 |
7. Given positive numbers $x_{1}, x_{2}, \cdots, x_{n}$ and $y_{1}, y_{2}, \cdots, y_{n}$ satisfying
(1) $x_{1}>x_{2}>\cdots>x_{n}, y_{1}>y_{2}>\cdots>y_{n}$
(2) $x_{1}>y_{1}, x_{1}+x_{2}>y_{1}+y_{2}, \cdots, x_{1}+x_{2}+\cdots+x_{n}>y_{1}+y_{2}+\cdots+y_{n}$.
Prove: For any positive integer $k$, we have $x_{1}^{k}+x_... | 7. Let $S_{i}=x_{1}+x_{2}+\cdots+x_{i}, i=1,2, \cdots, n$. $S_{0}=0, T_{i}=y_{1}+y_{2}+\cdots+y_{i}, i=$ $1,2, \cdots, n$. $T_{0}=0$, for any positive numbers $a_{1}>a_{2}>\cdots>a_{n}$ we have
$$\begin{aligned}
\sum_{k=1}^{n} a_{k} x_{k}= & \sum_{k=1}^{n} a_{k}\left(S_{k}-S_{k-1}\right)=\sum_{k=1}^{n} a_{k} S_{k}-\sum... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,135 |
8. If $x$ is a positive real number, prove that $[n x] \geqslant \frac{[x]}{1}+\frac{[2 x]}{2}+\frac{[3 x]}{3}+\cdots+\frac{[n x]}{n}$ where $[t]$ denotes the greatest integer not exceeding $t$. | 8. First, prove the following conclusion: If $f_{k}(x)=\sum_{i=1}^{k} \frac{[i x]}{i}$, then
$$n f_{n}(x)=\sum_{k=1}^{n}[k x]+\sum_{k=1}^{n-1} f_{k}(x)$$
In fact, let $f_{0}(x)=0, a_{k}=1, (k=0,1,2, \cdots, n)$, then $S_{k}=\sum_{i=1}^{k} a_{i}=k+1$. Thus, by Abel's summation formula, we get
$$\begin{aligned}
\sum_{k=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,136 |
10. Prove that for any real numbers $a_{1}, a_{2}, \cdots, a_{n}$, there exists a positive integer $k, 1 \leqslant k \leqslant n$, such that for any $1 \geqslant b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n} \geqslant 0$, we have $\left|\sum_{i=1}^{n} b_{i} a_{i}\right| \leqslant 1 \sum_{i=1}^{k} a_{i} \mid$. ... | 10. Let $S_{0}=0, S_{i}=a_{1}+a_{2}+\cdots+a_{i}, i=1,2, \cdots, n$. Then $a_{i}=S_{i}-S_{i-1}, i=1$, $2, \cdots, n$. Therefore, we have
$$\begin{aligned}
1 \sum_{i=1}^{n} b_{i} a_{i} \mid= & 1 \sum_{i=1}^{n} b_{i}\left(S_{i}-S_{i-1}\right)|=| \sum_{i=1}^{n} b_{i} S_{i}-\sum_{i=1}^{n} b_{i+1} S_{i} \mid= \\
& \left|\su... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,138 |
11. Let real numbers $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}$ satisfy $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n} \geqslant x_{n+1}=0$, prove: $\sqrt{x_{1}+x_{2}+\cdots+x_{n}} \geqslant \sum_{i=1}^{n} \sqrt{i}\left(\sqrt{x_{i}}-\sqrt{x_{i+1}}\right)$
(1996 Romanian Mathematical Olympiad) | 11. Let $c_{i}=\sqrt{i}-\sqrt{i-1}$ and $a_{i}=\sqrt{x_{i}}$, the inequality becomes to prove: $\left(a_{1} c_{1}+a_{2} c_{2}+\cdots+\right.$ $\left.a_{n} c_{n}\right)^{2} \geqslant a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}$.
If real numbers $b_{1}, b_{2}, \cdots, b_{n}$ satisfy $b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=1$, th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,139 |
12. Let real numbers $-1<x_{1}<x_{2}<\cdots<x_{n}<1, y_{1}<y_{2}<\cdots<y_{n}$, and satisfy $x_{1}+$ $x_{2}+\cdots+x_{n}=x_{1}^{13}+x_{2}^{13}+\cdots+x_{n}^{13}$, prove: $x_{1}^{13} y_{1}+x_{2}^{13} y_{2}+\cdots+x_{n}^{13} y_{n}<x_{1} y_{1}+$ $x_{2} y_{2}+\cdots+x_{n} y_{n} \cdot$ (2000 Russian Mathematical Olympiad Pr... | 12. By Abel's formula, we have
\[
\sum_{i=1}^{n} y_{i}\left(x_{i}^{13}-x_{i}\right)=\left(y_{1}-y_{2}\right)\left(x_{1}^{13}-x_{1}\right)+\left(y_{2}-y_{3}\right)\left(x_{1}^{13}+x_{2}^{13}-x_{1}-x_{2}\right)+\cdots+\left(y_{n-1}-y_{n}\right)\left(\sum_{i=1}^{n-1} x_{i}^{13}-\sum_{i=1}^{n-1} x_{i}\right)+y_{n}\left(\su... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,140 |
Example 2 Let $a, b, c$ be positive numbers, prove that: $\frac{a b}{(a+c)(b+c)}+\frac{b c}{(b+c)(c+a)}+$ $\frac{c a}{(c+b)(a+b)} \geqslant \frac{3}{4}$
(30th IMO Shortlist) | Prove that the original inequality is equivalent to
$$\begin{array}{l}
4[a b(a+b)+b c(b+c)+c a(c+a)] \geqslant 3(a+b)(b+c)(c+a) \Leftrightarrow \\
4\left[a\left(b^{2}+c^{2}\right)+b\left(c^{2}+a^{2}\right)+c\left(a^{2}+b^{2}\right)\right] \geqslant \\
3\left[a\left(b^{2}+c^{2}\right)+b\left(c^{2}+a^{2}\right)+c\left(a^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,142 |
Example 3 Positive numbers $a, b, c$ satisfy $a+b+c=1$, prove:
$$\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c} \leqslant 2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)$$ | Proof
$$\begin{aligned}
\text { Equation (1) } \Leftrightarrow & \frac{b}{a}+\frac{c}{b}+\frac{a}{c} \geqslant \frac{3}{2}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \Leftrightarrow \\
& \frac{b}{a}-\frac{b}{c+a}+\frac{c}{b}-\frac{c}{a+b}+\frac{a}{c}-\frac{a}{b+c} \geqslant \frac{3}{2} \Leftrightarrow \\
& \frac{b c}{a(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,143 |
Example 16 Given that $\alpha, \beta$ are two distinct real roots of the equation $4 x^{2}-4 t x-1=0(t \in \mathbf{R})$, and the domain of the function $f(x)=\frac{2 x-1}{x^{2}+1}$ is $[\bar{\alpha}, \beta]$
(1) $g(t)=\max f(x)-\min f(x)$;
(2) Prove: For $u_{i} \in\left(0, \frac{\pi}{2}\right)(i=1,2,3)$, if $\sin u_{1}... | Solve $(1) f^{2}(x)=\frac{2\left(x^{2}+1\right)-2 x(2 x-t)}{\left(x^{2}+1\right)^{2}}=\frac{2 x^{2}+2 t x+2}{\left(x^{2}+1\right)^{2}}$
From $\alpha \beta$ being the two distinct real roots of the equation $4 x^{2}-4 t x-1=0(t=\mathbf{R})$, we know $4 x^{2}-4 t x-1=4(x-\alpha)(x-\beta)$, and since $x \in [\alpha, \beta... | \frac{3}{4} \sqrt{6} | Algebra | proof | Yes | Yes | inequalities | false | 733,144 |
Example 4 Let $x, y$ be two distinct real numbers, $R=\sqrt{\frac{x^{2}+y^{2}}{2}}, A=\frac{x+y}{2}, G=\sqrt{x y}, H=$ $\frac{2 x y}{x+y}$, determine which of $R-A, A-G, G-H$ is the largest and which is the smallest: (30th IMO Canada) | Solve $A-G$ maximum, $G=H$ minimum because
$$\begin{array}{l}
-\frac{x+y}{2}=\sqrt{x y} \geqslant \sqrt{\frac{x^{2}+y^{2}}{2}}-\frac{x+y}{2} \Leftrightarrow \\
x+y \geqslant \sqrt{\frac{x^{2}+y^{2}}{2}+-\sqrt{x y} \Leftrightarrow} \\
-(x+y)^{z} \geqslant \frac{x^{2}+y^{2}}{2}+x y+\sqrt{2 x y\left(x^{2}+y^{2}\right)} \L... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,145 |
Example 5 Prove or disprove: If $x, y$ are real numbers, $y \geqslant 0, y(y+1) \leqslant(x+1)^{2}$, then $y(y-1)$ $\leqslant x^{2}$. (30th IMO Canadian Training Problem) | We prove that when
$$y \geqslant 0, y(y+1) \leqslant(x+1)^{2}$$
then,
$$y(y-1) \leqslant x^{2}$$
If $y \leqslant 1$, then $y(y-1) \leqslant 0$, and inequality (2) is obviously true. If $y>1$, from inequality (1) we get
$$y \leqslant \sqrt{\frac{1}{4}+(x+1)^{2}}-\frac{1}{2}$$
And (2) $\Leftrightarrow y-\frac{1}{2} \l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,146 |
Example 6 Prove that for all positive numbers $a, b, c$, we have $\frac{1}{a^{3}+b^{3}+a b c}+\frac{1}{b^{3}+c^{3}+a b c}+$ $\frac{1}{c^{3}+a^{3}+a b c} \leqslant \frac{1}{a b c}$. (26th USA Mathematical Olympiad Problem) | Prove that by eliminating the denominator and simplifying, the original inequality is equivalent to
$$a^{6}\left(b^{3}+c^{3}\right)+b^{6}\left(c^{3}+a^{3}\right)+c^{6}\left(a^{3}+b^{3}\right) \geqslant 2 a^{2} b^{2} c^{2}\left(a^{3}+b^{3}+c^{3}\right)$$
Because
$$2 a^{2} b^{2} c^{2}\left(a^{3}+b^{3}+c^{3}\right) \leqs... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,147 |
Example 9 Let $a, b, c, d$ be positive real numbers, and satisfy $a+b+c+d=1$, prove that: $(1-\sqrt{a})(1-$ $\sqrt{b})(1-\sqrt{c})(1-\sqrt{d}) \geqslant \sqrt{a b c d}$. (2005 Jiangsu Province Mathematical Winter Camp Lecture Question) | Prove the following inequality first:
$$(1-\sqrt{a})(1-\sqrt{b}) \geqslant \sqrt{c d}$$
It is sufficient to prove:
$$(1-\sqrt{a})(1-\sqrt{b}) \geqslant \frac{c+d}{2}$$
It is sufficient to prove:
$$2+2 \sqrt{a b}-2 \sqrt{a}-2 \sqrt{b} \geqslant 1-a-b$$
Equation (3) is equivalent to:
$$1+a+b-2 \sqrt{a}-2 \sqrt{b}+2 \s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,150 |
Example 11 Let positive real numbers $a, b, c$ satisfy $a b c \geqslant 2^{9}$, prove:
$$\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} \geqslant \frac{3}{\sqrt{1+\sqrt[3]{a b c}}}$$ | We prove its equivalent proposition: Let positive real numbers $a, b, c$ satisfy $abc = k^3$, and $k \geq 8$, then
$$\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} \geqslant \frac{3}{\sqrt{1+k}}$$
From the given, we have
$$\begin{array}{c}
a+b+c \geqslant 3 \cdot \sqrt[3]{abc}=3k \\
ab+bc+ca \geqslant ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,152 |
Example 12 If $a, b, c \in \mathbf{R},\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right) \geqslant(a b+$ $b c+c a)^{3}$. When does equality hold? (31st IMO Shortlist) | Proof
Since we have
$$\begin{array}{l}
a^{2}+a b+b^{2} \geqslant \frac{3}{4}(a+b)^{2} \\
b^{2}+b c+c^{2} \geqslant \frac{3}{4}(b+c)^{2} \\
c^{2}+c a+a^{2} \geqslant \frac{3}{4}(c+a)^{2}
\end{array}$$
It is sufficient to prove the following inequality:
$$27(a+b)^{2}(b+c)^{2}(c+a)^{2} \geqslant 64(a b+b c+c a)^{3}$$
Th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,153 |
Example 13 Given $x \geqslant 0, y \geqslant 0, z \geqslant 0$, prove the inequality $8\left(x^{3}+y^{3}+z^{3}\right)^{2} \geqslant 9\left(x^{2}+\right.$ $y z)\left(y^{2}+z x\right)\left(z^{2}+x y\right) \cdot(1982$ German National Team Question) | Prove that the original inequality is equivalent to
$$\begin{array}{l}
8\left[x^{6}+y^{6}+z^{6}+2\left(x^{3} y^{3}+y^{3} z^{3}+z^{3} x^{3}\right)\right] \\
9\left[2 x^{2} y^{2} z^{2}+\left(x^{3} y^{3}+y^{3} z^{3}+z^{3} x^{3}\right)+\left(x^{4} y z+x y^{4} z+x y z^{4}\right)\right] \geqslant 0 \Leftrightarrow \\
8\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,154 |
Example 14 Given that $a, b, c, d, k$ are all positive real numbers, and $a, b, c, d \leqslant k$, prove the inequality:
$$\begin{array}{l}
\frac{a^{4}+b^{4}+c^{4}+d^{4}}{(2 k-a)^{4}+(2 k-b)^{4}+(2 k-c)^{5}+(2 k-d)^{4}} \geqslant \\
\frac{a b c d}{(2 k-a)(2 k-b)(2 k-c)(2 k-d)} \cdot(2002 \text { Taiwan Mathematical Oly... | Prove that the original inequality is equivalent to
$$\begin{array}{l}
\frac{a^{4}+b^{4}+c^{4}+d^{4}}{a b c d} \geqslant \frac{(2 k-a)^{4}+(2 k-b)^{4}+(2 k-c)^{4}+(2 k-d)^{4}}{(2 k-a)(2 k-b)(2 k-c)(2 k-d)} \Leftrightarrow \\
\frac{\left(a^{2}-b^{2}\right)^{2}+\left(c^{2}-d^{2}\right)^{2}+2\left(a^{2} b^{2}+c^{2} d^{2}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,156 |
1. Prove that for any positive numbers $a, b, c$, we have $\sqrt{a b(a+b)}+\sqrt{b c(b+c)}$ $\sqrt{c a(c+a)}>\sqrt{(a+b)(b+c)(c+a)}$ (1990 All-Russian Mathematical Olympiad Problem) | 1. To prove the original inequality, it suffices to prove
$$\begin{array}{l}
a b(a+b)+b c(b+c)+c a(c+a)+2 \sqrt{a b^{2} c(a+b)(b+c)}+ \\
2 \sqrt{a b c^{2}(b+c)(a+c)}+2 \sqrt{a^{2} b c(a+b)(a+c)}> \\
(a+b)(b+c)(c+a)
\end{array}$$
i.e., $\square$
$$b \sqrt{a c(a+b)(b+c)}+c \sqrt{a b(b+c)(a+c)}+a \sqrt{b c(a+b)(a+c)}>a b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,159 |
4. Let $x, y, z \in [0,1]$, prove that: $(1+x)(1+y)(1+z) \geqslant \sqrt{8(x+y)(y+z)(z+x)}$. | $$\begin{array}{l}
\text { 4. }(1+x)(1+y) \geqslant 2(x+y) \Leftrightarrow 1+x+y+x y \geqslant 2(x+y) \Leftrightarrow 1-x-y+ \\
x y \geqslant 0 \Leftrightarrow(1-x)(1-y) \geqslant 0
\end{array}$$
Similarly,
$$\begin{array}{l}
(1+y)(1+z) \geqslant 2(y+z) \\
(1+z)(1+x) \geqslant 2(z+x)
\end{array}$$
Multiplying the abo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,162 |
5: Given that $x, y, z$ are positive numbers, prove that: $\left(x^{2}+\frac{3}{4}\right)\left(y^{2}+\frac{3}{4}\right)\left(z^{2}+\frac{3}{4}\right) \geqslant$ $\sqrt{(x+y)(y+z)(z+x)} \cdot(2005$ St. Petersburg Mathematical Olympiad Problem E 1 Grade) | 5. It is sufficient to prove that $\left(x^{2}+\frac{3}{4}\right)\left(y^{2}+\frac{3}{4}\right) \geqslant x+y$, which means proving
$$x^{2} y^{2}+\frac{3}{4} x^{2}+\frac{3}{4} y^{2}-x-y+\frac{9}{16} \geqslant 0$$
And
$$\begin{array}{l}
x^{2} y^{2}+\frac{3}{4} x^{2}+\frac{3}{4} y^{2}-x-y+\frac{9}{16}=\left(x y-\frac{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,163 |
6. Prove: The inequality $\frac{a^{2}}{(a+b)(a+c)}+\frac{b^{2}}{(b+c)(b+a)}+\frac{-c^{2}}{(c+b)(c+a)} \geqslant \frac{3}{4}$ holds for all positive real numbers $a, b, c$. (2004 Croatian Mathematical Olympiad Problem) | $$\begin{array}{l}
\text { 6. } \frac{a^{2}}{(a+b)(a+c)}+\frac{b^{2}}{(b+c)(b+a)}+\frac{c^{2}}{(c+b)(c+a)} \geqslant \frac{3}{4} \Leftrightarrow \\
\frac{a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)}{(a+b)(b+c)(c+a)} \geqslant \frac{3}{4} \Leftrightarrow \\
\frac{a^{2} b+a^{2} c+b^{2} c+b^{2} a+c^{2} a+c^{2} b}{2 a b c+a^{2} b+a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,164 |
8. Let $a, b, c, d$ be positive real numbers, and satisfy $a+b+c+d=1$. Prove that: $6\left(a^{2}+b^{2}+c^{2}+\right.$ $\left.d^{3}\right) \geqslant\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+\frac{1}{8}$ (8th China Hong Kong Mathematical Olympiad: Junior Problem) | 8. It is easy to see that the original inequality is equivalent to
$$\begin{array}{l}
48\left(a^{3}+b^{3}+c^{3}+d^{3}\right) \geqslant \\
8\left(a^{2}+b^{2}+c^{2}+d^{2}\right)(a+b+c+d)+(a+b+c+d)^{3}
\end{array}$$
Expanding and combining like terms, we get
$$\begin{array}{l}
39\left(a^{3}+b^{3}+c^{3}+d^{3}\right) \geqs... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,167 |
10. Given that $a, b, c$ are positive numbers, prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}$ | 10. It is easy to know that the inequality to be proved is equivalent to
$$2\left(a^{3}+b^{3}+c^{3}\right) \geqslant a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b$$
By the AM-GM inequality, we have
$$\begin{array}{l}
\frac{a^{3}+a^{3}+b^{3}}{3} \geqslant a^{2} b \\
\frac{b^{3}+b^{3}+c^{3}}{3} \geqslant b^{2} c \\
\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,169 |
11. For all positive real numbers $a, b, c$, prove: $\frac{-a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1$ | 11. The original inequality is transformed into
$$\frac{1}{\sqrt{1+\frac{8 b c}{a^{2}}}}+\frac{1}{\sqrt{1+\frac{8 a c}{b^{2}}}}+\frac{1}{\sqrt{1+\frac{8 a b}{c^{2}}}} \geqslant 1$$
Let $\alpha=\frac{b c}{a^{2}}, \beta=\frac{c a}{b^{2}}, \gamma=\frac{a b}{c^{2}}$, clearly $\alpha, \beta, \gamma$ are positive real numbe... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,170 |
14. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive numbers, $\min \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}=a_{1}, \max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}=$ $a_{n}$, prove the inequality: $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant \frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}+\frac{1}{2}\left(a_{1}-... | 14. The inequality $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant \frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}+\frac{1}{2}\left(a_{1}-a_{n}\right)^{2}$ is equivalent to
$$\begin{array}{l}
2 n \sum_{i=1}^{n} a_{i}^{2}-n\left(a_{1}-a_{n}\right)^{2}-2\left(\sum_{i=1}^{n} a_{i}\right)^{2} \geqslant 0 \Leftrightarr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,173 |
15. Prove that for any positive real numbers $a, b, c$, we have $\frac{a^{3}}{b c}+\frac{b^{3}}{c a}+\frac{c^{3}}{a b} \geqslant a+b+c$. | 15. It is only necessary to prove that $a^{2}\left(a^{2}-b c\right)+b^{2}\left(b^{2}-c a\right)+c^{2}\left(c^{2}-a b\right) \geqslant 0$.
Observing that the left side of the above expression is symmetric, we can assume $a \geqslant b \geqslant c>0$, thus $a^{2}-b c \geqslant 0, c^{2}-a b \leqslant 0$, so
$$\begin{arra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,174 |
17. Let $x, y, z$ be real numbers, $k_{1}, k_{2}, k_{3} \in\left(0, \frac{1}{2}\right)$. And $k_{1}+k_{2}+k_{3}=1$, prove: $k_{1} k_{2} k_{3}$ $(x+y+z)^{2} \geqslant x y k_{3}\left(1-2 k_{3}\right)+y z k_{1}\left(1-2 k_{1}\right)+z x k_{2}\left(1-2 k_{2}\right)$. 1990 National Team Selection | 17. In $\frac{x}{k_{1}}, \frac{y}{k_{2}}, \frac{z}{k_{3}}$ these three numbers, we can assume $\frac{x}{k_{1}}$ is the largest.
$$\text { If } \frac{y}{k_{2}} \geqslant \frac{z}{k_{3}}, \text { let } a=\frac{x}{k_{1}}-\frac{y}{k_{2}}, b=\frac{y}{k_{2}}-\frac{z}{k_{3}} \text {. }$$
The original inequality can be transf... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,176 |
For every positive integer $n$, let $p_{n}=\left(1+\frac{1}{n}\right)^{n}, P_{n}=\left(1+\frac{1}{n}\right)^{n+1}, h_{n}=$ $\frac{2 p_{n} P_{n}}{p_{n}+P_{n}}$, prove: $h_{1}<h_{2}<\cdots<h_{n}<h_{n+1}$ (6th Putnam Mathematical Competition Problem) | It is easy to derive that $h_{n}=\frac{2(n+1)^{n+1}}{n^{n}(2 n+1)}$. Consider the function $g(x)$ defined by:
$$g(x)=\ln 2+(x+1) \ln (x+1)-x \ln x-\ln (2 x+1)$$
Then, for $0<x<+\infty$, we have
$$\begin{array}{c}
g^{\prime}(x)=\ln (x+1)-\ln x-\frac{2}{2 x+1} \\
g^{\prime \prime}(x)=\frac{1}{x+1}-\frac{1}{x}+\frac{4}{(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,177 |
18. Given that $a, b, c$ are positive real numbers, and $abc=1$, prove: $\frac{a}{(a+1)(b+1)}+$ $\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)} \geqslant \frac{3}{4} \cdot$(2006 French Training Team Problem) | 18. $\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)} \geqslant \frac{3}{4} \Leftrightarrow$
$$\begin{array}{l}
4[a(c+1)+b(a+1)+c(b+1)] \geqslant 3(a+1)(b+1)(c+1) \Leftrightarrow \\
4(a b+b c+c a)+(a+b+c) \geqslant 3(a b+b c+c a)+3(a+b+c)+ \\
3 a b c+3
\end{array}$$
Since $a b c=1$, the inequality (1) is... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,178 |
19. Given that $a, b$ are positive real numbers, prove: $\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}} \leqslant \sqrt[3]{2\left(1+\frac{b}{a}\right)\left(1+\frac{a}{b}\right)}$. | 19. Let $x=\sqrt[3]{\frac{a}{b}}, y=\sqrt[3]{\frac{b}{a}}$, then $x y=1$,
$$\begin{array}{l}
\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}} \leqslant \sqrt[3]{2\left(1+\frac{b}{a}\right)\left(1+\frac{a}{b}\right)} \Leftrightarrow \\
x+y \leqslant \sqrt[3]{2\left(1+x^{3}\right)\left(1+y^{3}\right) \Leftrightarrow} \\
(x+y)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,179 |
21. Given that $a, b, c$ are positive real numbers, and $abc=1$, prove: $\frac{a+3}{(a+1)^{2}}+\frac{b+3}{(b+1)^{2}}+$ $\frac{c+3}{(c+1)^{2}} \geqslant 3 .(2005$ Moldova Mathematical Olympiad Problem) | 21. $\frac{a+3}{(a+1)^{2}}+\frac{b+3}{(b+1)^{2}}+\frac{c+3}{(c+1)^{2}} \geqslant 3 \Leftrightarrow$
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+2\left[\frac{1}{(a+1)^{2}}+\frac{1}{(b+1)^{2}}+\frac{1}{(c+1)^{2}}\right] \geqslant 3$$
Because
$$\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}-\frac{1}{1+a b}=\frac{a b(a-b)^{2}+(a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,181 |
25. Let $a, b, c$ be positive numbers, prove: $\frac{a^{3}}{(a+b)^{3}}+\frac{b^{3}}{(b+c)^{3}}+\frac{c^{3}}{(c+a)^{3}} \geqslant \frac{3}{8}$ | 25. By the power mean inequality, we have
$$\sqrt[3]{\frac{x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}}{n}} \geqslant \sqrt{\frac{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}{n}}$$
Thus, it suffices to prove
$$\begin{array}{c}
\frac{a^{2}}{(a+b)^{2}}+\frac{b^{2}}{(b+c)^{2}}+\frac{c^{2}}{(c+a)^{2}} \geqslant \frac{3}{4} \\
\Leftrigh... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,185 |
26. Let $a_{1}, a_{2}, \cdots, a_{n}$ and $x_{1}, x_{2}, \cdots, x_{n}$ be two sets of positive numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1$, $x_{1}+x_{2}+\cdots+x_{n}=1$. Prove the inequality: $2 \sum_{i<j} x_{i} x_{j} \leqslant \frac{n-2}{n-1}+\sum_{i=1}^{n} \frac{a_{i} x_{i}^{2}}{1-a_{i}}$. And determine the conditions... | 26. To prove $2 \sum_{i<j} x_{i} x_{j} \leqslant \frac{n-2}{n-1}+\sum_{i=1}^{n} \frac{a_{i} x_{i}^{2}}{1-a_{i}}$, it is sufficient to prove $1-\sum_{i=1}^{n} x_{i}^{2} \leqslant \frac{n-2}{n-1}+$ $\sum_{i=1}^{n} \frac{a_{i} x_{i}^{2}}{1-a_{i}}$
which is equivalent to proving
$$\frac{1}{n-1} \leqslant \sum_{i=1}^{n} \fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,186 |
Example 20 Let $n \in \mathbf{N}, x_{0}=0, x_{i}>0, i=1,2,3 \cdots, n$. And $\sum_{i=1}^{n} x_{i}=1$, prove: $1 \leqslant$ $\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1+x_{0}+x_{1}+x_{2}+\cdots+x_{i-1}} \cdot \sqrt{x_{i}+\cdots+x_{n}}}<\frac{\pi}{2}$. (1996 China Mathematical Olympiad) | Prove: First, state a fact. Let $f(x)$ be a non-negative function defined on $[a, b]$ and monotonic increasing on $[a, b]$. Let $x_{1}, x_{2}, \cdots, x_{n}$ be the lengths of $n$ subintervals, which form a partition of $[a, b]$. Corresponding to each subinterval is a rectangle, and the sum of the areas of these rectan... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,189 |
29. Let $a, b, c \in\left[\frac{1}{2}, 1\right]$, prove the inequality: $2 \leqslant \frac{a+b}{1+c}+\frac{b+c}{1+a}+\frac{c+a}{1+b} \leqslant 3$. | 29. $2 \leqslant \frac{a+b}{1+c}+\frac{b+c}{1+a}+\frac{c+a}{1+b} \leqslant 3$ is equivalent to $2 \leqslant\left(\frac{a}{1+c}+\frac{c}{1+a}\right)+\left(\frac{b}{1+a}+\right.$ $\left.\frac{a}{1+b}\right)+\left(\frac{c}{1+b}+\frac{b}{1+c}\right) \leqslant 3$
If under the condition $a, b, c \in\left[\frac{1}{2}, 1\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,191 |
31. Let $x, y, z$ be positive numbers, prove the inequality: $\frac{1}{x^{2}+y z}+\frac{1}{y^{2}+z x}+\frac{1}{z^{2}+x y} \leqslant \frac{1}{2}\left(\frac{1}{x y}+\right.$ $\left.\frac{1}{y z}+\frac{1}{z x}\right) \cdot(2006$ Romanian Mathematical Olympiad Problem) | $$\begin{array}{l}
\quad 31 . \frac{1}{x^{2}+y z}+\frac{1}{y^{2}+z x}+\frac{1}{z^{2}+x y} \leqslant \frac{1}{2}\left(\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}\right) \Leftrightarrow \frac{x y z}{x^{2}+y z}+\frac{x y z}{y^{2}+z x}+ \\
\frac{x y z}{z^{2}+x y} \leqslant \frac{x+y+z}{2}
\end{array}$$
By the AM-GM inequali... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,193 |
34. Let $a, b, c$ be positive real numbers, prove that: $\sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c})+(a+b+c)^{2} \geqslant$ $4 \sqrt{3 a b c(a+b+c)} \cdot$ (2004 China National Training Team Problem) | 34. By substituting $x=\sqrt{a}, y=\sqrt{b}, z=\sqrt{c}$, the original inequality becomes
$$x y z(x+y+z)+\left(x^{2}+y^{2}+z^{2}\right)^{2} \geqslant 4 x y z \sqrt{3\left(x^{2}+y^{2}+z^{2}\right)}$$
Expanding, it suffices to prove
$$\begin{array}{l}
x^{4}+y^{4}+z^{4}+2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,196 |
36. Prove: For any positive real numbers $a, b, c$, we have $\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geqslant 9(a b+$ $b c+c a)$. | 36. To prove $\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geqslant 9(a b+b c+c a)$.
That is to prove
$$a^{2} b^{2} c^{2}+2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+4\left(a^{2}+b^{2}+c^{2}\right)+8 \geqslant 9(a b+b c+c a)$$
By the AM-GM inequality, we have
$$\begin{array}{l}
a^{2}+b^{2} \geq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,198 |
37. Positive real numbers $x, y, z$ satisfy $x y z \geqslant 1$, prove: $\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+$ $\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \geqslant 0$. (46th IMO problem) | $$\begin{array}{l}
\text { 37. } \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 \frac{2 x^{2}-y^{2}-z^{2}}{x^{2}+y^{2}+z^{2}}+ \\
\frac{2 x^{2}-y^{2}-z^{2}}{x^{2}+y^{2}+z^{2}}+\frac{2 y^{2}-x^{2}-z^{2}}{x^{2}+y^{2}+z^{2}}+\frac{2 z^{2}-x^{2}-y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,199 |
1. Given $a, b, c \in(-1,1)$, prove: $a b c+2>a+b+c$. | 1. Treat $a$ as a variable, $b, c$ as constants, and let $f(a)=a b c+2-(a+b+c)=(b c-1) a+2-b-c, a \in(-1,1)$.
Now, we only need to prove $f(a)>0$ based on the monotonicity of a linear function, since $b, c \in(-1, 1)$, so $b c-1f(1)=1-b-c+b c=$ $(1-b)(1-c)>0$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,200 |
38. Let $a, b, c$ be real numbers, prove that: $\sqrt{2\left(a^{2}+b^{2}\right)}+\sqrt{2\left(b^{2}+c^{2}\right)}+\sqrt{2\left(c^{2}+a^{2}\right)} \geqslant$ $\sqrt{3(a+b)^{2}+3(b+c)^{2}+3(c+a)^{2}}$. (2004 Polish Mathematical Olympiad) | 38 .
$$\begin{array}{l}
\sqrt{2\left(a^{2}+b^{2}\right)}+\sqrt{2\left(b^{2}+c^{2}\right)}+\sqrt{2\left(c^{2}+a^{2}\right)} \geqslant \\
\sqrt{3(a+b)^{2}+3(b+c)^{2}+3(c+a)^{2}} \Leftrightarrow \\
2\left[\sqrt{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)}+\sqrt{\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)}+\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,201 |
41. Given $x, y, z \geqslant 0$, prove: $[(x+y+z) x-y z](y+z)^{2}+[(x+y+z) y- z x](z+x)^{2}+[(x+y+z) z-x y](x+y)^{2} \leqslant(x+y+z)(x+y)(y+z)(z+x)$. | 41.
$$\begin{array}{l}
{[(x+y+z) x-y z](y+z)^{2}+[(x+y+z) y-z x](z+x)^{2}+} \\
{[(x+y+z) z-x y](x+y)^{2} \leqslant(x+y+z)(x+y)(y+z)(z+x) \Leftrightarrow} \\
(x+y+z)\}\left[x(y+z)^{2}+y(z+x)^{2}+z(x+y)^{2}\right] \\
(x+y)(y+z)(z+x) \leqslant y z(y+z)^{2}+z x(z+x)^{2}+x y(x+y)^{2} \Leftrightarrow \\
4 x y z(x+y+z) \leqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,204 |
42. Let $a, b, c$ be positive numbers, and $a^{2}+b^{2}+c^{2}=1$, prove: $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant$ $\frac{2\left(a^{3}+b^{3}+c^{3}\right)}{a b c}+3 .($ Crux Problem 2532) | $$\begin{array}{l}
42 \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant \frac{2\left(a^{3}+b^{3}+c^{3}\right)}{a b c}+3 \Leftrightarrow \\
\frac{a^{2}+b^{2}+c^{2}}{a^{2}}+\frac{a^{2}+b^{2}+c^{2}}{b^{2}}+\frac{a^{2}+b^{2}+c^{2}}{c^{2}} \geqslant \frac{2\left(a^{3}+b^{3}+c^{3}\right)}{a b c}+3 \Leftrightarrow \\
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,205 |
2. Let $0 \leqslant a, b, c \leqslant 1$, prove that: $\frac{a}{b c+1}+\frac{b}{c a+1} \pm \frac{c}{a b+1} \leqslant 2$. | 2. Since $0 \leqslant a, b, c \leqslant 1$, we have
$$\frac{a}{b c+1}+\frac{b}{c a+1}+\frac{c}{a b+1} \leqslant \frac{a}{a b c+1}+\frac{b}{a b c+1}+$$
$$\frac{c}{a b c+1} \leqslant \frac{a+b+c}{a b c+1}$$
It suffices to prove that $\frac{a+b+c}{a b c+1} \leqslant 2$, which is equivalent to proving $a+b+c \leqslant 2(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,211 |
49. Let $x, y, z \geqslant 0$, and $x+y+z=1$, prove: $3 \leqslant \frac{1}{1-x y}+\frac{1}{1-y z}+\frac{1}{1-z x} \leqslant \frac{27}{8}$. | 49. Since $x, y, z \geqslant 0$, and $x+y+z=1$, therefore, $0 \leqslant xy, yz, zx < 1$, thus, $\frac{1}{1-xy} + \frac{1}{1-yz} + \frac{1}{1-zx} \geqslant 3$. Below, we prove that $\frac{1}{1-xy} + \frac{1}{1-yz} + \frac{1}{1-zx} \leqslant \frac{27}{8}$. Noting that $x+y+z=1$, the inequality is equivalent to:
$$\begin{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,213 |
50. Let $x, y, z$ be positive numbers, and $x+y+z=1$, prove: $\sqrt{3 x y z}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{1-x}+\right.$ $\left.\frac{1}{1-y}+\frac{1}{1-z}\right) \geqslant 4+\frac{4 x y z}{(1-x)(1-y)(1-z)} \cdot(2004$ Serbian Mathematical Olympiad) | 50. Considering the necessary and sufficient condition for the equality in the inequality is \(a=b=c=\frac{1}{3}\), we have, by the AM-GM inequality,
$$x y+x y+x y+\frac{1}{9}+y z+y z+y z+\frac{1}{9}+z x+z x+z x+\frac{1}{9} \geqslant 12 \sqrt[12]{\frac{x^{6} y^{6} z^{6}}{9^{3}}}$$
i.e., \(\square\)
$$\begin{array}{c}
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,214 |
51. Given $a, b, c > 0$, prove: $\sqrt{\left(a^{2} b+b^{2} c+c^{2} a\right)\left(a b^{2}+b c^{2}+c a^{2}\right)} \geqslant a b c + \sqrt[3]{\left(a^{3}+a b c\right)\left(b^{3}+a b c\right)\left(c^{3}+a b c\right)}$. (2001 Korean Mathematical Olympiad) | 51 .
$$\begin{array}{l}
\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)} \Leftrightarrow \\
\sqrt{\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)\left(\frac{c}{a}+\frac{a}{b}+\frac{b}{c}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,215 |
52. Let non-negative real numbers $a, b, c$ satisfy $a b + b c + c a = 1$, prove that $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \geqslant \frac{5}{2}$. | 52. Since $a b+b c+c a=1$, by homogenization we get $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \geqslant \frac{5}{2} \Leftrightarrow(a b+$
$$\begin{array}{l}
b c+c a)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)^{2} \geqslant \frac{25}{4} \Leftrightarrow 4 \sum_{s y m} a^{5} b+\sum_{s m} a^{4} b c+14 \sum_{s m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,216 |
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