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
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Example 2 Real numbers $a, b, c$ and a positive number $\boldsymbol{\lambda}$ make $f(x)=x^{3}+a x^{2}+b x+c$ have three real numbers $x_{1}, x_{2}, x_{3}$, and satisfy
(1) $x_{2}-x_{1}=\lambda$;
(2) $x_{3}>\frac{1}{2}\left(x_{1}+x_{2}\right)$;
Find: $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}$'s maximum value. (2002 Nati... | Let $x_{1}=m-\frac{\lambda}{2}, x_{2}=m+\frac{\lambda}{2}, x_{3}>\frac{1}{2}\left(x_{1}+x_{2}\right)=m$, let $x_{3}=m+t,(\lambda >0, t>0)$.
By Vieta's formulas, we have
$$\left\{\begin{array}{l}
a=-\left(x_{1}+x_{2}+x_{3}\right)=-(3 m+t) \\
b=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=3 m^{2}+2 m t-\frac{\lambda^{2}}{4} \\
c... | \frac{3 \sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,370 |
Example 5 For a fixed $\theta \in\left(0, \frac{\pi}{2}\right)$, find the smallest positive number $a$ that satisfies the following two conditions:
( i ) $\frac{\sqrt{a}}{\cos \theta}+\frac{\sqrt{a}}{\sin \theta}>1$;
( ii ) There exists $x \in\left[1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\right]$, s... | From (i) we get
$$\sqrt{a}>\frac{\sin \theta \cos \theta}{\sin \theta+\cos \theta}$$
Without loss of generality, assume
$$\frac{a}{\sin ^{2} \theta}+\frac{a}{\cos ^{2} \theta} \leqslant 1$$
(ii) is equivalent to: there exists $x \in\left[1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\right]$, such that $... | \frac{3 \sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,373 |
Example 7 Let $x, y, z$ be positive numbers, and $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $\frac{x^{5}}{y^{2}+z^{2}-y z}+\frac{y^{5}}{z^{2}+x^{2}-z x}+$ $\frac{z^{5}}{x^{2}+y^{2}-x y}$. | By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(-\frac{x^{6}}{x y^{2}+x z^{2}-x y z}+\frac{y^{6}}{z^{2} y+x^{2} y-x y z}+\frac{z^{6}}{x^{2} z+y^{2} z-x y z}\right) \\
\left(x y^{2}+x z^{2}+z^{2} y+x^{2} y+x^{2} z+y^{2} z-3 x y z\right) \geqslant \\
\left(x^{3}+y^{3}+z^{3}\right)^{2}
\end{array}$$
By... | \frac{\sqrt{3}}{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,375 |
Example 8 Given an integer $n > 3$, let real numbers $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}, x_{n+2}$ satisfy $0<x_{1}<x_{2}$ $<\cdots<x_{n}<x_{n+1}<x_{n+2}$. Try to find the minimum value of $\left.\frac{\left(\sum_{i=1}^{n} \frac{x_{i+1}}{x_{i}}\right)\left(\sum_{j=1}^{n} \frac{x_{j+2}}{x_{j+1}}\right)}{x_{k+1} x_{k+2... | Let $(\bar{I})$ denote $t_{i}=\frac{x_{i+1}}{x_{i}}(>1), 1 \leqslant i \leqslant n$, the expression in the problem can be written as
$$\frac{\left(\sum_{i=1}^{n} t_{i}\right):\left(\sum_{i=1}^{n} t_{i+1}\right)}{\left(\sum_{i=1}^{n} \frac{t_{i} t_{i+1}}{t_{i}+t_{i+1}}\right)=\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\righ... | 1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,376 |
17. Prove the inequality: $\frac{1}{3}<\sin \frac{\pi}{9}<\frac{7}{20}$. | 17. Let $\sin \frac{\pi}{9}=x$, then $\sin \frac{\pi}{3}=3 \sin \frac{\pi}{9}-4 \sin ^{3} \frac{\pi}{9}$, thus $4 x^{3}-3 x+\frac{\sqrt{3}}{2}=0$, let $f(x)=4 x^{3}-3 x+\frac{\sqrt{3}}{2}$, then $f(-1)=-1+\frac{\sqrt{3}}{2}<0, f\left(\frac{1}{3}\right)=\frac{4}{27}-1+\frac{\sqrt{3}}{2}>0, f\left(\frac{7}{20}\right)=4 \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,377 |
Example 10 Determine the smallest real number $M$ such that the inequality
$$\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2}$$
holds for all real numbers $a, b, c$. (47th IMO Problem) | Solving because
$$\begin{array}{l}
a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)= \\
a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left[\left(c^{2}-b^{2}\right)+\left(b^{2}-a^{2}\right)\right]= \\
\left(a^{2}-b^{2}\right)(a b-c a)+\left(b^{2}-c^{2}\right)(b c-c a)= \\... | \frac{9 \sqrt{2}}{32} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,379 |
1. $x, y, z$ are positive real numbers, and satisfy $x^{4}+y^{4}+z^{4}=1$, find the minimum value of $\frac{x^{3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}+\frac{z^{3}}{1-z^{8}}$. (1999 Jiangsu Province Mathematical Winter Camp Problem) | 1. $\frac{x^{3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}+\frac{z^{3}}{1-z^{8}} \geqslant \frac{x^{4}}{x\left(1-x^{8}\right)}+\frac{y^{4}}{y\left(1-y^{8}\right)}+\frac{z^{4}}{z\left(1-z^{8}\right)}$.
Let $f(u)=u\left(1-u^{8}\right), u \in(0,1)$, we find the maximum value of $f(u)$. By $A_{9} \geqslant G_{9}$ we get $8=8 u^{8}+\... | \frac{9}{8} \cdot \sqrt[4]{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,380 |
2. Let $x, y, z$ be positive real numbers, and $x y z + x + z = y$, find the maximum value of $P = \frac{2}{x^{2}+1} - \frac{2}{y^{2}+1} + \frac{3}{z^{2}+1}$. (1999 Vietnam Mathematical Olympiad) | 2. It is easy to prove that $a c \neq 1$. From $x y z + x + z = y$, we can deduce that $b = \frac{a + c}{1 - a c}$. Let $a = \tan \alpha, b = \tan \beta, c = \tan \gamma, \alpha, \beta, \gamma \in \left(0, \frac{\pi}{2}\right)$, then $\tan \beta = \frac{\tan \alpha + \tan \gamma}{1 - \tan \alpha \tan \gamma} = \tan (\a... | \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,381 |
4. Let the function $y=f(x)$ have a domain of $\mathbf{R}$, and for $x>1$, and for any real numbers $x, y \in \mathbf{R}$, $f(x+y)=f(x) f(y)$ holds. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=f(0)$, and $f\left(a_{n+1}\right)=\frac{1}{f\left(-2-a_{n}\right)}(n \in \mathbf{N})$. If the inequality $\left(1+\fra... | 4. Let $x=-1, y=0$, we get
$$f(-1)=f(-1) f(0)=f(0) \equiv 1$$
Thus $a_{1}=f(0)=1$.
When $x>0$, $-x<0$
\end{aligned}$$
Therefore, $f\left(x_{1}\right)>f\left(x_{2}\right)$, the function $y=f(x)$ is monotonically decreasing on $\mathbf{R}$.
From $f\left(a_{n+1}\right)=\frac{1}{f\left(-2-a_{n}\right)}$, we get
$$f\left(... | \frac{2}{3} \sqrt{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,383 |
5. Given the inequality $\sqrt{2}(2 a+3) \cos \left(\theta-\frac{\pi}{4}\right)+\frac{6}{\sin \theta+\cos \theta}-2 \sin 2 \theta<3 a+6$, for $\theta \in\left[0, \frac{\pi}{2}\right]$ to always hold. Find the range of $\theta$. (1st China Southeast Mathematical Olympiad) | 5. Let $\sin \theta+\cos \theta=x$, then $\cos \left(\theta-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} x, \sin 2 x=x^{2}-1, x \in[1, \sqrt{2}]$, thus the original inequality can be transformed into $(2 a+3) x+\frac{6}{x}-2\left(x^{2}-1\right)0$
Rearranging, we get
$$(2 x-3)\left(x+\frac{2}{x}-a\right)>0, x \in[1, \sqrt{2... | a>3 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,384 |
6. Let $a$ and $b$ be real numbers such that the equation $x^{4} + a x^{3} + b x^{2} + a x + 1 = 0$ has at least one real root. For all such pairs of real numbers $(a, b)$, find the minimum value of $a$. (15th IMO Problem) | 6. First consider the equation
$$x+\frac{1}{x}=y$$
where $y$ is a real number. This equation can be written as a quadratic equation in $x$:
$$x^{2}-y \bar{x}+1=0$$
It has real roots if and only if its discriminant is greater than or equal to 0, i.e.,
$$y^{2}-4 \geqslant 0,|y| \geqslant 2$$
The original equation can ... | -2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,385 |
7. Find the maximum value of the positive real number $M$ such that for all $x, y, z \in \mathbf{R}$, the inequality $x^{4}+y^{4}+z^{4}+$ $x y z(x+y+z) \geqslant M(x y+y z+z x)^{2}$ holds. (2004 Greek Mathematical Olympiad) | 7. When $x=y=z$, the inequality becomes $6 x^{4} \geqslant M \cdot 9 x^{4}, x \in \mathbf{R}$, i.e., $M \leqslant \frac{2}{3}$. We will now prove that the maximum value of $M$ is $\frac{2}{3}$.
$$\begin{array}{l}
\quad \text { For } x, y, z \in \mathbf{R}, x^{4}+y^{4}+z^{4}+x y z(x+y+z) \geqslant \frac{2}{3}(x y+y z+z ... | \frac{2}{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,386 |
8. Let $n$ be a positive integer. For positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have $a_{1}+a_{2}+\cdots+a_{n}=1$. Let $A$ be the minimum of the following $n$ numbers $\frac{a_{1}}{1+a_{1}}, \frac{a_{2}}{1+a_{1}+a_{2}}, \cdots, \frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n}}$. Find the maximum value of $A$ when ... | 8. From $A=\min _{1 \leqslant i \leqslant n}\left\{\frac{a_{i}}{1+\sum_{k=1}^{i} a_{k}}\right\}$, we know that $1-A=\max _{1 \leqslant i \leqslant n}\left\{\frac{1+\sum_{k=1}^{i-1}}{1+\sum_{k=1}^{i} a_{k}}\right\} \geqslant \sqrt[n]{\prod_{i=1}^{n} \frac{1+\sum_{k=1}^{i-1}}{1+\sum_{k=1}^{i} a_{k}}}=$ $\sqrt[n]{\frac{1}... | 1-\frac{1}{\sqrt[n]{2}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,387 |
18. Given that $a, b, c$ are positive numbers, prove:
(1) $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}$. (1963 Moscow Mathematical Olympiad Problem)
(2) $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{a+b+c}{2}$. (2nd World Friendship Cup Mathematics Competition Problem) | 18. (1) In $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$, replace $a, b, c$ with $\frac{a}{a+b+c}, \frac{b}{a+b+c}$, and $\frac{c}{a+b+c}$ respectively, and assume without loss of generality that $a+b+c=1$,
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}$$
Consider $f(x)=\frac{x}{1-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,388 |
9. Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}^{+}$, define $S_{n}=\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)^{2}$.
(1) Find the minimum value of $S_{n}$;
(2) Under the condition $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$, find the minimum value of $S_{n}$;
(3) Under the condition $x_{1}... | 9. (1) $S_{n}=\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)^{2} \geqslant \sum_{i=1}^{n}\left(2 \sqrt{x_{i} \cdot \frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}}\right)^{2}=$ $4 \sum_{i=1}^{n} \frac{n-1}{n^{2}}=\frac{4(n-1)}{n}$
Therefore, when $x_{i}=\frac{\sqrt{n-1}}{n}$, $S_{n}$ reaches its mini... | \left(1\right) \frac{4(n-1)}{n}, \left(2\right) \left(2-\frac{1}{n}\right)^{2}, \left(3\right) n | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,389 |
10. Non-negative numbers $a$ and $d$, and positive numbers $b$ and $c$ satisfy the condition $b+c \geqslant a+d$. What is the minimum value that the expression $\frac{b}{c+d} + \frac{c}{a+b}$ can take? (1988 22nd All-Soviet Union Mathematical Olympiad) | 10. Let's assume $a+b \geqslant c+d$.
Since $\frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}-c\left(\frac{1}{c+d}-\frac{1}{a+b}\right), b+c=\frac{1}{2}(b+c+b+c) \geqslant$ $\frac{1}{2}(a+b+c+d), c \leqslant c+d$, thus $\frac{b}{c+d}+\frac{c}{a+b} \geqslant \frac{1}{2} \cdot \frac{a+b+c+d}{c+d}-(c+$ d) $\left(\frac{1}{c+d}... | \sqrt{2}-\frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,390 |
12. Given that $x, y$ are real numbers, and $x+y=1$, find the maximum value of $\left(x^{3}+1\right)\left(y^{3}+1\right)$. | 12. Since $x+y=1$, then $\left(x^{3}+1\right)\left(y^{3}+1\right)=(x y)^{3}+x^{3}+y^{3}+1=(x y)^{3}+$
$$\begin{array}{l}
(x+y)\left(x^{2}-x y+y^{2}\right)+1=(x y)^{3}+(x+y)\left[(x+y)^{2}-3 x y\right]+1=(x y)^{3}- \\
3 x y+2=(x y-1)^{2}(x y+2)=\frac{1}{2}(1-x y)(1-x y)(2 x y+4) \leqslant \\
\frac{1}{2}\left(\frac{(1-x ... | 4 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,391 |
13. Let $a, b>0$, find the largest positive integer $c$ such that for any positive real number $x$, we have $c \leqslant \max \{a x+ \left.\frac{1}{a x}, b x+\frac{1}{b x}\right\}$. (2003 Irish Mathematical Olympiad Problem) | 13. Let $f(t)=t+\frac{1}{t}, t>0, g(x)=\max \{f(a x), f(b x)\}$. The problem is to find the minimum value of $g(x)$ for $x>0$.
If $a=b$, then $g(x)=a x+\frac{1}{a x} \geqslant 2$, with equality when $x=\frac{1}{a}$, in which case the desired $c=2$.
If $a \neq b$, without loss of generality, assume $0<f(t)$.
Therefore... | \sqrt{\frac{b}{a}}+\sqrt{\frac{a}{b}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,392 |
14. Let $x_{i} \geqslant 0, i=1,2, \cdots, n$, and $\sum_{i=1}^{n} x_{i}=1, n \geqslant 2$. Find the maximum value of $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)$. (32nd IMO Shortlist) | 14. The maximum value of $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)$ is $\frac{1}{4}$. The proof is as follows:
$$\begin{array}{l}
x_{1} x_{2}\left(x_{1}+x_{2}\right)+x_{1} x_{3}\left(x_{1}+x_{3}\right)+\cdots+ \\
x_{1} x_{n}\left(x_{1}+x_{n}\right)+x_{2} x_{3}\left(x_{2}+x_{3}\right)+\cdot... | \frac{1}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,393 |
15. Let \(a, b \in [0,1]\), find the maximum and minimum values of \(S = \frac{a}{1+b} + \frac{b}{1+a} + (1-a)(1-b)\). (2006 Shanghai High School Mathematics Competition Problem) | $$\begin{array}{l}
\quad \text { 15. } S=\frac{a}{1+b}+\frac{b}{1+a}+(1-a)(1-b)=\frac{1+a+b+a^{2} b^{2}}{(1+a)(1+b)}=1- \\
\frac{a b(1-a b)}{(1+a)(1+b)} \leqslant 1
\end{array}$$
When $a b=0$ or $a b=1$, the equality holds, so the maximum value of $S$ is 1.
$$\begin{array}{l}
\text { Let } T=\frac{a b(1-a b)}{(1+a)(1+... | \frac{11-5 \sqrt{5}}{2} \text{ and } 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,394 |
16. Given a positive integer $n$, find the smallest positive number $\lambda$, such that for any $\theta_{i}^{-} \in\left(\theta, \frac{\pi}{2}\right)(i=1$, $2, \cdots, n)$, if $\tan \theta_{1} \tan \theta_{2} \cdots \tan \theta_{n}=\sqrt{2^{n}}$, then $\cos \theta_{1}+\cos \theta_{2}+\cdots+\cos \theta_{n}$ is not gre... | When $n=1$, $\cos \theta_{1}=\frac{1}{\sqrt{1+\tan ^{2} \theta_{1}}}=\frac{\sqrt{3}}{3}$, so $\lambda=\frac{\sqrt{3}}{3}$.
When $n=2$, it can be proven that
$$\cos \theta_{1}+\cos \theta_{2} \leqslant \frac{2 \sqrt{3}}{3}$$
and the equality holds when $\theta_{1}=\theta_{2}=-\arctan \sqrt{2}$. In fact, equation (1) $... | n-1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,395 |
17. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, satisfying $\sum_{i=1}^{n} x_{i}^{2}=1$, find the minimum value of $\frac{x_{1}^{5}}{x_{2}+x_{3}+\cdots+x_{n}}+$ $\frac{x_{2}^{5}}{x_{3}+\cdots+x_{n}+x_{1}}+\cdots+\frac{x_{n}^{5}}{x_{1}+x_{2}+\cdots+x_{n-1}}$. (2006 Turkish National Team Selection) | $$\begin{array}{l}
\text { 17. From } \sum_{i \neq j}\left(x_{i}-x_{j}\right)^{2}=(n-1) \sum_{i=1}^{n} x_{i}^{2}-2 \sum_{i \neq j} x_{i} x_{j}=(n-1)-2 \sum_{i \neq j} x_{i} x_{j} \text { we get } \\
\sum_{i=1}^{n} x_{i} \sum_{j \neq i} x_{j} \leqslant n-1 \text { and by the Cauchy-Schwarz inequality, we have }\left(\fr... | \frac{1}{n(n-1)} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,396 |
18. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, satisfying $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}=n$, find the minimum value of $x_{1}+\frac{x_{2}^{2}}{2}+$ $\frac{x_{3}^{2}}{3}+\cdots+\frac{x_{n}^{n}}{n}$. (1995 Polish Mathematical Olympiad Problem) | 18. From the known $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}=n \geqslant n \sqrt[n]{\frac{1}{x_{1}} \cdot \frac{1}{x_{2}} \cdots \cdots \cdot \frac{1}{x_{n}}}$, we get $x_{1} x_{2} x_{3} \cdots x_{n} \geqslant 1$. To $\frac{x_{1}+\frac{x_{2}^{2}}{2}+\frac{x_{3}^{2}}{3}+\cdots+\frac{x_{n}^{n}}{n}}{1+\frac{... | 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,397 |
19. Let the number $a$ have the following property: For any four real numbers $x_{1}, x_{2}, x_{3}, x_{4}$, there always exist integers $k_{1}, k_{2}, k_{3}, k_{4}$, such that $\sum_{1 \leqslant i<j \leqslant 4}\left(\left(x_{i}-k_{i}\right)-\left(x_{j}-k_{j}\right)\right)^{2} \leqslant a$, find the minimum value of su... | 19. First, let $k_{i}=\left[x_{i}\right], \alpha_{i}=x_{i}-k_{i}$, then $\alpha_{i} \in[0,1), i=1,2,3,4$. Let $\left\{\alpha^{\prime}{ }_{1}, \alpha^{\prime}{ }_{2}, \alpha^{\prime}{ }_{3}, \alpha^{\prime}{ }_{4}\right\}$ be a permutation of $\left\{\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right\}$ in ascending o... | \frac{5}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,398 |
19. Find real numbers $A, B, C$ such that for any real numbers $x, y, z$,
$$A(x-y)(x-z)+B(y-z)(y-x)+C(z-x)(z-y) \geqslant 0$$
is satisfied.
(1988 China National Training Team Selection Examination Problem). | 19. In equation (1), let $x=y \neq z$, we get $C(z-x)^{2} \geqslant 0$, thus $C \geqslant 0$. By symmetry, we have
$$A \geqslant 0, B \geqslant 0, C \geqslant 0$$
Let $s=x-y, t=y-z$, then $x-z=s+t$, so (1) is equivalent to
$$A s(s+t)-B s t+C t(s+t) \geqslant 0$$
i.e., $\square$
$$A s^{2}+(A-B+C) s t+C t^{2} \geqslant... | A^{2}+B^{2}+C^{2} \leqslant 2(A B+B C+C A) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,399 |
20. Given 5 real numbers $u_{0}, u_{1}, u_{2}, u_{3}, u_{4}$, prove that there always exist 5 real numbers $v_{0}, v_{1}, v_{2}, v_{3}, v_{4}$ satisfying the following conditions:
(1) $u_{i}-v_{i} \in \mathbf{N}^{*}$;
(2) $\sum_{0 \in i<j \leqslant 4}\left(v_{i}-v_{j}\right)^{2}<4$. (28th IMO Shortlist) | 20. First prove the following inequality: for any real number $v$,
$$\sum_{0 \leqslant i<j \leqslant 4}\left(v_{i}-v_{j}\right)^{2} \leqslant 5 \sum_{i=0}^{4}\left(v_{i}-v\right)^{2}$$
In fact,
$$\begin{aligned}
\sum_{0 \leqslant i<j \leqslant 4}\left(v_{i}-v_{j}\right)^{2}= & \sum_{0 \leqslant i<j \leqslant 4}\left[\... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,400 |
21. (1) Given that $a, b, c, d, e, f$ are real numbers, and satisfy the following two equations: $a+b+c+d+e+f=10, (a-1)^{2}+(b-1)^{2}+(c-1)^{2}+(d-1)^{2}+(e-1)^{2}+(f-1)^{2}=6$, find the maximum value of $f$. (1993 Balkan Mathematical Olympiad)
(2) Given that $a, b, c, d \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,... | 21. (1) From $a+b+c+d+e+f=10, (a-1)^{2}+(b-1)^{2}+(c-1)^{2}+$ $(d-1)^{2}+(e-1)^{2}+(f-1)^{2}=6$, we get
$$\begin{array}{c}
a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}=20 \\
a+b+c+d+e+f=10
\end{array}$$
By the Cauchy-Schwarz inequality, we have
$$5\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\right) \geqslant(a+b+c+d+e)^{2}$$
That is, ... | \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,401 |
22. Let $n$ be a given positive integer, $n \geqslant 2, a_{1}, a_{2}, \cdots, a_{n} \in(0,1)$, find the maximum value of $\sum_{i=1}^{n} \sqrt[6]{a_{i}\left(1-a_{i}\right)}$, where $a_{n+1}=a_{1}$. (2006 China Western Mathematical Olympiad) | 22. By the AM-GM inequality, we have
$$\begin{aligned}
\sqrt[6]{a_{i}\left(1-a_{i+1}\right)}= & 2^{\frac{4}{6}} \sqrt[6]{a_{i}\left(1-a_{i+1}\right) \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}} \leqslant \\
& 2^{\frac{2}{3}} \cdot \frac{1}{6}\left(a_{i}+\left(1-a_{i+1}\right)+\frac{1}{2}+\fr... | \frac{n}{\sqrt[3]{2}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,402 |
23. Given $n \in \mathbf{N}$ and $a \in[0, n]$, under the condition $\sum_{i=1}^{n} \sin ^{2} x_{i}=a$, find the maximum value of $\sum_{i=1}^{n} \sin 2 x_{i} \mid$. (1983 Czechoslovak Mathematical Olympiad Problem) | 23. Since $\sum_{i=1}^{n} \sin ^{2} x_{i}=a$, then $\sum_{i=1}^{n} \cos 2 x_{i}=\sum_{i=1}^{n}\left(1-2 \sin ^{2} x_{i}\right)=n-2 a$. Consider $n$ unit vectors in the plane $\left(\cos -2 x_{i}, \sin 2 x_{i}\right), i=1,2, \cdots, n$. By the Cauchy-Schwarz inequality, the length of their sum does not exceed $n$, i.e.,... | 2 \sqrt{a(n-a)} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,403 |
24. Let $a, b, c>0$, and $a+b+c=1$, find the range of $\sqrt{\frac{a b}{a b+c}}+\sqrt{\frac{b c}{b c+a}}+\sqrt{\frac{c a}{c a+b}}$. (2006 Jiangsu Province Mathematical Olympiad Winter Camp Problem) | $$\begin{array}{l}
\text { 24, from } a+b+c=1 \text { we get } \\
\sqrt{\frac{a b}{a b+c}}+\sqrt{\frac{b c}{b c+a}}+\sqrt{\frac{-c a}{c a+b}}=\sqrt{\frac{a b}{(a+c)(b+c)}}+\sqrt{\frac{b c}{(a+b)(a+c)}}+ \\
\sqrt{\frac{c a}{(a+b)(b+c)}}
\end{array}$$
It can be proven that
$$\begin{array}{l}
\sqrt{\frac{a b}{(a+c)(b+c)}... | \left(0, \frac{3}{2}\right] | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,404 |
25. Let $a, b, c$ be positive numbers, and let $d$ be the minimum of $(a-b)^{2},(b-c)^{2},(c-a)^{2}$.
(1) Prove that there exists $\lambda(0<\lambda<1)$, such that
$$d \leqslant \lambda\left(a^{2}+b^{2}+c^{2}\right)$$
(2) Find the smallest positive number $\lambda$ for which the inequality (1) holds and provide a proof... | 25. (1) By the definition of $d$, we know that $d \leqslant(a-b)^{2}, d \leqslant(b-c)^{2}, d \leqslant(c-a)^{2}$. Adding these three inequalities, we get
$$\begin{aligned}
-3 d \leqslant & (a-b)^{2}+(b-c)^{2}+(c-a)^{2}= \\
& 2\left(a^{2}+b^{2}+c^{2}\right)-2(a b+b c+c a)
\end{aligned}$$
If $\frac{a+c}{2} \leqslant 2 b... | \frac{1}{5} | Inequalities | proof | Yes | Yes | inequalities | false | 733,405 |
26. Let $x_{1}, x_{2}, x_{3}, x_{4}$ all be positive numbers, and $x_{1}+x_{2}+x_{3}+x_{4}=\pi$, find the minimum value of the expression $\left(2 \sin ^{2} x_{1}+\right.$ $\left.\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right)\left(2 \sin ^{2} x_{3}+\frac{1}{\sin ^{2} x_{3}}\rig... | 26. By the AM-GM inequality, we have
$$2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x i}=2 \sin ^{2} x_{i}+\frac{1}{2 \sin ^{2} x_{i}}+\frac{1}{2 \sin ^{2} x_{i}} \geqslant 3 \sqrt[3]{\frac{1}{2 \sin ^{2} x_{i}}}$$
Therefore,
$$\prod_{i=1}^{4}\left(2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}\right) \geqslant 81\left(4 \sin x_... | 81 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,406 |
27. Let $0<p \leqslant x_{i} \leqslant q\left(i=1,2, \cdots, n, p, q\right.$ be given constants), try to find the maximum value of $F\left(x_{4}, x_{2}, \cdots\right.$, $\left.x_{n}\right)=\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}\right)$. $(n=5$ is from the 1977 ... | 27. (1) First prove: when $x_{i}(i=1,2, \cdots, n-1)$ are fixed, $x_{n}=p$ (or $\left.q\right), F\left(x_{1}\right.$, $\left.x_{2}, \cdots, x_{n}\right)$ is maximized.
In fact, let $u=x_{1}+x_{2}+\cdots+x_{n-1}, v=\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n-1}}$,
since $0<p \leqslant x_{i} \leqslant q(i=1,2, ... | 25 + 6\left(\sqrt{\frac{p}{q}} - \sqrt{\frac{q}{p}}\right)^2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,407 |
28. Let $a_{1}, a_{2}, \cdots, a_{6} ; b_{1}, b_{2}, \cdots, b_{6} ; c_{1}, c_{2}, \cdots, c_{6}$ all be permutations of $1,2,3,4,5,6$. Find the minimum value of $\sum_{i=1}^{6} a_{i} b_{i} c_{i}$. (2005 China National Training Team Selection Exam Problem) | 28- Let $S=\sum_{i=1}^{6} a_{i} b_{i} c_{i}$. By the AM-GM inequality, we have
$$\left.S \geqslant 6 \sqrt[6]{\prod_{i=1}^{6} a_{i} b_{i} c_{i}}=6 \sqrt[6]{\prod_{i=1}^{6}(6!}\right)^{3}=6 \sqrt{6!}=72 \sqrt{5} \geq 160$$
Next, we prove that $S>161$.
Since the geometric mean of $a_{1} b_{1} c_{1}, a_{2} b_{2} c_{2}, a... | 162 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 733,408 |
20. Let $a<b<c, a+b+c=6, ab+bc+ca=9$, prove: $0<a<1<b<3<c<4$. (1995 British Mathematical Olympiad Problem) | 20. Let $t=a b c$, and set $f(x)=(x-a)(x-b)(x-c)$, then $f(x)=x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c=x^{3}-6 x^{2}+9 x-t, f(x)=0$ has three roots $a$, $b$, and $c$. Since $f^{\prime}(x)=3 x^{2}-12 x+9=3(x-1)(x-3)$, $f(x)$ is monotonically increasing on $(-\infty, 1)$, monotonically decreasing on $(1,3)$, and monoton... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,411 |
30. Let $f(x, y, z)=\sin ^{2}(x-y)+\sin ^{2}(y-z)+\sin ^{2}(z-x), x, y, z \in \mathbf{R}$, find the maximum value of $f(x, y, z)$. (2007 Zhejiang Province Mathematics Competition Problem) | 30.
$$\begin{array}{l}
f(x, y, z)=\sin ^{2}(x-y)+\sin ^{2}(y-z)+\sin ^{2}(z-x)= \\
\frac{1}{2}[1-\cos 2(x-y)+1-\cos 2(y-z)+1-\cos 2(z-x)]= \\
\frac{3}{2}-\frac{1}{2}[(\cos 2 x \cos 2 y+\sin 2 x \sin 2 y)+ \\
(\cos 2 y \cos 2 z+\sin 2 y \sin 2 z)+ \\
(\cos 2 z \cos 2 x+\sin 2 z \sin 2 x)] = \\
\frac{3}{2}-\frac{1}{4}\le... | \frac{9}{4} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,412 |
31. Let positive real numbers $a, b, c$ and non-negative real numbers $x, y, z$ satisfy $a^{6}+b^{6}+c^{6}=3,(x+1)^{2}+y^{2} \leqslant 2$, find the minimum value of $I=\frac{1}{2 a^{3} x+b^{3} y^{2}}+\frac{1}{2 b^{3} x+c^{3} y^{2}}+\frac{1}{2 c^{3} x+a^{3} y^{2}}$, and prove it: (2007 year | 31. According to the Cauchy-Schwarz inequality, we have $\left(a^{3}+b^{3}+c^{3}\right)^{2} \leqslant 3\left(a^{6}+b^{6}+c^{6}\right)$, and by the transformed form of the Cauchy-Schwarz inequality $\sum_{k=1}^{n} \frac{a_{k}^{2}}{b_{k}} \geqslant \frac{\left(\sum_{k=1}^{n} a_{k}\right)^{2}}{\sum_{k=1}^{n} b_{k}}$, we g... | 3 | Inequalities | proof | Yes | Yes | inequalities | false | 733,413 |
32. Let $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, find the maximum value of $A=\frac{\left(1-\sqrt{\tan \frac{\alpha}{2} \tan \frac{\beta}{2}}\right)^{2}}{\cot \alpha+\cot \beta}$. | 32. Let $x=\tan \frac{\alpha}{2}, y=\tan \frac{\beta}{2}, x, y \in(0,1)$, then $\cot \alpha+\cot \beta=\frac{1-x^{2}}{2 x}+\frac{1-y^{2}}{2 y} =\frac{(x+y)(1-x y)}{2 x y}$, so, $A=\frac{2 x y(1-\sqrt{x y})^{2}}{(x+y)(1-x y)}=\frac{2 x y(1-\sqrt{x y})}{(x+y)(1+\sqrt{x y})} \leqslant$ $\frac{2 x y(1-\sqrt{x y})}{2 \sqrt{... | 3-2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,414 |
33. Let $n>3$ be an integer, and let $a_{1}, a_{2}, \cdots, a_{n}$ be non-negative real numbers satisfying $a_{1}+a_{2}+\cdots+a_{n}=2$. Find the minimum value of $\frac{a_{1}}{a_{2}^{2}+1}+\frac{a_{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n}}{a_{1}^{2}+1}$. 2007 China Girls' Mathematical Olympiad | 33. From $a_{1}+a_{2}+\cdots+a_{n}=2$, the problem is equivalent to finding the maximum value of:
$$\begin{array}{l}
a_{1}-\frac{a_{1}}{a_{2}^{2}+1}+a_{2}-\frac{a_{2}}{a_{3}^{2}+1}+\cdots+a_{n}-\frac{a_{n}}{a_{1}^{2}+1}= \\
\frac{a_{1} a_{2}^{2}}{a_{2}^{2}+1}+\frac{a_{2} a_{3}^{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n} a_{1}... | \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,415 |
34. Given the equation $x^{3}-a x^{2}+b x-c=0$ has three positive roots (which can be equal), find the minimum value of $\frac{1+a+b+c}{3+2 a+b}-\frac{c}{b}$. (2008 Turkey Training Team Problem) | 34. Considering the equation $x^{3}-a x^{2}+b x-c=0$ has three equal positive roots $p, q, r$, when $\frac{1+a+b+c}{3+2 a+b}-\frac{c}{b}=\frac{1}{3}$, it is only necessary to prove $\frac{1+a+b+c}{3+2 a+b}-\frac{c}{b} \geqslant \frac{1}{3}$. This is equivalent to proving $a b+$ $2 b^{2} \geqslant 9 c+6 a c$.
By Vieta'... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,416 |
35. (1) Given positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $a_{1} a_{2} \cdots a_{n}=1$, find the smallest constant $c_{n}$ such that the inequality $\frac{1}{1+a_{1}}+\frac{1}{1+a_{2}}+\cdots+\frac{1}{1+a_{n}} \geqslant c_{n}$ holds for any positive integer $n \geqslant 2$.
(2) Given positive real nu... | 35. (1) Let $a_{1}=\varepsilon^{n-1}, a_{k}=\frac{1}{\varepsilon}(k=2,3, \cdots, n)$, and let $\varepsilon \rightarrow 0$. We easily get $c_{n} \leqslant 1$.
Thus, we only need to prove $\frac{1}{1+a_{1}}+\frac{1}{1+a_{2}}+\cdots+\frac{1}{1+a_{n}} \geqslant 1$.
Without loss of generality, assume $a_{1} \leqslant a_{2}... | proof | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,417 |
36. Given that $x, y, z$ are positive numbers, find the minimum value of the following two expressions:
(1) $\frac{x^{2}+y^{2}+z^{2}}{x y+y z}$; $\frac{x^{2}+y^{2}+2 z^{2}}{x y+y z}$. (2008 Croatian Mathematical Olympiad) | 36. (1) By the AM-GM inequality, $\frac{x^{2}+y^{2}+z^{2}}{x y+y z}=\frac{x^{2}+\frac{1}{2} y^{2}+\frac{1}{2} y^{2}+z^{2}}{x y+y z} \geqslant$ $\frac{\sqrt{2} x y+\sqrt{2} y z}{x y+y z}=\sqrt{2}$.
(2) $\frac{x^{2}+y^{2}+2 z^{2}}{x y+y z}=\frac{x^{2}+\frac{2}{3} y^{2}+\frac{1}{3} y^{2}+2 z^{2}}{x y+y z} \geqslant \frac{... | \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,418 |
38. Let $x, y, z \in(0,1)$, and satisfy $\sqrt{\frac{1-x}{y z}}+\sqrt{\frac{1-y}{z x}}+\sqrt{\frac{1-z}{x y}}=2$, find the maximum value of $x y z$. (2008 China Western Mathematical Olympiad) | 38. Since $\sqrt{\frac{1-x}{y z}}+\sqrt{\frac{1-y}{z x}}+\sqrt{\frac{1-z}{x y}}=2$, by the AM-GM inequality we have
$$\begin{aligned}
2 \sqrt{x y z}= & \sqrt{x(1-x)}+\sqrt{y(1-y)}+\sqrt{z(1-z)}= \\
& \frac{1}{\sqrt{3}}[\sqrt{x(3-3 x)}+\sqrt{y(3-3 y)}+\sqrt{z(3-3 z)}] \leqslant \\
& \frac{1}{\sqrt{3}}\left[\frac{x+(3-3 ... | \frac{27}{64} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,420 |
39. Let $x_{i}$ be positive real numbers, $(i=1,2, \cdots, 2010)$ and $\sum_{i=1}^{2010} x_{i}^{2009}=1$, find $\min \sum_{i=1}^{2010}$ $\left.\frac{x_{i}^{2008}}{1-x_{i}^{2009}}\right\}$, and prove it. (2009 Zhejiang Province Mathematics Competition Problem) | 39. Given $\left.\sum_{i=1}^{2010} x_{i}^{2008} \sum_{i=1}^{2010} x_{i} x_{i}^{2009}-x_{i}^{2009}\right)$, let $y_{i}=x_{i}\left(1-x_{i}^{200}\right)$, then for any $i=$
$$\begin{array}{l}
\left.1,2, \ldots, 2,001, x^{2009}-\frac{1}{2009}\left[20009 x_{i}^{2009}(1)-x^{2009}\right)^{2009}\right] \leqslant \\
\frac{1}{20... | \frac{2010}{2009} \cdot \sqrt[2009]{2010} | Inequalities | proof | Yes | Yes | inequalities | false | 733,421 |
21. Given that $a$ is a real number, satisfying $a^{5}-a^{3}+a=2$, prove: $3<a^{6}<4$. | 21. Clearly $a \neq 0$, and $a \neq 1$, $a^{6}+1=\left(a^{2}+1\right)\left(a^{4}-a^{2}+1\right)=\left(a^{2}+1\right)$. $\frac{a^{5}-a^{3}+a}{a}=\frac{2\left(a^{2}+1\right)}{a} \geqslant 4$, since $a \neq 1$, the equality does not hold. Thus, $a^{6}>3$.
On the other hand, let $f(x)=x^{5}-x^{3}+x$, then $f^{\prime}(x)=5... | 3<a^{6}<4 | Algebra | proof | Yes | Yes | inequalities | false | 733,422 |
40. Let $n \geqslant 2$ be a positive integer. Find the maximum value of the constant $C(n)$ such that for all real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{i} \in(0, 1)$ $(i=1,2, \cdots, n)$, and $\left(1-x_{i}\right)\left(1-x_{j}\right) \geqslant \frac{1}{4}(1 \leqslant i<j \leqslant n)$, we have $\sum_{i... | 40. First, take $x_{i}=\frac{\mathrm{F}}{2}(i=1,2, \cdots, n)$, substitute into $\sum_{i=1}^{n} x_{i} \geqslant C(n)\left(2 x_{i} x_{j}+\sqrt{x_{i} x_{j}}\right)$ to get $\frac{n}{2} \geqslant G(n) C_{n}^{2}\left(\frac{1}{2}+\frac{1}{2}\right)$
Then, $C(n) \leqslant \frac{\mathrm{F}}{n-1}$. Below, we prove that $C(n)=... | \frac{1}{n-1} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,423 |
41. Find the smallest constant $c$, such that for all real numbers $x, y$, we have $1+(x+y)^{2} \leqslant c\left(1+x^{2}\right)$ ( $\left.1+y^{2}\right)$ )(2008 German Mathematical Olympiad problem) | 41. Take $x=y=\frac{\sqrt{2}}{}$, substitute into the original inequality to get $3 \leqslant \frac{9}{4} c \Rightarrow c \geqslant \frac{4}{3}$. Now we prove that $c=\frac{4}{3}$ satisfies the condition.
We only need to prove: $1+(x+y)^{2} \leqslant \frac{4}{3}\left(1+x^{2}\right)\left(1+y^{2}\right)$. Simplifying an... | \frac{4}{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,424 |
42. For any non-negative real numbers $x, y, z$, we have $x^{3}+y^{3}+z^{3}+c\left(x y^{2}+y z^{2}+z x^{2}\right) \geqslant(c+$ 1) $\left(x^{2} y+y^{2} z+z^{2} x\right)$. Find the maximum value of the real number $c$. (2008 Mongolian National Training Team Exam) | 42 Assume $x=\min |x, y, z|, y=x+p, z=x+q(p, q \geqslant 0)$, substitute into the inequality
get
$$x^{3}+y^{3}+z^{3}+c\left(x y+y z+z x^{2}\right) \geqslant(c+1)\left(x^{2} y+y z+z^{2} x\right)$$
$$\begin{array}{l}
2\left(p^{2}-p q+q^{2}\right) x+p^{3}+q\left(q^{2}-p^{2}\right)+c p q(q-p) \geqslant \\
0(x, p, q \geqsl... | 2.4844 \cdots | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,425 |
43. Let $f(x, y, z)=\frac{x(2 y-z)}{1+x+3 y}+\frac{y(2 z-x)}{1+y+3 z}+\frac{z(2 x-y)}{1+z+3 x}$, where $x, y, z \geqslant 0, x+$ $y+z=1$, find the maximum and minimum values of $f(x, y, z)$. (2009 China Southeast Mathematical Olympiad | 43. Proof: $f(x, y, z) \leqslant \frac{1}{7}$. Equality holds if and only if $x=y=z=\frac{1}{3}$.
Since $\frac{x(2 y-z)}{1+x+3 y}=\frac{x(-2 y-(1-x-y))}{1+x+3 y}=\frac{x(-1+x+3 y)}{1+x+3 y}=x- \frac{2 x}{1+x+3 y}$, we have
$$\begin{aligned}
f(x, y, z)= & (x+y+z)-2\left(\frac{x}{1+x+3 y}+\frac{y}{1+y+3 z}+\frac{z}{1+z+... | f(x, y, z)_{\text {max }}=\frac{1}{7}, \, f(x, y, z)_{\text {min }}=0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,426 |
44. Given an integer $n(n \geqslant 2)$. Find the largest constant $\lambda(n)$ such that: if the real number sequence $a_{0}, a_{1}, a_{2}, \cdots, a_{n}$ satisfies $0=a_{0} \leqslant a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$ and $a_{i} \geqslant \frac{1}{2}\left(a_{i-1}+a_{i+1}\right)(i=$ $1,2, \cdots, ... | 44. $\lambda(n)=\frac{n(n+1)^{2}}{4}$.
First, let $a_{1}=a_{2}=\cdots=a_{n}=1$. Then $\lambda(n) \leqslant \frac{n(n+1)^{2}}{4}$.
Next, we prove that for any sequence $a_{0}, a_{1}, a_{2}, \cdots, a_{n}$ satisfying the conditions, we have
$$\left(\sum_{i=1}^{n} i a_{i}\right)^{2} \geqslant \frac{n(n+1)^{2}}{4} \sum_{i... | \frac{n(n+1)^{2}}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,427 |
45. Find the maximum and minimum values of the function $y=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$. | 45. The domain of the function is $[0,13]$, because
$$\begin{aligned}
y= & \sqrt{x+27}+\sqrt{13-x}+\sqrt{x}=\sqrt{x+27}+\sqrt{13+2 \sqrt{x(13-x)}} \geqslant \\
& \sqrt{27}+\sqrt{13}=3 \sqrt{3}+\sqrt{13}
\end{aligned}$$
When $x=0$, the equality holds. Therefore, the minimum value of $y$ is $3 \sqrt{3}+\sqrt{13}$.
By th... | 11 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,428 |
46. Given that $x, y, z$ are real numbers, and satisfy $x+y+z=xy+yz+zx$, find the minimum value of $\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+$ $\frac{z}{z^{2}+1}$. (2008 Brazilian Mathematical Olympiad Problem) | 46. Let $x=1, y=z=-1$. Then $\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1}=-\frac{1}{2}$. Conjecture that the minimum value of $\frac{x}{x^{2}+1}+$ $\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1}$ is $-\frac{1}{2}$.
It is sufficient to prove
$$\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1} \geqslant-\frac{1}{2} \L... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,429 |
47. Given an integer $n \geqslant 2$ and a positive integer $a$, positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1} x_{2} \cdots x_{n}=1$, find the smallest real number $M=M(n, a)$, such that $\sum_{i=1}^{n} \frac{1}{a+S-x_{i}} \leqslant M$ always holds, where $S=x_{1}+x_{2}+\cdots+x_{n} \cdot(2010$ Chi... | 47. First consider the case $a \geqslant 1$, let $x_{i}=y_{i}^{n}, y_{i}>0$, then $y_{1} y_{2} \cdots y_{n}=1$, we have
$$\begin{aligned}
S-x_{i}= & \sum_{j \neq i} y_{j}^{n} \geqslant(n-1)\left(\frac{\sum_{j \neq i} y_{j}}{n-1}\right)^{n}(\text { by the AM-GM inequality }) \geqslant \\
& (n-1)\left(\frac{\sum_{j \neq ... | M=\left\{\begin{array}{l}
\frac{n}{a-1+n}, \text { if } a \geqslant 1 \\
\frac{1}{a}, \text { if } 0<a<1
\end{array}\right.} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,430 |
48. Find all positive real numbers $\lambda$ such that for any integer $n \geqslant 2$ and positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $\sum_{i=1}^{n} a_{i}=n$, we always have $\sum_{i=1}^{n} \frac{1}{a_{i}}-\lambda \prod_{i=1}^{n} \frac{1}{a_{i}} \leqslant n-\lambda$. (2010 China National Training T... | 48. We need the following conclusion: the sequence $\left(1+\frac{1}{n}\right)^{n}$ is strictly monotonically increasing, and $\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}=$ e:
We first prove that the inequality always holds when $\lambda \geqslant e$.
Assume $a_{n-1}=\min _{1 \leqslant i \leqslant n} a_... | \lambda \geqslant e | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,431 |
49. Let $x, y, z \in [0,1]$, and $|x-y| \leqslant \frac{1}{2}, |y-z| \leqslant \frac{1}{2}, |z-x| \leqslant \frac{1}{2}$, find the maximum and minimum values of $W=x+y+z-xy-yz-zx$. (2010 China Southeast Mathematical Olympiad) | 49. Since $x, y, z \in [0,1]$, we have $x+y+z \geqslant xy + yz + zx$, i.e., $W \geqslant 0$. Therefore, $W_{\text{min}} = 0$ if and only if $x = y = z = 0$ or $1$. It is easy to see that at least two of $x, y, z$ are no greater than $\frac{1}{2}$ or no less than $\frac{1}{2}$. Also, from $W = x + y + z - xy - yz - zx ... | \frac{5}{6} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,432 |
22. Let $a, b, c$ be positive numbers, prove that: $\frac{a+b+c}{3}-\sqrt[3]{a b c} \leqslant \max (\sqrt{a}-\sqrt{b})^{2},(\sqrt{b}-$ $\sqrt{c})^{2},(\sqrt{c}-\sqrt{a})^{2} \cdot$ (2002 USA National Training Team Exam) | 22. Without loss of generality, assume $b$ is between $a$ and $c$, and $a \leqslant b \leqslant c$, i.e., $b \in [a, c]$. The inequality becomes
$$a+b+c-3 \sqrt[3]{a b c} \leqslant 3(c+a-2 \sqrt{c a})$$
Consider the left side as a function of $b$, $f(b)=a+b+c-3 \sqrt[3]{a b c}$. It is a continuous function on $[a, c]$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,433 |
50. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$. Try to find the integer part of $E=x_{1}+$ $\frac{x_{2}}{\sqrt{1-x_{1}^{2}}}+\frac{x_{3}}{\sqrt{1-\left(x_{1}+x_{2}\right)^{2}}}+\cdots+\frac{x_{n}}{\sqrt{1-\left(x_{1}+x_{2}+\cdots+x_{n-1}\right)^{2}}}$. | 50. Since $\sqrt{1-\left(x_{1}+x_{2}+\cdots+x_{i}\right)^{2}} \leqslant 1(1 \leqslant i \leqslant n)$, we have $E \geqslant x_{1}+ x_{2}+\cdots+x_{n}=1$
Let $\alpha_{1}=\frac{\pi}{2}, \alpha_{i}=\arccos \left(x_{1}+x_{2}+\cdots+x_{i-1}\right) \quad(2 \leqslant i \leqslant n)$, which means $x_{i}=\cos \alpha_{i+1}-\cos... | 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,434 |
51. Let complex numbers $a, b, c$ satisfy that for any complex number $z$ with modulus not exceeding 1, $\left|a z^{2}+b z+c\right| \leqslant 1$. Find the maximum value of $|bc|$. (2010 China Mathematical Olympiad Problem) | $$\begin{array}{l}
51 \text { Let } f(z)=a z^{2}+b z+c, g(z)=z^{2} f(z)=a+b z^{+1}+c z^{-2}, h(-z)= \\
\mathrm{e}^{i \alpha} g\left(\mathrm{e}^{i \beta} \bar{z}\right)=c^{\prime} z^{-2}+b^{\prime} z^{-1}+a^{\prime}
\end{array}$$
Take appropriate real numbers $\alpha, \beta$, such that $c^{\prime}, b^{\prime} \geqslant... | \frac{3 \sqrt{3}}{16} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,435 |
53. Real numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=0$, and $2 a_{k} \leqslant a_{k-1}+$ $a_{k+1}, k=2,3, \cdots, n-1$. Find the smallest $\lambda(n)$, such that for all $k \in\{1,2, \cdots, n\}$, we have $|a_{k}| \leqslant \lambda(n) \cdot \max \left\{|a_{1}|,|a_{n}|\right\}... | 53. $\lambda(n)_{m n}=\frac{n+1}{n-1}$.
First, take $a_{1}=1, a_{2}=-\frac{n+1}{n-1}, a_{k}=-\frac{n+1}{n-1}+\frac{2 n(k-2)}{2(n-1)(n-2)}, k=3,4, \cdots$,
n. Then it satisfies $a_{1}+a_{2}+\cdots+a_{n}=0$ and $2 a_{k} \leqslant a_{k-1}+a_{k+1}, k=2,3, \cdots, n-1$. At this point, $\lambda(n) \geqslant \frac{n+1}{n-1}$... | \frac{n+1}{n-1} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,437 |
54. Let $x_{1}, x_{2}, \cdots, x_{5}$ be real numbers, and $\sum_{i=1}^{5} x_{i}=0$, let $x_{6}=x_{1}$, find the smallest real number $k$ such that the inequality $\sum_{i=1}^{5} x_{i} x_{i+1} \leqslant k\left(\sum_{i=1}^{5} a_{i}^{2}\right) .(2001$ Czech and Slovak Olympiad problem) | 54. Let $x_{6}=x_{1}, x_{7}=x_{2}, 0=\left(\sum_{i=1}^{5} x_{i}\right)^{2}=\sum_{i=1}^{5} x_{i}^{2}+2 \sum_{i=1}^{5} x_{i} x_{i+1}+2 \sum_{i=1}^{5} x_{i} x_{i+2}$. For any real number $\lambda$,
$$\begin{aligned}
0 \leqslant & \sum_{i=2}^{6}\left(x_{i+1}+x_{i-1}-\lambda x_{i}\right)^{2}= \\
& \left(2+\lambda^{2}\right)... | \frac{\sqrt{5}-1}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,438 |
55. Let $x, y, z$ be positive numbers, and $(x+y+z)^{3}=32 x y z$, find the maximum and minimum values of $\frac{x^{4}+y^{4}+z^{4}}{(x+y+z)^{4}}$. (2004 Vietnam Mathematical Olympiad) | 55. Let's assume $x+y+z=4, xyz=2$, so we only need to find the maximum and minimum values of $\frac{x^{4}+y^{4}+z^{4}}{4^{4}}$. Now we have $x^{4}+y^{4}+z^{4}=\left(x^{2}+y^{2}+z^{2}\right)^{2}-2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)=$ $[16-2(xy+yz+zx)]^{2}-2(xy+yz+zx)^{2}+4xyz(x+y+z)=a^{2}-64a+288$, where $a... | \frac{383-165 \sqrt{5}}{256}, \frac{9}{128} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,439 |
56. Determine the largest real number $C$, for all real numbers $x, y, x \neq y$, and $xy=2$, such that the inequality $\frac{\left[(x+y)^{2}-6\right]\left[(x-y)^{2}+8\right]}{(x-y)^{2}} \geqslant C$ holds. (2002 Austrian Mathematical Olympiad Problem) | $\begin{array}{l}\text { 56. Since } \frac{\left[(x+y)^{2}-6\right]\left[(x-y)^{2}+8\right]}{(x-y)^{2}}=\frac{\left[(x-y)^{2}+4 x y-6\right]\left[(x-y)^{2}+8\right]}{(x-y)^{2}}= \\ \frac{\left[(x-y)^{2}+2\right]\left[(x-y)^{2}+8\right]}{(x-y)^{2}}=(x-y)^{2}+\frac{16}{(x-y)^{2}}+10 \geqslant 18 \text {, therefore } C_{\... | 18 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,440 |
57. If $x, y, z$ are all positive real numbers, and $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $S=\frac{(z+1)^{2}}{2 x y z}$. | $\begin{array}{l}\text { 57. } S=\frac{(z+1)^{2}}{2 x y z} \geqslant \frac{(z+1)^{2}}{\left(x^{2}+y^{2}\right) z}=\frac{(z+1)^{2}}{\left(1-z^{2}\right) z}=\frac{z+1}{(1-z) z}=\frac{1-z+2 z}{(1-z) z}=\frac{1}{z}+\frac{2}{1-z}= \\ {[z+(1-z)] \cdot\left(\frac{1}{z}+\frac{2}{1-z}\right)=3+\frac{2 z}{1-z}+\frac{1-z}{z} \geq... | 3+2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,441 |
58. If $a, b, c$ are all positive real numbers, and $a+b+c=1$, find the largest real number $M$ such that the inequality $(a+b c)(b+c a)(c+a b)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant M a b c$. (2011 Uzbekistan Mathematics | 58. Since $a+b+c=1$, we have $a+bc=a(a+b+c)+bc=(a+b)(c+a)$, so $(a+bc)(b+ca)(c+ab)=(a+b)^{2}(b+c)^{2}(c+a)^{2}$. The problem is reduced to finding the largest real number $M$ such that the inequality
$$(a+b)^{2}(b+c)^{2}(c+a)^{2}(ab+bc+ca) \geqslant M(abc)^{2}(a+b+c)^{2}$$
holds. Given that $(a+b)(b+c)(c+a) \geqslant ... | \frac{64}{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,442 |
23. Given that $a, b, c$ are all positive numbers, and $a b c=1$, prove: $(a+b)(b+c)(c+a) \geqslant 4(a+$ $b+c-1) \cdot(2001$ USA Mathematical Olympiad MOSP Problem) | 23. Treat $a$ as a constant, and consider it as a quadratic equation in $b+c$, and prove it using the discriminant method. $\square$
Since $a b c=1$, we may assume without loss of generality that $a \geqslant 1$. The original inequality is equivalent to
$$a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)+6 \geqslant 4(a+b+c)$$
which ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,444 |
Example 3 Let $n(n \geqslant 2)$ be an integer, prove that $\sum_{i=1}^{n-1} \frac{n}{n-k} \frac{F}{2^{k-1}} < 4$ | Prove that for $a_{n}=\sum_{i=1}^{n=1} n \cdot \frac{1}{2^{k-1}}$, we have
$$\begin{aligned}
a_{n+1}= & \sum_{i=1}^{n} \frac{n+1}{n+1-k} \cdot \frac{1}{2^{k-1}}=\frac{n+1}{n}+\frac{n+1}{n-1} \cdot \frac{1}{2}+\cdots+\frac{n+1}{2^{n}}= \\
& \frac{n+1}{n}+\frac{1}{2}\left(\frac{n}{n-1}+\cdots+\frac{n}{2^{n-2}}\right)+\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,448 |
Example 4 Given that $a, b$ are positive numbers, $n \in \mathbf{N}^{*}$ and $n \geqslant 2$, prove: $\frac{a^{n}+a^{n-1} b+a^{n-2} b^{2}+\cdots+a b^{n-1}+b^{n}}{n+1} \geqslant$ $\left(\frac{a+b}{2}\right)^{n} .(1988$ Hunan Province Middle School Mathematics Summer Camp Mathematics Competition Problem) | Let $P_{n}=a^{n}+a^{n-1} b+a^{n-2} b^{2}+\cdots+a b^{n-1}+b^{n}, Q_{n}=\left(\frac{a+b}{2}\right)^{n}$. We need to prove that $P_{n} \geqslant(n+1) Q_{n}$.
Since $(a+b) P_{n-1}+a^{n}+b^{n}=2 P_{n}$, i.e., $P_{n}=\frac{a+b}{2} P_{n-1}+\frac{a^{n}+b^{n}}{2}$.
By the power mean inequality, we have
$$P_{n} \geqslant \frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,449 |
Example 6 Let $a_{0}, a_{1}, a_{2}, \cdots, a_{n}, \cdots$ be a sequence of positive real numbers, satisfying $a_{i=1} a_{i+1} \leqslant a_{i}^{2}(i=1,$
$2, \cdots)$, prove that for all $n>1$, $\frac{a_{0}+a_{1}+a_{2}+\cdots+a_{n}}{n+1} \cdot \frac{a_{1}+a_{2}+\cdots+a_{n-1}}{n-1} \geqslant$
$\frac{a_{0}+a_{1}+a_{2}+\c... | Given the conditions, we have
$$\frac{a_{0}}{a_{1}} \leqslant \frac{a_{1}}{a_{2}} \leqslant \cdots \leqslant \frac{a_{n-2}}{a_{n-1}} \leqslant \frac{a_{n-1}}{a_{n}}$$
Thus,
$$a_{0} a_{n} \leqslant a_{1} a_{n-1} \leqslant a_{2} a_{n-2} \leqslant \cdots$$
Let \( S = a_{1} + a_{2} + \cdots + a_{n-1} \), the inequality t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,451 |
Example 8 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{21}{16}$,
$$2 a_{n}-3 a_{n-1}=\frac{3}{2^{n+1}}, n \geqslant 2$$
Let $m$ be a positive integer, $m \geqslant 2$. Prove: when $n \leqslant m$, we have
$$\left(a_{n}+\frac{3}{2^{n+3}}\right)^{\frac{1}{m}}\left[m-\left(\frac{2}{3}\right)^{\frac{n(... | Prove that from equation (1) we get $2^{n} a_{n}=3 \cdot 2^{n-1} a_{n-1}+\frac{3}{4}$, let $b_{n}=2^{n} a_{n}, n=1,2,3, \cdots$
$$\begin{array}{c}
b_{n}=3 b_{n-1}+\frac{3}{4} \\
b_{n}+\frac{3}{8}=3\left(b_{n-1}+\frac{3}{8}\right)
\end{array}$$
Since $b_{1}=2 a_{1}=\frac{21}{8}$, we have $b_{n}+\frac{3}{8}=3^{n-1}\left... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,453 |
1. $n$ is a positive integer, prove: $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n-1}>n(\sqrt[n]{2}-1)$. | $\begin{array}{l}\text { 1. Since } n+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n-1}=\left(1+\frac{1}{n}\right)+\left(1+\frac{1}{n+1}\right)+ \\ \left(1+\frac{1}{n+2}\right)+\cdots+\left(1+\frac{1}{2 n-1}\right)=\frac{n+1}{n}+\frac{n+2}{n+1}+\frac{n+3}{n+2}+\cdots+\frac{2 n}{2 n-1}> \\ n \sqrt[n]{\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,454 |
24. Let $x, y, z$ be non-negative real numbers, and $x+y+z=1$, prove: $0 \leqslant y z+z x+x y-2 x y z \leqslant \frac{7}{27}$. (25th IMO problem) | 24. Only prove the right inequality. Since $x, y, z$ are non-negative real numbers, and $x+y+z=1$, we have $y+z=1-x$, $0 \leqslant yz \leqslant \frac{(1-x)^{2}}{4}$, and
$$\begin{array}{l}
yz + zx + xy - 2xyz = (1-2x)yz + x(y+z) = \\
(1-2x)yz + x(1-x)
\end{array}$$
Let $w = yz$, then $0 \leqslant yz \leqslant \frac{(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,455 |
2. Given $a_{0}=1, a_{1}=2, a_{n+1}=a_{n}+\frac{a_{n-1}}{1+\left(a_{n-1}\right)^{2}}, n \geqslant 1$, prove: $52<a_{1371}<65$. | 2. We prove the general case
$$\sqrt{2 n+1} \leqslant a_{n} \leqslant \sqrt{3 n+2} \text { (for all } n \geqslant 0 \text { ) }$$
Taking $n=1371, \sqrt{2 n+1}=\sqrt{2743} \approx 52.37, \sqrt{3 n+2}=\sqrt{4115} \approx 64.148$, then we have $520$. From equation (2), we know the sequence $\left\{a_{n}\right\}$ is stric... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,456 |
3. Let \(a_{1}=1, a_{n}=\frac{a_{n-1}}{2}+\frac{1}{a_{n-1}}\), where \(n=1,2,3, \cdots\), prove: \(0<a_{10}-\sqrt{2}<10^{-370}\). (49th Moscow Mathematical Olympiad Problem) | 3. $a_{n}-\sqrt{2}=\frac{a_{n-1}}{2}+\frac{1}{a_{n-1}}-\sqrt{2}=\frac{\left(a_{n-1}-\sqrt{2}\right)^{2}}{2 a_{n-1}}, a_{n}+\sqrt{2}=\frac{a_{n-1}}{2}+\frac{1}{a_{n-1}}+\sqrt{2}=$ $\frac{\left(a_{n-1}+\sqrt{2}\right)^{2}}{2 a_{n-1}}$, therefore, $\frac{a_{n}-\sqrt{2}}{a_{n}+\sqrt{2}}=\left(\frac{a_{n-1}-\sqrt{2}}{a_{n-1... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,457 |
4. Let $n \geqslant 2$ be a natural number, prove: $\frac{1}{n+1}\left(1+\frac{1}{3}+\cdots+\frac{1}{2 n-1}\right)>\frac{1}{n}\left(\frac{1}{2}+\right.$ $\left.\frac{1}{4}+\cdots+\frac{1}{2 n}\right) \cdot(1998$ Canadian Mathematical Olympiad problem) | 4. $\frac{1}{n+1}\left(1+\frac{1}{3}+\cdots+\frac{1}{2 n-1}\right)>\frac{1}{n}\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right)$ is equivalent to
$$\begin{array}{l}
\frac{1}{n+1}\left(1+\frac{1}{3}+\cdots+\frac{1}{2 n-1}\right)+\frac{1}{n+1}\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right)> \\
\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,458 |
5. Given that $n>1$ is a positive integer, $x>y>1$, prove: $\frac{x^{n+1}-1}{x^{n}-x}>\frac{y^{n+1}-1}{y^{n}-y}$. | 5. Since $x>y>1$, the inequality $\frac{x^{n+1}-1}{x^{n}-x}>\frac{y^{n+1}-1}{y^{n}-y}$ is equivalent to $\left(x^{n+1}-1\right)\left(y^{n}-y\right)>\left(y^{n+1}-1\right)\left(x^{n}-x\right) \Leftrightarrow x^{n+1} y^{n}-x^{n} y^{n+1}+x y^{n+1}-x^{n+1} y+x^{n}-y^{n}-x+y>0 \Leftrightarrow$ $(x-y)\left(x^{n} y^{n}-1\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,459 |
6. Given the sequence $\left\{a_{n}\right\}$ with the general term formula $a_{n}=\frac{n^{2}+1}{\sqrt{n^{4}+4}}$, and the sequence $\left\{b_{n}\right\}$ with the general term $b_{n}=$ $a_{1} a_{2} \cdots a_{n}$, prove:
(1) $b_{n}=\frac{\sqrt{2\left(n^{2}+1\right)}}{\sqrt{n^{2}+2 n+2}}$;
(2) $\frac{1}{(n+1)^{3}}<\frac... | 6. (1) Since $a_{n}=\frac{n^{2}+1}{\sqrt{n^{4}+4}}=\frac{n^{2}+1}{\sqrt{n^{4}+4 n^{2}+4-4 n^{2}}}=\frac{n^{2}+1}{\sqrt{\left(n^{2}+2\right)^{2}-4 n^{2}}}=$ $\frac{n^{2}+1}{\sqrt{\left(n^{2}+2 n+2\right)\left(n^{2}-2 n+2\right)}}=\frac{n^{2}+1}{\sqrt{\left((n-1)^{2}+1\right)\left((n+1)^{2}+1\right)}}$, therefore $b_{n}=... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,460 |
7. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1} a_{n}-1=a_{n}^{2}$.
(1) Prove: $\sqrt{2 n-1} \leqslant a_{n} \leqslant \sqrt{3 n-2}$;
(2) Find the integer $m$, such that $\left|a_{2005}-m\right|$ is minimized. (2005 Hebei Province High School Mathematics Competition Problem) | 7. (1) From the given information, it is easy to see that $\{a_n\}$ is an increasing sequence, and all its terms are positive.
Since $a_{n+1} a_{n} - 1 = a_{n}^2$, we have, for $k \geq 2$, $a_{k} = a_{k-1} + \frac{1}{a_{k-1}}$.
Thus, $a_{k}^2 = \left(a_{k-1} + \frac{1}{a_{k-1}}\right)^2 = a_{k-1}^2 + \frac{1}{a_{k-1}^... | 63 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,461 |
8. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=a_{n}+2 n(n=1,2, \cdots), b_{n+1}=b_{n}+$ $\frac{b_{n}^{2}}{n}(n=1,2, \cdots)$. Prove that $: \frac{1}{2} \leqslant \sum_{k=1}^{n} \frac{1}{\sqrt{a_{k+1} b_{k}+k a_{k+1}-b_{k}-k}}<1$. (2006 Zhejiang Province | 8. Let $I_{n}=\sum_{k=1}^{n} \frac{1}{\sqrt{a_{k+1} b_{k}+k a_{k+1}-b_{k}-k}}$, then $I_{1}=\frac{1}{2}<I_{2}<\cdots<I_{n}$.
And
$$\begin{aligned}
I_{n}= & \sum_{k=1}^{n} \frac{1}{\sqrt{a_{k+1} b_{k}+k a_{k+1}-b_{k}-k}}= \\
& \sum_{k=1}^{n} \frac{1}{\sqrt{\left(a_{k+1}-1\right)\left(b_{k}+k\right)}} \leqslant \\
& \sq... | \frac{1}{2} \leqslant I_{n}<1 | Inequalities | proof | Yes | Yes | inequalities | false | 733,462 |
9. Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n} \geqslant a_{n+1}=0$ be a sequence of real numbers, prove: $\sqrt{\sum_{k=1}^{n} a_{k}} \leqslant \sum_{k=1}^{n}$ $\sqrt{a_{k}}\left(\sqrt{a_{k}}-\sqrt{a_{k+1}}\right)$. (38th IMO Shortlist Problem) | 9. Let $x_{k}=\sqrt{a_{k}}-\sqrt{a_{k+1}}, k=1,2, \cdots, n$. Then we have
$$a_{1}=\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2}, a_{2}=\left(x_{2}+x_{3}+\cdots+x_{n}\right)^{2}, \cdots, a_{n}=x_{n}^{2}$$
Expanding the right-hand side of these equations and adding them, we get
$$\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} k x_{k}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,463 |
10. (1) Let $\left\{b_{n}\right\}$ be a sequence of positive integers, and for all $n \geqslant 1$ we have $b_{n+1}^{2} \geqslant \frac{b_{1}^{2}}{1^{3}}+\frac{b_{2}^{2}}{2^{3}}+\cdots+\frac{b_{n}^{2}}{n^{3}}$. Prove: There exists a positive integer $k$, such that $\sum_{n=1}^{k} \frac{b_{n+1}}{b_{1}+b_{2}+\cdots+b_{n}... | 10. (1) By the Cauchy inequality, we have
$$\left(1^{3}+2^{3}+\cdots+n^{3}\right) b_{n+1}^{2} \geqslant\left(1^{3}+2^{3}+\cdots+n^{3}\right)$$
$$\left(\frac{b_{1}^{2}}{1^{3}}+\frac{b_{2}^{2}}{2^{3}}+\cdots+\frac{b_{n}^{2}}{n^{3}}\right)=\left(b_{1}+b_{2}+\cdots+b_{n}\right)^{2}$$
Since $1^{3}+2^{3}+\cdots+n^{3}=\left... | \frac{1998}{1000}>\frac{1993}{1000} | Inequalities | proof | Yes | Yes | inequalities | false | 733,464 |
11. Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{0}=\frac{\sqrt{2}}{2}, a_{n+1}=\frac{\sqrt{2}}{2} \sqrt{1-\sqrt{1-a_{n}^{2}}}, n=0,1$,
$2, \cdots . b_{0}=1, b_{n+1}=\frac{\sqrt{1+b_{n}^{2}}-1}{b_{n}}, n=0,1,2, \cdots$. Prove that for every $n=0,1$, $2, \cdots$, we have $2^{n+2} a_{n}... | 11. Let $a_{0}=\frac{\sqrt{2}}{2}=\sin \frac{\pi}{2^{2}}$, then $a_{1}=\frac{\sqrt{2}}{2} \sqrt{1-\cos \frac{\pi}{4}}=\sin \frac{\pi}{2^{3}}$, if $a_{n}=\sin \frac{\pi}{2^{n+2}}$, then $a_{n+1}=\frac{\sqrt{2}}{2} \sqrt{1-\cos \frac{\pi}{2^{n+2}}}=\sin \frac{\pi}{2^{n+3}}$.
Thus, by mathematical induction, we know that... | 2^{n+2} a_{n}<\pi<2^{n+1} b_{n} | Algebra | proof | Yes | Yes | inequalities | false | 733,465 |
25. Let $x, y, z$ be non-negative real numbers, and $x+y+z=1$, prove: $Z(y z+z x+x y) \leqslant 2+9 x y z$.
Let $x, y, z$ be non-negative real numbers, and $x+y+z=1$, prove: $Z(y z+z x+x y) \leqslant 2+9 x y z$. | 25. Since $x, y, z$ are non-negative real numbers, and $x+y+z=1$, we have $y+z=1-x, 0 \leqslant y z \leqslant \frac{(1-x)^{2}}{4}$.
$$\begin{array}{c}
7(y z+z x+x y)-2-9 x y z=(7-9 x) y z+7 x(y+z)-2= \\
(7-9 x) y z+7 x(1-x)-2
\end{array}$$
Let $w=y z$, then
$$0 \leqslant y z \leqslant \frac{(y+z)^{2}}{4}=\frac{(1-x)^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,466 |
12. Let $x_{0}=10^{9}, x_{n}=\frac{x_{n-1}^{2}+2}{2 x_{n-1}}, n=1,2, \cdots$. Prove: $0<x_{36}-\sqrt{2}<10^{-9}$. | 12. It is easy to prove by mathematical induction that
$$x_{n}>\sqrt{2}, n=0,1,2, \cdots$$
From the recursive formula, we get
$$0<x_{n}-\sqrt{2}=\frac{x_{n-1}^{2}+2}{2 x_{n-1}}-\sqrt{2}=\frac{\left(x_{n-1}-\sqrt{2}\right)^{2}}{2 x_{n-1}}$$
From equation (1), we get
$$0<x_{n}-\sqrt{2}<\frac{\left(x_{n-1}-\sqrt{2}\righ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,467 |
13. For any positive integer $n$, let $a_{n}$ be the real root of the equation $x^{3}+\frac{x}{n}=1$, prove that: (1) $a_{n+1}>a_{n}$;
(2) $\sum_{i=1}^{n} \frac{1}{(i+1)^{2} a_{i}}<a_{n}$. (2006 China Southeast Mathematical Olympiad) | 13. From $a_{n}^{3}+\frac{a_{n}}{n}=1$, we get $0 < a_{n} < 1$. Since $a_{n}^{3}+\frac{a_{n}}{n}=1$, it follows that $a_{n+1}^{3}+\frac{a_{n+1}}{n+1}=1$. Because $a_{n}^{3}+\frac{a_{n}}{n} > a_{n+1}^{3}+\frac{a_{n+1}}{n+1}$ for $n > 0$, we have $a_{n+1}-a_{n}>0$, which means $a_{n+1}>a_{n}$.
(2) Since $a_{n}\left(a_{n}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,468 |
14. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive numbers, and for any $1 \leqslant k \leqslant n$, we have $a_{1} a_{2} \cdots a_{k} \geqslant 1$. Prove:
$$\frac{1}{1+a_{1}}+\frac{2}{\left(1+a_{1}\right)\left(1+a_{2}\right)}+\cdots+\frac{n}{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)}<2 . \text... | 14. For any $1 \leqslant k \leqslant n$, since $1+a_{1} \geqslant 2 \sqrt{a_{1}}, 1+a_{2} \geqslant 2 \sqrt{a_{2}}, \cdots, 1+$ $a_{k} \geqslant 2 \sqrt{a_{k}}$, noting that $a_{1} a_{2} \cdots a_{k} \geqslant 1$ we get
$$\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{k}\right) \geqslant 2^{k} \sqrt{a_{1} a_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,469 |
15. Given a sequence of real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $a_{1}=1, a_{2}=\frac{1}{2}, a_{k+2}=a_{k}+\frac{1}{2} a_{k+1}+ \frac{1}{4 a_{k} a_{k+1}}, k \geqslant 1$. Prove that: $\frac{1}{a_{1} a_{3}}+\frac{1}{a_{2} a_{4}}+\frac{1}{a_{3} a_{5}}+\cdots+\frac{1}{a_{98} a_{100}}<4$. (2005 Baltic Way Mat... | 15. From the known condition $a_{k+2}=a_{k}+\frac{1}{2} a_{k+1}+\frac{1}{4 a_{k} a_{k+1}}$, we get $2\left(a_{k+2}-a_{k}\right)>a_{k+1}$, and knowing that all terms of the sequence are positive, dividing both sides by $a_{k} a_{k+1} a_{k+2}$ yields $\frac{F}{a_{k} a_{k+2}}<\frac{2}{a_{k} a_{k+1}}-\frac{2}{a_{k+1} a_{k+... | 4 | Algebra | proof | Yes | Yes | inequalities | false | 733,470 |
17. The terms of the sequence $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ are non-negative, and satisfy (1) $a_{n}+a_{2 n} \geqslant 3 n$; (2) $a_{n+1}+n \leqslant 2 \sqrt{a_{n}(n+1)}, n=1,2, \cdots$, try to prove $a_{n} \geqslant n$, and give the sequences that satisfy the conditions. (2004 Baltic Way Mathematical Competiti... | 17. By the mean inequality, we have $a_{n+1}+n \geqslant 2 \sqrt{n a_{n+1}}$, and $a_{n+1}+n \leqslant 2 \sqrt{a_{n}(n+1)}, n=$ $1,2, \cdots$, , so $a_{n}(n+1) \geqslant n a_{n+1}$, hence, $\frac{a_{n+1}}{a_{n}} \leqslant \frac{n+1}{n}$. Replacing $n$ with $n+1, n+2, \cdots$, $2 n$, we get $\frac{a_{n+2}}{a_{n+1}} \leq... | a_{n} \geqslant n | Inequalities | proof | Yes | Yes | inequalities | false | 733,472 |
18. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive numbers, prove: $\sqrt{a_{1}+a_{2}+\cdots+a_{n}}+\sqrt{a_{2}+a_{3}+\cdots+a_{n}}+$ $\sqrt{a_{3}+\cdots+a_{n}}+\cdots+\sqrt{a_{n}} \geqslant \sqrt{a_{1}+4 a_{2}+9 a_{3}+\cdots+n^{2} a_{n}}$. (2006 Jiangxi Province Mathematics Competition) | 18. Induction on $n$.
When $n=1$, the inequality obviously holds.
Assume that when $n=k$, the conclusion holds for any $k$ positive numbers. When $n=k+1$, for any $k+1$ positive numbers $a_{1}, a_{2}, \cdots, a_{k}, a_{k+1}$, according to the induction hypothesis, we have $\sqrt{a_{2}+a_{3}+\cdots+a_{k+1}}+\sqrt{a_{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,473 |
19. Let $x_{1}, x_{2}, x_{3}, \cdots$ be a decreasing sequence of positive numbers and for any positive integer $n$ we have $x_{1}+\frac{x_{4}}{2}+$ $\frac{x_{9}}{3}+\cdots+\frac{x_{n}^{2}}{n} \leqslant 1$. Prove that for any positive integer $n$ we have $x_{1}+\frac{x_{2}}{2}+\frac{x_{3}}{3}+\cdots+\frac{x_{n}}{n}<3$. | 19. Since $x_{1}, x_{2}, x_{3}, \cdots$ is a decreasing sequence of positive numbers, for any positive integer $k$ we have
$$\frac{x_{k}^{2}}{k^{2}}+\frac{x_{k+1}^{2}}{k^{2}+1}+\cdots+\frac{x_{(k+1)^{2}}-1}{(k+1)^{2}-1}<(2 k+1) \frac{x_{k}^{2}}{k^{2}}=\left(2+\frac{1}{k}\right) \frac{x_{k}^{2}}{k} \leqslant 3 \cdot \fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,474 |
20. Given a strictly increasing unbounded sequence of positive numbers $a_{1}, a_{2}, \cdots$. Prove:
(1) There exists a positive integer $k_{0}$ such that for all $k \geqslant k_{0}$, we have $\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{k}}{a_{k+1}}<k-1$.
(2) When $k$ is sufficiently large, we have $\frac{... | 20. Let $S_{k}=\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{k}}{a_{k+1}}$, it can be proven that for any $M>0$, there exists a positive integer $k_{0}$ such that when $k \geqslant k_{0}$, we have
$$S_{k}<k-M$$
From this, we can conclude that (1) and (2) both hold.
In fact, since $a_{1}, a_{2}, \cdots$ is st... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,475 |
21. 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 CMO) | 21. Let $n=m q+r, q \in \mathbf{N}, 0 \leqslant r<m$. Then, by the condition $a_{m+n} \leqslant a_{n}+a_{m}$, we have
$$\begin{aligned}
a_{n} \leqslant & a_{m q}+a_{r} \leqslant q a_{m}+a_{r}=\frac{n-r}{m} a_{m}+a_{r}= \\
& \left(\frac{n}{m}-1\right) a_{m}+\frac{m-r}{m} a_{m}+a_{r} \leqslant \\
& \left(\frac{n}{m}-1\ri... | a_{n} \leqslant m a_{1}+\left(\frac{n}{m}-1\right) a_{m} | Inequalities | proof | Yes | Yes | inequalities | false | 733,476 |
22. The sequence $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfies the condition $a_{1}=1, a_{n+1}=\sqrt{a_{n}^{2}+\frac{1}{a_{n}}}, n=1,2,3, \cdots$. Prove that there exists a positive number $\alpha$, such that $\frac{1}{2} \leqslant \frac{a_{n}}{n^{\alpha}} \leqslant 2$. (1988 Swedish Mathematical Olympiad problem) | 22. Prove by mathematical induction that
$$\frac{1}{2} n^{\frac{1}{3}} \leqslant a_{n} \leqslant 2 n^{\frac{1}{3}}, n=1,2,3, \cdots$$
That is, $\alpha=\frac{1}{3}$ is the desired value.
In fact, when $n=1$, by $a_{1}=1$ we know that (1) obviously holds. Assume that when $n=k$, (1) holds. Then, when $n=k+1$, we have
$$... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,478 |
23. Given that $n$ is a positive integer, prove that $\sum_{k=1}^{n} \frac{1}{k^{3}}<\frac{5}{4}$. | 23. When $n=1,2$, direct verification shows that the inequality obviously holds. When $k \geqslant 3$,
$$\frac{1}{k^{3}}<\frac{1}{(k-1) k(k+1)}=\frac{1}{2}\left[\frac{1}{(k-1) k}-\frac{1}{k(k+1)}\right], k=3,4, \cdots, n$$
Adding up, we get
$$\sum_{k=3}^{n} \frac{1}{k^{3}}<\frac{1}{2}\left[\frac{1}{6}-\frac{1}{n(n+1)}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,479 |
24- Given the sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}$ satisfy $a_{1}>\theta, b_{1}>0, a_{n+1}=a_{n}+\frac{1}{b_{n}}, b_{n+1}=b_{n}+\frac{1}{a_{n}}$, prove: $a_{25}+b_{25}>10 \sqrt{2}$. (1996 Lithuanian Mathematical Olympiad Problem) | 24. We consider a general problem:
$u, v, w$ are all positive numbers, and the sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}$ satisfy $a_{1}>0, b_{1}>0, a_{n+1}=u a_{n}+\frac{v}{b_{n}}, b_{n+1}=$ $\frac{b_{n}}{u}+\frac{w}{a_{n}}$, we prove that when $n \geqslant 3$, $a_{n} b_{n}>(n-1)\left(\frac{v}{u}+u w\right)... | a_{25}+b_{25}>10 \sqrt{2} | Algebra | proof | Yes | Yes | inequalities | false | 733,480 |
25. For each positive integer $n$ define $f(n)=\left\{\begin{array}{l}1, n=1 \\ \frac{n}{f(n-1)}, n \geqslant 2\end{array}\right.$ Prove: $\sqrt{1992}<$ $f(1992)<\frac{4}{3} \sqrt{1992}$ (1992: Danish Mathematical Olympiad Problem) | 25. We prove that when $n$ is even and $n \geqslant 6$,
$$\sqrt{n+1}\frac{2 k}{2 k-1} \cdot \frac{2 k-2}{2 k-3} \cdots \cdots \cdot \frac{6}{5} \cdot \frac{4}{3} \cdot \frac{2}{1} \\
f \cdot \frac{2 k-1}{2 k-2} \cdot \cdots \cdot \frac{7}{6} \cdot \frac{5}{4} \cdot \frac{3}{2}
\end{array}$$
Multiplying (1) and (2) giv... | \sqrt{1992}<f(1992)<\frac{4}{3} \sqrt{1992} | Inequalities | proof | Yes | Yes | inequalities | false | 733,481 |
26. The sequence $\left\{a_{n}\right\}(n=0,1,2, \cdots)$ is a sequence of real numbers, and satisfies $a_{n+1} \geqslant a_{n}^{2}+\frac{1}{5}(n \geqslant 0)$. Prove that when $n \geqslant 5$, $\sqrt{a_{n+5}} \geqslant a_{n-5}^{2}$. (2001 USA National Training Team Problem) | 26. From the known $a_{n+1} \geqslant a_{n}^{2}+\frac{1}{5}$, we can deduce:
$$\begin{array}{l}
a_{n+1} \geqslant a_{n}^{2}+\frac{1}{5} \\
a_{n+2} \geqslant a_{n+1}^{2}+\frac{1}{5} \\
a_{n+3} \geqslant a_{n+2}^{2}+\frac{1}{5} \\
a_{n+4} \geqslant a_{n+3}^{2}+\frac{1}{5} \\
a_{n+5} \geqslant a_{n+4}^{2}+\frac{1}{5}
\end... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,482 |
27. Given the sequence $\left\{a_{n}\right\}$ defined as follows: $a_{k}=\frac{1}{2} \cdot \frac{3}{4} \cdot \cdots \cdot \frac{2 k-3}{2(k-1)} \cdot \frac{2 k-1}{2 k}(k=1$, $2, \cdots)$. Prove:
(1) $a_{n+1}<\frac{1}{\sqrt{2 n+3}}$;
(2) For any positive integer $n$, $\sum_{k=1}^{n} a_{k}<\sqrt{2(n+1)}-$
1. (2006 Sichuan... | 27. (1) Let $A=\frac{1}{2} \cdot \frac{3}{4} \cdot \cdots \cdot \frac{2 n-1}{2 n} \cdot \frac{2 n+1}{2(n+1)}, B=\frac{2}{3} \cdot \frac{4}{5} \cdot \cdots \cdot \frac{2 n}{2 n+1} \cdot$ $\frac{2(n+1)}{2 n+3}$, then $A<B$. Therefore, $A^{2}<A B=\frac{1}{2 n+3}$.
So $A<\frac{1}{\sqrt{2 n+3}}$, which means $a_{n+1}<\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,483 |
28. Given $x_{n}=\frac{(2 n+1)(2 n+3) \cdots(4 n-1)(4 n+1)}{(2 n)(2 n+2) \cdots(4 n-2)(4 n)}$, prove: $\frac{1}{4 n}<x_{n}-$ $\sqrt{2}<\frac{2}{n}$. (2001 Polish Mathematical Olympiad Problem) | 28. Using the inequality $(x+1)(-x-1)-\frac{(4 n+1)^{2}}{(2 n)(4 n)}=\frac{16 n^{2}+8 n+1}{8 n^{2}}>\frac{16 n^{2}+8 n}{8 n^{2}}=2+\frac{1}{n}$
From (2) we get $x_{n}^{2}-2>0$, so,
$$x_{n}-\sqrt{2}=\frac{x_{n}^{2}-2}{x_{n}+\sqrt{2}}\frac{x_{n}^{2}-2}{2+\sqrt{2}}>\frac{1}{(2+\sqrt{2}) n}>\frac{1}{4 n}$$
In summary, $\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,484 |
29. Given that $a_{1}, a_{2}, \cdots, a_{n}$ are all positive numbers, prove:
$$\frac{\left(a_{1}+a_{2}+a_{3}\right)\left(a_{2}+a_{3}+a_{4}\right) \cdots\left(a_{n}+a_{1}+a_{2}\right)}{\left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \cdots\left(a_{n}+a_{1}\right)}>(\sqrt{2})^{n}$$ | 29.
$$\begin{array}{l}
\left(a_{1}+a_{2}+a_{3}\right)\left(a_{2}+a_{3}+a_{4}\right) \cdots\left(a_{n}+a_{1}+a_{2}\right)= \\
\left(a_{1}+\frac{1}{2} a_{2}+\frac{1}{2} a_{2}+a_{3}\right)\left(a_{2}+\frac{1}{2} a_{3}+\frac{1}{2} a_{3}+a_{4}\right) \cdots\left(a_{n}+\frac{1}{2} a_{1}+\frac{1}{2} a_{1}+a_{2}\right) \geqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,485 |
27. Let $a, b, c$ be real numbers, and $a+b+c=0$, prove: $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+3 \geqslant 6 a b c$. | 27. Since $a+b+c=0$, the inequality to be proved is symmetric with respect to $a, b, c$, so we can assume $a \geqslant b \geqslant c$, hence $c \leqslant 0$,
$$\begin{aligned}
a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+3-6 a b c= & a^{2} b^{2}+c^{2}\left[(a+b)^{2}-2 a b\right]+3-6 a b c= \\
& a^{2} b^{2}+c^{2}\left(c^{2}-2 a ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,488 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.