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20. Let $a, b, c$ be non-negative real numbers. Prove that: $a^{4}+b^{4}+c^{4}-2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+$ $a^{2} b c+b^{2} c a+c^{2} a b \geqslant 0$. (45th Moscow Mathematical Olympiad Problem) | 20. Using Schur's inequality when $r=2$, we get
$$a^{2}(a-b)(a-c)+b^{2}(b-a)(b-c)+c^{2}(c-a)(c-b) \geqslant 0$$
Expanding, we get
$$a^{4}+b^{4}+c^{4}-\left(a^{3} b+a b^{3}\right)-\left(b^{3} c+b c^{3}\right)-\left(a^{3} c+a c^{3}\right)+a^{2} b c+b^{2} c a+c^{2} a b \geqslant 0$$
Therefore,
$$a^{4}+b^{4}+c^{4}+a^{2} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,778 |
22. Let $x, y, z$ be positive numbers, and $x+y+z=3$, prove: $\frac{x^{2}}{x+y^{2}}+\frac{y^{2}}{y+z^{2}}+\frac{z^{2}}{z+x^{2}} \geqslant \frac{3}{2}$. (2010 Croatian Mathematical Olympiad Problem) | 22. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
\left(\frac{x^{2}}{x+y^{2}}+\frac{y^{2}}{y+z^{2}}+\frac{z^{2}}{z+x^{2}}\right)\left[x^{2}\left(x+y^{2}\right)+y^{2}\left(y+z^{2}\right) z^{2}+z^{2}\left(z+x^{2}\right)\right] \geqslant \\
\left(x^{2}+y^{2}+z^{2}\right)^{2}
\end{array}$$
It suffices to prove that
$$\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,780 |
Example 2 Let $a, b, \epsilon$ be positive real numbers, and satisfy $abc=1$. Try to prove: $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+$ $\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}$. (Problem from the 36th IMO) | Prove that starting from the condition $a b c=1$, the left side of the original inequality can be transformed. Let the left side be $S$, then
$$\begin{aligned}
S= & \frac{(a b c)^{2}}{a^{3}(b+c)}+\frac{(a b c)^{2}}{b^{3}(c+a)}+\frac{(a b c)^{2}}{c^{3}(a+b)}= \\
& \frac{b c}{a(b+c)} \cdot b c+\frac{a c}{b(c+a)} \cdot a ... | S \geqslant \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,782 |
52. Let $x, y, z$ be positive real numbers, prove that $x^{3}\left(y^{2}+z^{2}\right)^{2}+y^{3}\left(z^{2}+x^{2}\right)^{2}+z^{3}\left(x^{2}+y^{2}\right)^{2} \geqslant$ $x y z\left[x y(x+y)^{2}+y z(y+z)^{2}+z x(z+x)^{2}\right$. (2009 USA National Training Team Problem) | 52. Proof
$$\begin{array}{l}
x^{3}\left(y^{2}+z^{2}\right)^{2}+y^{3}\left(z^{2}+x^{2}\right)^{2}+z^{3}\left(x^{2}+y^{2}\right)^{2} \geqslant \\
x y z\left[x y(x+y)^{2}+y z(y+z)^{2}+z x(z+x)^{2}\right] \Leftrightarrow \\
x\left[\left(x y^{2}+x z^{2}\right)^{2}-\left(y^{2} z+y z^{2}\right)^{2}\right]+y\left[\left(y z^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,784 |
Example 6 Let $\frac{1}{2} \leqslant p \leqslant 1, a_{i} \geqslant 0,0 \leqslant b_{i} \leqslant p(i=1,2, \cdots, n, n \geqslant 2)$, if $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}=1$, prove: $\sum_{i=1}^{n} b_{i} \prod_{1 \leqslant j \leqslant n} a_{j} \leqslant \frac{p}{(n-1)^{n-1}}$. (32nd IMO Shortlist Problem) | Let $A_{i}=a_{1} a_{2} \cdots a_{i-1} a_{i+1} \cdots a_{n}$, by the rearrangement inequality, without loss of generality, assume $b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}, A_{1} \geqslant A_{2} \geqslant \cdots \geqslant A_{n}$. Since $0 \leqslant b_{i} \leqslant p$, and $\sum_{i=1}^{n} b_{i}=1, \frac{1}{... | \sum_{i=1}^{n} b_{i} A_{i} \leqslant p\left(\frac{1}{n-1}\right)^{n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 732,786 |
Example 7 Let $x, y, z \in \mathbf{R}^{+}$, and $x+y+z=1$, prove that:
$$\frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+x z}}+\frac{x z}{\sqrt{x z+x y}} \leqslant \frac{\sqrt{2}}{2}$$
(2006 China National Training Team Exam Question) | Prove that since $(x+y)(y+z)(z+x) \leqslant\left[\frac{2(x+y+z)}{3}\right]^{3}=\frac{8}{27}$, we only need to prove a slightly stronger conclusion:
$$\begin{array}{l}
\frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+x z}}+\frac{x z}{\sqrt{x z+x y}} \leqslant \frac{3 \sqrt{3}}{4} \sqrt{(x+y)(y+z)(z+x)} \Leftrightarrow \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,787 |
Example 8 Let $a_{1}, a_{2}, \cdots, a_{N}$ be positive real numbers, and $2 S=a_{1}+a_{2}+\cdots+a_{N}, n \geqslant m \geqslant 1$, then
$$\sum_{k=1}^{N} \frac{a_{k}^{n}}{\left(2 S-a_{k}\right)^{m}} \geqslant \frac{(2 S)^{n-m}}{N^{n-m-1}(N-1)^{m}}$$ | Assume without loss of generality that $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{N}$, then
$$\frac{1}{2 S-a_{1}} \geqslant \frac{1}{2 S-a_{2}} \geqslant \cdots \geqslant \frac{1}{2 S-a_{N}}$$
By Chebyshev's inequality, we have
$$\sum_{k=1}^{N} \frac{a_{k}^{n}}{\left(2 S-a_{k}\right)^{m}} \geqslant \frac{1}{... | \frac{(2 S)^{n-m}}{N^{n-m-1}(N-1)^{m}} | Inequalities | proof | Yes | Yes | inequalities | false | 732,788 |
Example 11 Let $0<x_{1} \leqslant \frac{x_{2}}{2} \leqslant \cdots \leqslant \frac{x_{n}}{n}, 0<y_{n} \leqslant y_{n-1} \leqslant \cdots \leqslant y_{1}$, prove:
$$\left(\sum_{k=1}^{n} x_{k} y_{k}\right)^{2} \leqslant\left(\sum_{k=1}^{n} y_{k}\right)\left[\sum_{k=1}^{n}\left(x_{k}^{2}-\frac{1}{4} x_{k} x_{k-1}\right) y... | Prove by mathematical induction on $n$.
When $n=1$, inequality (1) becomes an equality.
Assume that (1) holds for $n-1$, i.e.,
$$\left(\sum_{k=1}^{n-1} x_{k} y_{k}\right)^{2} \leqslant\left(\sum_{k=1}^{n-1} y_{k}\right)\left[\sum_{k=1}^{n-1}\left(x_{k}^{2}-\frac{1}{4} x_{k} x_{k-1}\right) y_{k}\right]$$
To prove (1), ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,791 |
1. Given $x_{i}, y_{i}(i=1,2, \cdots, n)$ are real numbers, and $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}, y_{1} \geqslant y_{2} \geqslant \cdots \geqslant$ $y_{n}$, and $z_{1}, z_{2}, \cdots, z_{n}$ is any permutation of $y_{i}, y_{2}, \cdots, y_{n}$. Prove: $\sum_{i=1}^{n}\left(x_{i}-y_{i}\right)^{2} \l... | 1. Since $z_{1}, z_{2}, \cdots, z_{n}$ is any permutation of $y_{1}, y_{2}, \cdots, y_{n}$, we have $\sum_{i=1}^{n} y_{i}^{2}=\sum_{i=1}^{n} z_{i}^{2}$, thus $\sum_{i=1}^{n}\left(x_{i}-y_{i}\right)^{2} \leqslant \sum_{i=1}^{n}\left(x_{i}-z_{i}^{-}\right)^{2}$ is equivalent to $\sum_{i=1}^{n} x_{i} y_{i} \geqslant \sum_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,792 |
2. (1) Let $a, b, c$ be the sides of a triangle, prove that: $a^{2}(b+c-a)+b^{2}(c+a-b)+$ $c^{2}(a+b-c) \leqslant 3 a b c$. (6th IMO problem) | 2. $(1)$ Let $a \geqslant b \geqslant c$, first prove the inequality
$$c(a+b-c) \geqslant b(c+a-b) \geqslant a(b+c-a)$$
In fact
$$\begin{array}{l}
c(a+b-c)-b(c+a-b)=a c+b c=c^{2}-a b-b c+b^{2}= \\
b^{2}-c^{2}+a c-a b=(b-c)(b+c-a)
\end{array}$$
Since $b \geqslant c, b+c>a$, we have $b-c \geqslant 0, b+c-a>0$, thus
$$c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,793 |
Example 7 Positive real numbers $x, y, z$ satisfy $x y z \geqslant$ 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 IM $\Theta$ problem) | Prove that because $x y z \geqslant 1$, we have
$$\begin{array}{l}
\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}} \geqslant \frac{x^{5}-x^{2} \cdot x y z}{x^{5}+\left(y^{2}+z^{2}\right) \cdot x y z}=\frac{x^{4}-x^{2} y z}{x^{4}+y z\left(y^{2}+z^{2}\right)} \geqslant \\
\frac{2 x^{4}-x^{2}\left(y^{2}+z^{2}\right)}{2 x^{4}+\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,795 |
53. Let $x, y, z$ be positive real numbers, and satisfy the equation: $x^{2}+y^{2}+z^{2}+9=4(x+y+z)$, prove: $x^{4}+y^{4}+z^{4}+16\left(x^{2}+y^{2}+z^{2}\right) \geqslant 8\left(x^{3}+y^{3}+z^{3}\right)+27$. (2009 Central European Mathematical Olympiad Problem) | 53. Notice that
$$x^{4}-8 x^{3}+16 x^{2}-9+6\left(x^{2}-4 x+3\right)=(x-1)^{2}(x-3)^{2} \geqslant 0$$
Therefore,
$$x^{4}-8 x^{3}+16 x^{2}-9 \geqslant-6\left(x^{2}-4 x+3\right)$$
So,
$$\begin{array}{l}
x^{4}+y^{4}+z^{4}+16\left(x^{2}+y^{2}+z^{2}\right)-8\left(x^{3}+y^{3}+z^{3}\right)-27= \\
\left(x^{4}-8 x^{3}+16 x^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,796 |
4. 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) | 4. (1)-Since the inequality is symmetric with respect to $a, b, c$, without loss of generality, assume $0<a \leqslant b \leqslant c$, and let $x=$ $\frac{1}{b+c}, y=\frac{1}{c+a}, z=\frac{1}{a+b}$, then we have $x \leqslant y \leqslant z$. By the rearrangement inequality, we have
$x a+y b+z c \geqslant z b+y c+z a($ ou... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,798 |
(2) $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{a+b+c}{2}$
(Problem from the 2nd World Friendship Cup Mathematics Competition) | (2) Since the inequality is symmetric with respect to $a, b, c$, without loss of generality, assume $0<a \leqslant b \leqslant c$, and let $x=\frac{a}{b+c}$, $y=\frac{b}{c+a}$, $z=\frac{c}{a+b}$, then we have $0<x \leqslant y \leqslant z$. By the rearrangement inequality, we have
$$\begin{array}{l}
x a+y b+z c \geqslan... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,799 |
5. (1) Given that $a_{1}, a_{2}, \cdots, a_{n}$ are positive numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1$, prove that: $\frac{a_{1}}{2-a_{1}}+$ $\frac{a_{2}}{2-a_{2}}+\cdots+\frac{a_{n}}{2-a_{n}} \geqslant \frac{n}{2 n-1}$ (1984 Balkan Mathematical Olympiad Problem) | 5. (1) Since $a_{1}+a_{2}+\cdots+a_{n}=1$, then $2\left(a_{1}+a_{2}+\cdots+a_{n}\right)=2$. Since the inequality is a symmetric inequality, without loss of generality, assume $0<a_{1}<a_{2}<\cdots<a_{n}$, and let $A_{1}=\frac{1}{2-a_{1}}, A_{2}=\frac{a_{2}}{2-a_{2}}, \cdots, A_{n}=\frac{1}{2-a_{n}}$, then $0<A_{1}<A_{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,800 |
6. Let $0 \leqslant a \leqslant b \leqslant c \leqslant d \leqslant e$, and $a+b+c+d+e=1$. Prove: $a d+d c+c b+$ $b e+e a \leqslant \frac{1}{5}$. (1994 Chinese National Training Team Test Question) | 6. Since $\theta \leqslant a \leqslant b \leqslant c \leqslant d \leqslant e$, therefore, $d+e \geqslant c+e \geqslant b+d \geqslant a+c \geqslant a+b$. By Chebyshev's inequality, we have
$$\begin{array}{l}
a(d+e)+b(c+e)+c(b+d)+d(a+c)+e(a+b) \leqslant \\
\frac{1}{5}(a+b+c+d+e)[(d+e)+(c+e)+(b+d)+(a+c)+(a+b)]= \\
\frac{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,802 |
7. Prove: For any positive real numbers $a, b, c$, we have $\frac{a^{3}}{b c}+\frac{b^{3}}{c a}+\frac{c^{3}}{a b} \geqslant a+b+c$. (2002 Canadian Mathematical Olympiad) | 7- Let's assume $a \leqslant b \leqslant c$, then
$$\begin{array}{l}
\frac{a^{3}}{b c}+\frac{b^{3}}{c a}+\frac{c^{3}}{a b}=\frac{a}{b c} \cdot a^{2}+\frac{b}{c a} \cdot b^{2}+\frac{c}{a b} \cdot c^{2} \geqslant \\
\frac{a}{b c} \cdot b^{2}+\frac{b}{c a} \cdot c^{2}+\frac{c}{a b} \cdot a^{2}=\frac{a b}{c}+\frac{b c}{a}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,803 |
8. Let $a, b, c$ be positive numbers, prove that: $\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right) \geqslant 2\left(1+\frac{a+b+c}{\sqrt[3]{a b c}}\right)$. (1998 Asia Pacific Mathematical Olympiad Problem) | 8. The original inequality is equivalent to
$$\begin{array}{l}
(a+b)(b+c)(c+a) \geqslant 2 a b c+2(a+b+c) \sqrt[3]{(a b c)^{2}} \Leftrightarrow \\
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \geqslant 2(a+b+c) \sqrt[3]{(a b c)^{2}}
\end{array}$$
Since
$$\sum a^{2}(b+c) \geqslant 2\left(\sqrt{a^{3}}+\sqrt{b^{3}}+\sqrt{c^{3}}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,804 |
9. Given that $x, y, z$ are positive numbers, prove that $\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}} \geqslant \sqrt{\frac{3}{2}(x+y+z)}$. (2005 Serbian Mathematical Olympiad Problem) | 9. Suppose $x \geqslant y \geqslant z>0$, then
$$\frac{1}{\sqrt{y+z}} \geqslant \frac{1}{\sqrt{z+x}} \geqslant \frac{1}{\sqrt{x+y}}$$
By the Cauchy-Schwarz inequality, we have
$$(\sqrt{y+z}+\sqrt{z+x}+\sqrt{x+y})\left(\frac{1}{\sqrt{y+z}}+\frac{1}{\sqrt{z+x}}+\frac{1}{\sqrt{x+y}}\right) \geqslant 9$$
Thus,
$$\frac{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,805 |
10. Let $a_{1}, a_{2}, \cdots, a_{N}$ be positive real numbers, and $2 S=a_{1}+a_{2}+\cdots+a_{N}, n \geqslant m \geqslant 1$, denote $P=\sum_{k=1}^{N} a_{k}^{m}$, then $\sum_{k=1}^{N} \frac{a_{k}^{n}}{P-a_{k}^{m}} \geqslant \frac{(2 S)^{n-m}}{(N-1) N^{n-m-1}}$. (A generalization of another problem from the 28th IMO Sh... | 10. Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{N}$, then $a_{1}^{n} \geqslant a_{2}^{n} \geqslant \cdots \geqslant a_{N}^{n}$, so
$$\begin{array}{c}
0 \leqslant P-a_{1}^{n} \leqslant P-a_{2}^{n} \leqslant \cdots \leqslant P-a_{N}^{n} \\
1-a_{1}^{m} \geqslant P-a_{2}^{m} \geqslant \cdots \geqslant \frac{1}... | \frac{(2 S)^{n-m}}{(N-1) N^{n-m-1}} | Inequalities | proof | Yes | Yes | inequalities | false | 732,806 |
11. Let $a, b, c$ be the lengths of the sides of a triangle. Prove that: $a^{2} b(a-b)+b^{2} c(b-c)+$ $c^{2} a(c-a) \geqslant 0$, and determine the conditions under which equality holds. (24th IMO Problem) | $$^{15}$$ Since the conditions of the problem are cyclically symmetric with respect to $a, b, c$, we can assume without loss of generality that $a$ is the largest side. Let $a \geqslant b \geqslant c$, and $s=\frac{1}{2}(a+b+c)$. Then we have:
$$2(s-a) \leqslant 2(s-b) \leqslant 2(s-c)$$
Because
$$2 a(s-a)-2 b(s-b)=(a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,808 |
14. Let $x, y, z$ be positive numbers, and $x+y+z \geqslant 1$, prove: $\frac{x \sqrt{x}}{y+z}+\frac{y \sqrt{y}}{z+x}+\frac{z \sqrt{z}}{x+y} \geqslant \frac{\sqrt{3}}{2}$. (2003 Moldova Mathematical Olympiad Problem) | 14. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
\left(\frac{x \sqrt{x}}{y+z}+\frac{y \sqrt{y}}{z+x}+\frac{z \sqrt{z}}{x+y}\right)[\sqrt{x}(y+z)+\sqrt{y}(z+x)+\sqrt{z}(x+y)] \geqslant \\
(x+y+z)^{2}=1
\end{array}$$
Assume without loss of generality that $x \geqslant y \geqslant z$, then by Chebyshev's inequality,
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,811 |
15. Let \(x, y, z \in \left(0, \frac{\pi}{2}\right)\), prove: \(x + y + z \geqslant x\left(\frac{\sin y}{\sin x}\right) + y\left(\frac{\sin z}{\sin y}\right) + z\left(\frac{\sin x}{\sin z}\right)\).
(2005 Ukrainian Mathematical Olympiad Problem) | 15. Let $x \geqslant y \geqslant z$, then since $x, y, z \in\left(0, \frac{\pi}{2}\right)$, we have $\sin x \geqslant \sin y \geqslant \sin z$. Since $\left(\frac{x}{\sin x}\right)^{\prime}=\frac{\cos x(\tan x-x)}{\sin ^{2} x}>0$, it follows that $\frac{x}{\sin x} \geqslant \frac{y}{\sin y} \geqslant \frac{z}{\sin z}$.... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,812 |
16. Let $\alpha, \beta, \gamma$ be the three interior angles of a triangle. Prove that: $2\left(\frac{\sin \alpha}{\alpha}+\frac{\sin \beta}{\beta}+\frac{\sin \gamma}{\gamma}\right) \leqslant$ $\left(\frac{1}{\beta}+\frac{1}{\gamma}\right) \sin \alpha+\left(\frac{1}{\gamma}+\frac{1}{\alpha}\right) \sin \beta+\left(\fra... | 16. Let $\alpha \leqslant \beta \leqslant \gamma$, since $\alpha, \beta, \gamma$ are the three interior angles of a triangle, therefore, $\sin \alpha \leqslant \sin \beta \leqslant \sin \gamma$.
By the rearrangement inequality, we have
$$\begin{array}{l}
\frac{\sin \alpha}{\alpha}+\frac{\sin \beta}{\beta}+\frac{\sin \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,813 |
17. Let $x_{i}>0, i=1,2, \cdots, n, k \geqslant 1$, prove: $\sum_{i=1}^{n} \frac{1}{1+x_{i}} \sum_{i=1}^{n} x_{i} \leqslant \sum_{i=1}^{n} \frac{x_{i}^{k+1}}{1+x_{i}}$. $\sum_{i=1}^{n} \frac{1}{x_{i}^{k}}$. (2006 China Girls Mathematical Olympiad Problem) | 17. Suppose $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}>0$, then
$$\frac{1}{x_{1}^{k}} \leqslant \frac{1}{x_{2}^{k}} \leqslant \cdots \leqslant \frac{1}{x_{n}^{k}}, \frac{x_{1}^{k}}{1+x_{1}} \geqslant \frac{x_{2}^{k}}{1+x_{2}} \geqslant \cdots \geqslant \frac{x_{n}^{k}}{1+x_{n}}, k \geqslant 1$$
Thus, acco... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,814 |
18. Let $0<p \leqslant a_{i} \leqslant q(i=1,2, \cdots, n), b_{1}, b_{2}, \cdots, b_{n}$ be a permutation of $a_{1}, a_{2}, \cdots, a_{n}$, prove that: $n \leqslant \frac{a_{1}}{b_{1}}+\frac{a_{2}}{b_{2}}+\cdots+\frac{a_{n}}{b_{n}} \leqslant n+\left[\frac{n}{2}\right]\left(\sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}\right)^{... | 18. By the mean inequality, we have $\frac{a_{1}}{b_{1}}+\frac{a_{2}}{b_{2}}+\cdots+\frac{a_{n}}{b_{n}} \geqslant n$. It is only necessary to prove the inequality on the right. Without loss of generality, assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, by the rearrangement inequality we get
$$\frac{a_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,815 |
19. Given a positive integer $n \geqslant 2$. Find the smallest positive number $\lambda$, such that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ and any $n$ positive numbers $b_{1}, b_{2}, \cdots, b_{n}$ in $[0, \left.\frac{1}{2}\right]$, if $a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n}=1$, then $a_{1} ... | 19. By Cauchy-Schwarz inequality,
$$1=\sum_{i=1}^{n} b_{i}=\sum_{i=1}^{n}\left(\frac{\sqrt{b_{i}}}{\sqrt{a_{i}}} \cdot \sqrt{a_{i} b_{i}}\right) \leqslant\left(\sum_{i=1}^{n} \frac{b_{i}}{a_{i}}\right)^{\frac{1}{2}}\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{\frac{1}{2}}$$
Thus,
$$\frac{1}{\sum_{i=1}^{n} a_{i} b_{i}} \le... | \frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,816 |
55. Let $x, y, z \in \left[\frac{1}{2}, 2 \right]$, and $a, b, c$ be a permutation of them. Prove: $\frac{60 a^{2}-1}{4 x y+5 z}+$ $\frac{60 b^{2}-1}{4 y z+5 x}+\frac{60 c^{2}-1}{4 z x+5 y} \geqslant 12 .(2009$ Moldova Training Team Problem) | 55. Since $x, y \in\left[\frac{1}{2}, 2\right]$, we have
$$\left\{\begin{array} { l }
{ ( x - \frac { 1 } { 2 } ) ( y - 2 ) \leqslant 0 } \\
{ ( y - \frac { 1 } { 2 } ) ( x - 2 ) \leqslant 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x y-2 x-\frac{1}{2} y+1 \leqslant 0 \\
x y-2 y-\frac{1}{2} x+1 \leqslant 0
\en... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,818 |
21. Given that $a, b, c, d$ are positive real numbers, and satisfy $a+b+c+d=4$, prove: $a^{2} b c+b^{2} c d+$ $c^{2} d a+d^{2} a b \leqslant 4$. (2007 Middle European Mathematical Olympiad Problem) | 21_ Let $p=q, r, s$ be a permutation of $a, b, c, d$, and $p=q \geqslant \varepsilon \geqslant s$, then $p q r \geqslant p q s \geqslant p r s \geqslant q r s$. By the rearrangement inequality, we have
$$\begin{array}{l}
a^{2} b c+b^{2} c d+c^{2} d a+d^{2} a b= \\
a(a b c)+b(b c d)+c(c d a) \pm d(d a b) \leqslant \\
p(... | 4 | Inequalities | proof | Yes | Yes | inequalities | false | 732,819 |
22. Let $\alpha, \beta, \gamma$ be the three interior angles of a triangle, and their opposite sides be $a, b, c$, respectively. Prove: $\frac{a}{\alpha(b+c-a)}+\frac{b}{\beta(c+a-b)}+\frac{c}{\gamma(a+b-c)} \geqslant \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$. (1999 Polish Mathematical Olympiad Problem) | 22. We first prove $a \geqslant b \Leftrightarrow \frac{1}{\alpha(p-a)} \geqslant \frac{1}{\beta(p-b)}$.
Let $f(x)=\frac{\tan x}{x}$, where $x \in\left(\theta, \frac{\pi}{2}\right)$. Since
$$f^{\prime}(x)=\frac{x-\sin x \cos x}{x^{2} \cos ^{2} x}=\frac{2 x-\sin 2 x}{2 x^{2} \cos ^{2} x}>0$$
From $a \geqslant b$ we ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,820 |
23. The equation $x^{3}-a x^{2}+b x-c=0$ has three positive roots (which can be equal), find the minimum value of $\frac{1+a+b+c}{3+2 a+b}-\frac{c}{b}$. (2008 Turkey Training Team Problem) | 23. Let the three positive roots (which can be equal) of the equation $x^{3}-a x^{2}+b x-c=0$ be $p, q, r$. By Vieta's formulas, we have $a=p+q+r, b=p q+q r+r p, c=p q r$, so
$$\begin{array}{c}
\frac{1+a+b+c}{3+2 a+b}=\frac{(1+p)(1+q)(1+r)}{(1+p)(1+q)+(1+q)(1+r)+(1+r)(1+p)} \\
\frac{c}{b}=\frac{p q r}{p q+q r+r p}
\end... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,821 |
25. Let $0<x, y, z<1$, prove: $\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z} \geqslant \frac{3 \sqrt[3]{x y z}}{1-\sqrt[3]{x y z}}$. 2002 Irish Mathematical Olympiad Problem) | 25. Suppose $0<z \leqslant y \leqslant x<1$, then $\frac{1}{1-z} \leqslant \frac{1}{1-y} \leqslant \frac{1}{1-x}$. By Chebyshev's inequality and the mean value inequality, we get
$$\begin{array}{l}
\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z} \geqslant \frac{1}{3}(x+y+z)\left(\frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,824 |
26. Given that $a, b, c$ are positive real numbers, prove: $\sqrt[4]{\frac{\left(a^{2}+b^{2}\right)\left(a^{2}-a b+b^{2}\right)}{2}}+$
$$\sqrt[4]{\frac{\left(b^{2}+c^{2}\right)\left(b^{2}-b c+c^{2}\right)}{2}}+\sqrt[4]{\frac{\left(c^{2}+a^{2}\right)\left(c^{2}-c a+a^{2}\right)}{2}} \leqslant \frac{2}{3}\left(a^{2}+b^{2... | 26. By Chebyshev's inequality, we have
$$\begin{array}{l}
\frac{2}{3}\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)= \\
\frac{1}{3}\left[\left(a^{2}+b^{2}\right)+\left(b^{2}+c^{2}\right)+\left(c^{2}+a^{2}\right)\right]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right) \geqslant... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,825 |
27. Given $c \in\left(0, \frac{1}{2}\right)$, find the smallest constant $M$, such that for any positive integer $n \geqslant 2$ and real numbers $0 < a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, if
$$\frac{1}{n} \sum_{k=1}^{n} k a_{k}=c \sum_{k=1}^{q} a_{k}$$
then
$$\frac{1}{n} \sum_{k=1}^{n} a_{k} \leqsl... | 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 | math-word-problem | Yes | Yes | inequalities | false | 732,826 |
28. Let $a, b, c$ be positive real numbers, and $a+b+c=3$, prove: $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant a^{2}+b^{2}+c^{2}$ (2006 Romanian National Training Team Problem) | 28. The inequality is equivalent to proving
$$\begin{aligned}
\sum_{a} a^{2} b^{2} \geqslant & a^{2} b^{2} c^{2} \sum_{a r} a^{2} \Leftrightarrow \sum_{a c} a^{2} b^{2}\left(1-c^{4}\right)= \\
& \sum_{c, c} a^{2} b^{2}(1-c)\left(1+c+c^{2}+c^{3}\right) \geqslant 0
\end{aligned}$$
If $a b \leqslant 2$, and $a \geqslant ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,827 |
29. Given that $a, b, c$ are positive numbers, and $a^{4}+b^{4}+c^{4}=3$, prove: $\frac{1}{4-a b}+\frac{1}{4-b c}+\frac{1}{4-c a} \leqslant$
1. (2005 Moldova Mathematical Olympiad Problem) | 29. Let $x=ab, y=bc, z=ca$, then $\frac{1}{4-ab}+\frac{1}{4-bc}+\frac{1}{4-ca} \leqslant 1$ is equivalent to
$$\frac{1+x}{4-x}+\frac{1+y}{4-y}+\frac{1+z}{4-z} \geqslant 0 \Leftrightarrow \frac{1-x^{2}}{4+3 x-x^{2}}+\frac{1-y^{2}}{4+3 y-y^{2}}+\frac{1-z^{2}}{4+3 z-z^{2}} \geqslant 0$$
Notice that $a^{4}+b^{4}+c^{4}=3$,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,828 |
30. Given that $x, y$ are positive real numbers, and $m, n$ are positive integers, prove:
$$\begin{array}{l}
(n-1)(m-1)\left(x^{m+n}+y^{m+n}\right)+(-m+n-1)\left(x^{m} y^{n}+y^{m} x^{n}\right) \geqslant \\
m n\left(x^{m+n} y+y^{m+n-1}-x\right)
\end{array}$$
(1995 Austria-Poland Mathematical Competition Problem) | 30. When $x=y$, the inequality obviously becomes an equality. Without loss of generality, let's assume $x>y$,
$$\begin{array}{l}
(n-1)(m-1)\left(x^{m+n}+y^{m+n}\right)+(m+n-1)\left(x^{m} y^{n}+y^{m} x^{n}\right) \geqslant \\
m n\left(x^{m+n-1} y+y^{m+n-1} x\right)
\end{array}$$
This can be transformed into
$$\begin{ar... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,830 |
Example 2 Given a tetrahedron $O-ABC$ with three lateral edges $OA, OB, OC$ mutually perpendicular, $P$ is a point within the base $ABC$, and $OP$ forms angles $\alpha, \beta, \gamma$ with the three lateral faces, respectively. Prove that $\frac{\pi}{2}<\alpha+\beta+\gamma \leqslant 3 \arcsin \frac{\sqrt{3}}{3}$. (2004... | Given the problem, we have $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=1$, and $\alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$, so
$$\begin{aligned}
\sin ^{2} \alpha= & 1-\left(\sin ^{2} \beta+\sin ^{2} \gamma\right)=\cos ^{2} \beta-\sin ^{2} \gamma= \\
& \frac{1}{2}(1+\cos 2 \beta)-\frac{1}{2}(1-\cos 2... | \frac{\pi}{2}<\alpha+\beta+\gamma \leqslant 3 \arcsin \frac{\sqrt{3}}{3} | Geometry | proof | Yes | Yes | inequalities | false | 732,834 |
Example 3 Given $5 n$ real numbers $r_{i}, s_{i}, t_{i}, u_{i}, v_{i}(\mathrm{~F} \leqslant i \leqslant n)$ all greater than 1, let $R=$ $\left(\frac{1}{n} \sum_{i=1}^{n} r_{i}\right), S=\left(\frac{1}{n} \sum_{i=1}^{n} s_{i}\right), T=\left(\frac{1}{n} \sum_{i=1}^{n} t_{i}\right), U=\left(\frac{1}{n} \sum_{i=1}^{n} u_... | Prove that the function $y=\ln \frac{\mathrm{e}^{x}+1}{\mathrm{e}^{x}-1}$ is a convex function $(x>0)$.
For this, it suffices to prove that for real numbers $a, b>1$, we have $\left(\frac{a+1}{a-1}\right)\left(\frac{b+1}{b-1}\right) \geqslant\left(\frac{\sqrt{a b}+1}{\sqrt{a b}-1}\right)^{2}$, and this inequality is eq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,835 |
Example 4 Let $0<a, b, c<1$, and $a b+b c+c a=1$, prove that:
$$\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\frac{c}{1-c^{2}} \geqslant \frac{3 \sqrt{2}}{2}$$
(2004 Singapore National Team Selection Examination Question) | Prove that because $a b+b c+c a=1$, we can set $a=\cot A, b=\cot B, c=\cot C$, where $A, B, C$ are the three interior angles of a triangle, and from $0<a, b, c<1$ we know $\frac{\pi}{4}<A, B, C<\frac{\pi}{2}$. Therefore,
$$\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\frac{c}{1-c^{2}}=-(\tan 2 A+\tan 2 B+\tan 2 C)$$
We only ne... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,836 |
For example, let $5 \alpha, \beta, \gamma$ be the three interior angles of a given triangle. Prove that: $\csc ^{2} \frac{\alpha}{2}+\csc ^{2} \frac{\beta}{2}+$ $\csc ^{2} \frac{\gamma}{2} \geqslant 12$, and find the condition for equality. (1994 Korean Mathematical Olympiad) | Prove that by the arithmetic-geometric mean inequality,
$$\csc ^{2} \frac{\alpha}{2}+\csc ^{2} \frac{\beta}{2}+\csc ^{2} \frac{\gamma}{2} \geqslant 3 \sqrt[3]{\csc ^{2} \frac{\alpha}{2} \csc ^{2} \frac{\beta}{2} \csc ^{2} \frac{\gamma}{2}}$$
where equality holds if and only if $\alpha=\beta=\gamma$. By the arithmetic-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,837 |
Example 6 Let $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 2)$ all be positive numbers, and $\sum_{i=1}^{n} x_{i}=1$, prove that:
$$\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1-x_{i}}} \geqslant \frac{\sum_{i=1}^{n} \sqrt{x_{i}}}{\sqrt{n-1}}$$
(4th CMO Problem) | Prove that the function $f(x)=\frac{x}{\sqrt{1-x}}(0<x<1)$ is convex. For $0<x_{1}<1, 0<x_{2}<1$, and $x_{i}=1-\frac{1}{t_{i}^{2}}$ $(i=1,2)$, because
$$\begin{array}{l}
\frac{1}{2}\left(t_{1}+t_{2}\right) \geqslant \sqrt{t_{1} t_{2}} \\
\sqrt{t_{1}^{2}+t_{2}^{2}} \geqslant \sqrt{2 t_{1} t_{2}}
\end{array}$$
Thus,
$$\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,838 |
Example 7 Let $x, y, z \in \mathbf{R}^{+}$, and $x+y+z=1$, prove: $\frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+x z}}+$ $\frac{x z}{\sqrt{x z+x y}} \leqslant \frac{\sqrt{2}}{2} .(2006$ China National Training Team Exam Question) | Prove that since $\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=1$, by the generalized Jensen's inequality we have
$$\begin{aligned}
\sum_{c y c} \frac{x y}{\sqrt{x y+y z}}= & \sum_{c y c} \sqrt{\frac{x^{2} y}{x+z}}=\sum_{c y c} \frac{x+y}{2} \sqrt{\frac{4 x^{2} y}{(x+y)^{2}(x+z)}} \leqslant \\
& \sqrt{\sum_{c y c} \frac{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,839 |
57. Let $x, y, z$ be positive real numbers, and satisfy $x y z=1, \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geqslant x+y+z$, prove: for any positive integer $k$, we have $\frac{1}{x^{k}}+\frac{1}{y^{k}}+\frac{1}{z^{k}} \geqslant x^{k}+y^{k}+z^{k} \cdot(1999$ Russian Mathematical Olympiad problem) | 57. Since $x y z=1, \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geqslant x+y+z \Leftrightarrow(x-1)(y-1)(z-1) \leqslant 0$. For any positive integer $k, t-1$ has the same sign as $t^{k}-1$, so
$$\begin{array}{l}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geqslant x+y+z \Leftrightarrow(x-1)(y-1)(z-1) \leqslant 0 \Leftrightarrow \\
\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,840 |
(2) When $\lambda>\frac{5}{4}$, $\frac{a}{\sqrt{a^{2}+\lambda b^{2}}}+\frac{b}{\sqrt{b^{2}+\lambda c^{2}}}+\frac{c}{\sqrt{c^{2}+\lambda a^{2}}}<2$. (2005 China National Training Team Problem) | (2) If $p, q, r$ have at least two not greater than $\frac{2}{\lambda}$, without loss of generality, let $p, q \leqslant \frac{2}{\lambda}$. By Jensen's inequality, we have
$$f(\ln p)+f(\ln q) \leqslant 2 f\left(\frac{\ln p+\ln q}{2}\right)=2 f(\ln \sqrt{p q})$$
i.e.,
$$\frac{1}{\sqrt{1+\lambda p}}+\frac{1}{\sqrt{1+\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,842 |
1. (1) Let $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ be positive numbers. Prove: $\sqrt[n]{a_{1} a_{2} \cdots a_{n}}+$ $\sqrt[n]{b_{1} b_{2} \cdots b_{n}} \leqslant \sqrt[n]{\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right)\left(a_{n}+b_{n}\right)}$. (1992 Irish Mathematical Olympiad, 64th Putnam M... | 1. (1) First, prove that $f(\bar{x})=\Gamma n\left(1+\mathrm{e}^{x}\right)$ is a mountain function on $\mathbf{R}$. In fact, for any $x_{1}, x_{2} \in \mathbf{R}$, since $\mathrm{e}^{x_{1}}+\mathrm{e}^{x_{2}} \geqslant 2 \mathrm{e}^{\frac{x_{1}+x_{2}}{2}}$, we have
$$\left(1+\mathrm{e}^{x_{1}}\right)\left(\mathrm{F}+\m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,843 |
(2) Given that $a_{1}, a_{2}, \cdots, a_{n}$ are positive numbers, and $a_{1} a_{2} \cdots a_{n}=1$, prove: $\left(2+a_{1}\right)\left(2+a_{2}\right) \cdots$. $\left(2+a_{n}\right) \geqslant 3^{n}$. (1989 National High School Mathematics Competition Problem) | (2) Let $f(x)=\ln \left(2+\mathrm{e}^{x}\right)$, then
$$\begin{array}{c}
f^{\prime}(x)=\frac{\mathrm{e}^{x}}{2+\mathrm{e}^{x}}=1-\frac{2}{2+\mathrm{e}^{x}} \\
f^{\prime \prime}(x)=\frac{2 \mathrm{e}^{x}}{\left(2+\mathrm{e}^{x}\right)^{2}}>0
\end{array}$$
Therefore, $f(x)$ is a concave function, so
$$\begin{array}{l}
... | \left(2+a_{1}\right)\left(2+a_{2}\right) \cdots\left(2+a_{n}\right) \geqslant 3^{n} | Inequalities | proof | Yes | Yes | inequalities | false | 732,844 |
2. Given that $a, b, c$ are positive numbers, and $a^{4}+b^{4}+c^{4}=3$, prove: $\frac{1}{4-a b}+\frac{1}{4-b c}+\frac{1}{4-c a} \leqslant$
1. (2005 Moldova Mathematical Olympiad Problem) | 2. Let $f(x)=\frac{1}{4-\sqrt{x}}\left(0 \leqslant x \leqslant \frac{16}{9}\right)$. We prove that it is a convex function on its domain, i.e., we need to prove
that
i.e.,
i.e.,
$$\begin{array}{c}
f(x)+f(y) \leqslant 2 f\left(\frac{x+y}{2}\right) \\
\frac{1}{4-\sqrt{x}}+\frac{1}{4-\sqrt{y}} \leqslant \frac{2}{4-\sqr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,845 |
3. If positive real numbers $a, b, c$ satisfy $a+b+c=1$, prove that: $a \sqrt[3]{1+b-c}+b \sqrt[3]{1+c-a}+$ $c \sqrt[3]{1+a-b} \leqslant 1$. (2005 Japan Mathematical Olympiad Problem) | 3. $f(x)=\sqrt[3]{x}$ is a convex function on $(0,+\infty)$, by the general form of Jensen's inequality we get
$$\begin{array}{l}
\frac{a \sqrt[3]{1+b-c}+b \sqrt[3]{1+c-a}+c \sqrt[3]{1+c-a}}{a+b+c} \leqslant \\
\sqrt[3]{\frac{a(1+b-c)+b(1+c-a)+c(1+a-b)}{a+b+c}}=1
\end{array}$$
Therefore,
$$a \sqrt[3]{1+b-c}+b \sqrt[3]... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,846 |
4. If positive real numbers $a, b, c$ satisfy $a b c=1$, prove that: $a^{b+c} b^{c+a} c^{a+b} \leqslant 1$. (2006 Indian Mathematical Olympiad) | 4.
$$\begin{aligned}
a^{b+c} b^{c+a} c^{a+b} \leqslant 1 \Leftrightarrow & a^{a+b+c} b^{a+b+c} c^{a+b+c} \leqslant a^{a} b^{b} c^{c} \Leftrightarrow \\
& a^{a} b^{b} c^{c} \geqslant 1 \Leftrightarrow a \ln a+b \ln b+c \ln c \geqslant 0
\end{aligned}$$
Let $f(x)=x \ln x (x>0)$, then $f''(x)=\frac{1}{x}>0$. Therefore, $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,847 |
5. Given $0<x_{j}<\frac{1}{2}(j=1,2, \cdots, n)$, prove:
$$\frac{\prod_{j=1}^{n} x_{j}}{\left(\sum_{k=1}^{n} x_{j}\right)^{n}} \leqslant \frac{\prod_{j=1}^{n}\left(1-x_{j}\right)}{\left[\sum_{k=1}^{n}\left(1-x_{j}\right)\right]}$$
(2004 Indian Mathematical Olympiad Problem) | 5.
$$\frac{\prod_{j=1}^{n} x_{j}}{\left(\sum_{j=1}^{n} x_{j}\right)^{n}} \leqslant \frac{\prod_{j=1}^{n}\left(1-x_{j}\right)}{\left[\sum_{j=1}^{n}\left(1-x_{j}\right)\right]^{n}}$$
is equivalent to
$$\frac{\prod_{j=1}^{n} x_{j}}{\prod_{j=1}^{n}\left(1-x_{j}\right)} \leqslant \frac{\left(\sum_{j=1}^{n} x_{j}\right)^{n}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,848 |
7. Non-negative real numbers $a_{1}, a_{2}, \ldots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$, prove that
$$\frac{a_{1}}{1+a_{2}+a_{3}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots+\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}} \text { has }$$
a minimum value and calculate it. (1982 Federal German Intern... | $$\begin{array}{c}
\text { 7. Let } y=\frac{a_{1}}{1+a_{2}+a_{3}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots+ \\
\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}} \text {, so } y=\frac{a_{1}}{2-a_{1}}+\frac{a_{2}}{2-a_{2}}+\cdots+\frac{a_{n}}{2-a_{n}} . \\
\text { Let } f(x)=\frac{x}{2-x}=-1+\frac{2}{2-x}, x... | \frac{n}{2n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 732,850 |
9. (1) For all positive real numbers $a, b, c$, prove: $\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant$ F. (42nd IMO problem) | 9. (1) Suppose $a+b+c=1, f(x)=\frac{1}{\sqrt{x}}$, then
$$\begin{array}{c}
f^{\prime}(x)=-\frac{1}{2 \sqrt{x^{3}}} \\
f^{\prime \prime}(x)=\frac{3}{4 \sqrt{x^{5}}}>0
\end{array}$$
Thus, $f(x)$ is a convex function on the interval $[0,+\infty)$, and by Jensen's inequality we have
$$a ̈ f\left(x_{1}\right)+b f\left(x_{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,853 |
10. (1) Prove that for any positive real numbers $a, b, c$, we have $1<\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+$ $\frac{c}{\sqrt{c^{2}+a^{2}}} \leqslant \frac{3 \sqrt{2}}{2} \quad(2004$ China Western Mathematical Olympiad Problem) | 10. (1) $\frac{a}{\sqrt{a^{2}+b^{2}}}>\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}+2 a b-+2 b c+2 c a}} \neq \frac{a}{a+b+c}$
Similarly,
$$\frac{b}{\sqrt{b^{2}+c^{2}}}>\frac{b}{a+b+c}, \frac{c}{\sqrt{c^{2}+a^{2}}}>\frac{c}{a+b+c}$$
Therefore,
$$\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,855 |
E3. Let $x, y, z$ be positive real numbers, and satisfy $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$. Prove the inequality:
$$\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{-z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \geqslant \mathrm{r}$$
(2007 Asia Pacific Mathematical Olympiad Problem) | 13. The original inequality is transformed into
$$\begin{array}{l}
\frac{x}{\sqrt{2(y+z)}}+\frac{y}{\sqrt{2(z+x)}}+\frac{z}{\sqrt{2(x+y)}}+ \\
\frac{y z}{x \sqrt{2(y+z)}}+\frac{z x}{y \sqrt{2(z+x)}}+\frac{x y}{z \sqrt{2(x+y)}} \geqslant 1
\end{array}$$
We will prove the following separately:
$$\frac{x}{\sqrt{2(y+z)}}+... | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 732,859 |
15. Given that $a, b, c$ are positive numbers, and $a^{2}+b^{2}+c^{2}=1$, prove: $\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c} \geqslant \frac{3 \sqrt{3}+3}{2} .(2004$ Polish Mathematical Olympiad problem) | 15. Proof one: Let $S=a+b-c, f(x)=\frac{1}{1-x}, x \in(0, \mathrm{~F})$. The original inequality is equivalent to
$$\frac{a}{S} f(a)+\frac{b}{S} f(b)+\frac{c}{S} f(c) \geqslant \frac{3 \sqrt{3}+3}{2 S}$$
By Jensen's inequality, we have
$$\frac{a}{S} f(a)+\frac{b}{S} f(b)+\frac{c}{S} f(c) \geqslant f\left(\frac{a^{2}+b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,861 |
16. Given that $a, b, c$ are all positive numbers, and $a b+b c+c a=1$, prove: $\sqrt{a^{3}+a}+\sqrt{-b^{3}+b}+$ $\sqrt{c^{3}+c} \geqslant 2 \sqrt{a+b+c}$. (2008 Iran National Training Team Problem) | 16. The inequality is equivalent to
$$\begin{array}{l}
\sqrt{a^{3}+a(a b+b c+c a)}+\sqrt{b^{3}+b(a b+b c+c a)}+\sqrt{c^{3}+c(a b+b c+c a)} \geqslant \\
2 \sqrt{(a+b+c)(a b+b c+c a)} \Leftrightarrow \\
\sqrt{a(a+b)(c+a)}+\sqrt{b(a+b)(b+c)}+\sqrt{c(c+a)(b+c)} \geqslant \\
2 \sqrt{(a+b+c)(a b+b c+c a)} \Leftrightarrow \\
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,863 |
17. Let $x, y$ be positive numbers, prove: $\frac{1}{(1+\sqrt{x})^{2}}+\frac{1}{(1+\sqrt{y})^{2}} \geqslant \frac{2}{x+y+2}$. (2008 Indonesia Mathematical Olympiad) | 17. Let $f(x)=\frac{1}{(1+x)^{2}}$, by Jensen's inequality we have $f(a)+f(b) \geqslant 2 f\left(\frac{a+b}{2}\right)$, that is
$$\frac{1}{\left(1+a\right)^{2}}+\frac{1}{(1+b)^{2}} \geqslant \frac{1}{\left(1+\frac{a+b}{2}\right)^{2}}$$
By Cauchy-Schwarz inequality, we get
$$\begin{array}{c}
\left(1+\frac{a+b}{2}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,864 |
19. Given positive real numbers $a, b, c$ satisfying $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$, prove: $\frac{1}{(2 a+b+c)^{2}}+$ $\frac{1}{(a+2 b+c)^{2}}+\frac{1}{(2 a+b+2 c)^{2}} \leqslant \frac{3}{16} \cdot(2009$ IMO Shortlist $)$ | 19. Since $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$, we have
$$\frac{1}{(2 a+b+c)^{2}}+\frac{1}{(a+2 b+c)^{2}}+\frac{1}{(2 a+b+2 c)^{2}} \leqslant \frac{3}{16}$$
which is equivalent to
$$\begin{array}{l}
\frac{(a+b+c)^{2}}{(2 a+b+c)^{2}}+\frac{(a+b+c)^{2}}{(a+2 b+c)^{2}}+\frac{(a+b+c)^{2}}{(2 a+b+2 c)^{2}} \leqslant... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,866 |
Example 3 Given that $\alpha, \beta$ are two distinct real roots of the equation $4 x^{2}-4 t x-1=\theta(t \in \mathbf{R})$, and the domain of the function $f(x)=\frac{2 x-t}{x^{2}+1}$ is $[\alpha, \beta]$.
(1) Find $g(t)=\max f(x)-\min f(x)$. | (1) Let $\alpha \leqslant x_{1}t\left(x_{1}+x_{2}\right)-2 x_{1} x_{2}+\frac{1}{2}>0$
So $f\left(x_{2}\right)-f\left(x_{\mathrm{r}}\right)>0$, hence $f(x)$ is an increasing function on the interval $[\alpha, \beta]$.
Since
$$\alpha+\beta=t, \alpha \beta=-\frac{t}{4}, \beta-\alpha=\sqrt{(\alpha+\beta)^{2}-4 \alpha \bet... | \frac{8 \sqrt{t^{2}+1}\left(2 t^{2}+5\right)}{16 t^{2}+25} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,869 |
(2) Prove that for $u_{r} \operatorname{te}\left(0, \frac{\pi}{2}\right)(i-1,2,3)$, if $\sin u_{1}+\sin u_{2}+\sin u_{3}=1$, then $\frac{1}{g\left(\tan u_{1}\right)}+\frac{1}{g\left(\tan u_{2}\right)}+\frac{1}{g\left(\tan u_{3}\right)}<\frac{3}{4} \sqrt{6}(2004$ National High School Mathematics Competition Problem) | $$\begin{array}{c}
g\left(\tan u_{i}\right)=\frac{\frac{8}{\cos u_{i}}\left(\frac{2}{\cos ^{2} u_{i}}+3\right)}{\frac{16}{\cos ^{2} u_{i}}+9}=\frac{\frac{16}{\cos u_{i}}+24 \cos u_{i}}{16+9 \cos ^{2} u_{i}} \geqslant \\
\frac{16 \sqrt{6}}{16+9 \cos ^{2} u_{i}}, i=1,2,3 \\
\frac{1}{g\left(\tan u_{1}\right)}+\frac{1}{g\l... | \frac{3}{4} \sqrt{6} | Inequalities | proof | Yes | Yes | inequalities | false | 732,870 |
Example 4 In the Cartesian coordinate system $x O y$, the sequence of points $\left\{A_{n}\right\}$ on the positive $y$-axis and the sequence of points $\left\{B_{n}\right\}$ on the curve $y=\sqrt{2 x}$ $(x \geqslant 0)$ satisfy $\left|O A_{n}\right|=\left|O B_{n}\right|=\frac{1}{n}$, the intercept of the line $A_{n} B... | Proof (1) It is easy to know that $A_{n}\left(0, \frac{1}{n}\right), B_{n}\left(b_{n}, \sqrt{2 b_{n}}\right)\left(b_{n}>0\right)$. From $| O B_{n} \left\lvert\,=\frac{1}{n}\right.$ we get $b_{n}^{2}+$ $2 b_{n}=\left(\frac{1}{n}\right)^{2}$, hence
$$b_{n}=\sqrt{\left(\frac{1}{n}\right)^{2}+1}-1, n \in \mathbf{N}^{*}$$
... | proof | Geometry | proof | Yes | Yes | inequalities | false | 732,871 |
(2) There exists $n_{0} \in \mathbf{N}^{*}$, such that for all $n>n_{0}$ we have $\frac{b_{2}}{b_{1}}+\frac{b_{3}}{b_{2}}+\cdots+\frac{b_{n+1}}{b_{n}}<n-2004$. (2004 National High School Mathematics League Additional Problem) | (2) Let $c_{n}=1-\frac{b_{n+1}}{b_{n}}, n \in \mathbf{N}^{*}$, then
$$\begin{aligned}
c_{n}= & \frac{\sqrt{\left(\frac{1}{n}\right)^{2}+1}-\sqrt{\left(\frac{1}{n+1}\right)^{2}+1}}{\sqrt{\left(\frac{1}{n}\right)^{2}+1}-1}= \\
& n^{2}\left(\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right) \frac{\sqrt{\left(\frac{1}{n}\right)^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,872 |
60. Positive real numbers $a, b$ satisfy $a^{5}+b^{5}=a^{3}+b^{3}$, prove: $a^{2}+b^{2} \leqslant 1+a b$. (2003 Czech and Slovak Mathematical Olympiad problem) | \begin{aligned} 60.1+a b-a^{2}-b^{2}= & \frac{a^{5}+b^{5}}{a^{3}+b^{3}}+a b-a^{2}-b^{2}= \\ & \frac{a^{4} b+a b^{4}-a^{3} b^{2}-a^{2} b^{3}}{a^{3}+b^{3}}=\frac{a b(a+b)(a-b)^{2}}{a^{3}+b^{3}} \geqslant 0\end{aligned} | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,873 |
Example 5 Let $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n} \in[1,2]$, and $\sum_{i=1}^{n} a_{i}^{2}=\sum_{i=1}^{n} b_{i}^{2}$, prove that: $\sum_{i=1}^{n} \frac{a_{i}^{3}}{b_{i}} \leqslant \frac{17}{10} \sum_{i=1}^{n} a_{i}^{2}$. And ask: the necessary and sufficient condition for the equality to hold. (1... | Prove that since $a_{i}, b_{i} \in [1,2], i=1,2, \cdots, n$, therefore
$$\frac{1}{2} \leqslant \frac{\sqrt{\frac{a_{i}^{3}}{b_{i}}}}{\sqrt{a_{i} b_{i}}}=\frac{a_{i}}{b_{i}} \leqslant 2$$
Thus,
$$\left(\frac{1}{2} \sqrt{a_{i} b_{i}}-\sqrt{\frac{a_{i}^{3}}{b_{i}}}\right)\left(2 \sqrt{a_{i} b_{i}}-\sqrt{\frac{a_{i}^{3}}{... | \sum_{i=1}^{n} \frac{a_{i}^{3}}{b_{i}} \leqslant \frac{17}{10} \sum_{i=1}^{n} a_{i}^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,874 |
Example 7 Let $0 \leqslant a_{1}, a_{2}, \cdots, a_{n} \leqslant 1$, then
$$\begin{array}{l}
\frac{a_{1}}{a_{2}+a_{3}+\cdots+a_{n}+1}+\frac{a_{2}}{a_{1}+a_{3}+\cdots+a_{n}+1}+\cdots+ \\
\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n-1}+1}+\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right) \leqslant 1
\end{array}... | Suppose $0 \leqslant a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} \leqslant 1$, then
$$\begin{array}{l}
\frac{a_{1}}{a_{2}+a_{3}+\cdots+a_{n}+1}+\frac{a_{2}}{a_{1}+a_{3}+\cdots+a_{n}+1}+\cdots+\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n-1}+1}+ \\
\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right) \l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,876 |
Example 8 Let $n \in \mathbf{N}, x_{0}=0, x_{i}>0, i=1,2,3 \cdots, n$. And $\sum_{i=1}^{n} x_{i}=1$, prove that:
$$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}$$
(1996 China Mathematical Olympiad Problem) | Prove that from $\sum_{i=1}^{n} x_{i}=1$ and the arithmetic mean inequality, we have
$$\begin{array}{l}
\sqrt{1+x_{0}+x_{1}+x_{2}+\cdots+x_{i-1}} \cdot \sqrt{x_{i}+\cdots+x_{n}} \leqslant \\
\frac{1+x_{0}+x_{1}+x_{2}+\cdots+x_{i-1}+x_{i}+\cdots+x_{n}}{2}=1
\end{array}$$
Thus, the left inequality holds.
Since $0 \leqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,877 |
Example 10 Let $a, b, c$ be positive numbers. Their sum equals 1. Prove: $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geqslant$ $\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$ (29th Russian Mathematical Olympiad Problem) | Without loss of generality, let $a \geqslant b \geqslant c$, then, $1-c^{2} \geqslant 1-b^{2} \geqslant 1-a^{2}$, hence $\frac{1}{1-a^{2}} \geqslant \frac{1}{1-b^{2}} \geqslant \frac{1}{1-c^{2}}$. Note that $\frac{1}{1-a}-\frac{2}{1+a}=\frac{3 a-1}{1-a^{2}}$, so it suffices to prove
$$\frac{3 a-1}{1-a^{2}}+\frac{3 b-1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,879 |
Example 11 Let $n$ be a given natural number, $n \geq 3$. For $n$ given real numbers $a_{1}, a_{2}, \cdots, a_{n}$, the minimum value of $a_{i}-a_{j}$ $(1 \leqslant i, j \leqslant n)$ is $m$. Find the maximum value of $m$ under the condition $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$ (1992 Shanghai Mathematics Competitio... | $$\begin{array}{l}
a_{1} \geqslant 2 m, \cdots, a_{n}-a_{n-1} \geqslant m, a_{j}-a_{i} \geqslant(j-i) m\left(\mathrm{I}_{-} \leqslant i<j \leqslant n\right) . \\
\sum_{i \leqslant i<j \leqslant u}\left(a_{i}-a_{j}\right)^{2} \geqslant m^{2} \sum_{i \leqslant i<j \leq n}(i-j)^{2}=m^{2} \frac{h(k+1)(2 k+1)}{6}= \\
\frac{... | \sqrt{\frac{12}{n\left(n^{2}-1\right)}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,880 |
Example 13 Let $x, y, z$ be positive real numbers, and $x \geqslant y \geqslant z$, prove that: $\frac{x^{2} y}{z}+\frac{y^{2} z}{x}+\frac{z^{2} x}{y} \geqslant x^{2}+y^{2}+$ $z^{2}$. (31st IMO Preliminary Problem) | $$\begin{array}{l}
\frac{x^{2} y}{z}+\frac{y^{2} z}{x}+\frac{z^{2} x}{y}-\left(x^{2}+y^{2}+z^{2}\right)=\frac{x^{2}}{z}(y-z)+\frac{y^{2} z}{x}+\frac{z^{2} x}{y}-\left(y^{2}+z^{2}\right) \geqslant \\
\frac{y^{2}}{z}(y-z)+2 \sqrt{y z} \cdot z-\left(y^{2}+z^{2}\right)= \\
\frac{y^{2}}{z}(y-z)-(y-z)^{2}-2 z \sqrt{y}(\sqrt{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,881 |
Example 14 Let $n (n>3)$ be a natural number, and $x_{1}, x_{2}, \cdots, x_{n}$ be $n$ positive numbers, such that $x_{1} x_{2} \cdots x_{n}=1$. Prove: $\frac{1}{1+x_{1}+x_{1} x_{2}}+\frac{1}{1+x_{2}+x_{2} x_{3}}+\cdots+\frac{1}{1+x_{n}+x_{n} x_{1}}>1$. (2004 Russian Mathematical Olympiad Problem) | Prove that if $x_{n+k}=x_{k}(k=1,2, \cdots, n)$, then
$$\begin{array}{l}
\sum_{k=1}^{n} \frac{1}{1+x_{k}+x_{k} x_{k+1}}>\sum_{i=1}^{n} \frac{1}{1+\sum_{i=0}^{n-2} \prod_{j=k}^{k+i} x_{j}}= \\
\sum_{i=1}^{n} \frac{x_{n} \prod_{j=1}^{k-1} x_{j}}{x_{n} \prod_{j=1}^{k-1} x_{j}\left(1+\sum_{i=0}^{n-2} \prod_{j=k}^{k+i} x_{j... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,882 |
61. Given that $a, b, c$ are real numbers, and $a+b+c=0$, prove: $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+3 \geqslant 6 a b c$. | 61. It is sufficient to prove
$$\left(a^{2} b^{2}-2 a b c+1\right)+\left(b^{2} c^{2}-2 a b c+1\right)+\left(c^{2} a^{2}-2 a b c+1\right) \geqslant 0$$
It is easy to get
$$\begin{array}{l}
a^{2} b^{2}-2 a b c+1=(a b-c+1)^{2}-\left(c^{2}+2 a b-2 c\right) \\
b^{2} c^{2}-2 a b c+1=(b c-a+1)^{2}-\left(a^{2}+2 b c-2 a\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,883 |
For example, the sequence $\{a_{n}\}$ is defined as follows: $a_{1}=\frac{1}{2}$, and $a_{n+1}=\frac{a_{n}^{2}}{a^{2}-a_{n}+1}, n=1,2, \cdots$. Prove that for each positive integer $n$, $\sum_{i=1}^{n} a_{i}<1$. (2004 China National Training Team Problem, 2003 Romania National Training Team Problem) | Prove that taking the reciprocal of both sides of the original recurrence relation yields
$$\begin{array}{c}
\frac{1}{a_{n+1}}=\frac{1}{a_{n}^{2}}-\frac{1}{a_{n}}+1 \\
\frac{1}{a_{n+1}}-1=\frac{1}{a_{n}}\left(\frac{1}{a_{n}}-1\right)
\end{array}$$
Let \( x_{n}=\frac{1}{a_{n}}-1 \), then \(\frac{1}{x_{n+1}}=\frac{1}{x_... | proof | Algebra | proof | Yes | Yes | inequalities | false | 732,884 |
Example 16 Let $a, b, c$ be the lengths of the three sides of a triangle, and $a+b+c=1$. If the positive integer $n \geqslant 2$, prove that: $\sqrt[n]{a^{n}+b^{n}}+\sqrt[n]{b^{n}+c^{n}}+\sqrt[n]{c^{n}+a^{n}}<1+\frac{\sqrt[n]{2}}{2}$. (2003 Asia Pacific Mathematical Olympiad Problem) | Proof: Let $a \geqslant b \geqslant c > 0$, and since $a + b + c = 1$, then
$$\begin{array}{l}
\left(b + \frac{c}{2}\right)^{n} = b^{n} + \mathrm{C}_{n}^{1} b^{n-1} \frac{c}{2} + \mathrm{C}_{n}^{2} b^{n-2} \left(\frac{c}{2}\right)^{2} + \cdots + \mathrm{C}_{n}^{n} \left(\frac{c}{2}\right)^{n} \geqslant \\
b^{n} + \left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,885 |
Example 18 Prove: If two given positive numbers $p<q$, then for any $a, b, c, d, e \in [p, q]$, we have
$$(a+b+c+d+e)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}\right) \leqslant 25+6\left(\sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}\right)^{2}$$
and determine the necessary and sufficient conditions for ... | Prove from Lagrange's identity:
$$\sum_{i=1}^{n} a_{i}^{2} \sum_{i=1}^{n} b_{i}^{2}=\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2}+\sum_{1 \leqslant i<j \leqslant n}\left(-a_{i} b_{j}-a_{j} b_{i}\right)^{2}$$
$$(a+b+c+d+e)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}\right)=25+\sum\left(\sqrt{\frac{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,887 |
Example 19 Given that $a, b, c, d$ are positive real numbers, prove:
$$\begin{array}{l}
\sqrt{(a+c)^{2}+(b+d)^{2}} \leqslant \sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}} \leqslant \\
\sqrt{(a+c)^{2}+(b+d)^{2}}+\frac{2|a d-b c|}{\sqrt{(a+c)^{2}+(b+d)^{2}}}
\end{array}$$
(52nd Belarusian Mathematical Olympiad) | Proof: Let $P(a, b), Q(-c,-d)$ be in the first and third quadrants of the Cartesian coordinate system $x O y$, then $\square$
$$|O P|=\sqrt{a^{2}+b^{2}},|O Q|=\sqrt{c^{2}+d^{2}},|P Q|=\sqrt{(a+c)^{2}+(b+d)^{2}}$$
In $\triangle O P Q$, by $|O P|+|O Q| \geqslant|P Q|$ we get
$$\begin{array}{l}
\sqrt{(a+c)^{2}+(b+d)^{2}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,888 |
Example 20 Let $a \geqslant b \geqslant c \geqslant 0$, and $a+b+c=3$, prove: $a b^{2}+b c^{2}+c a^{2} \leqslant \frac{27}{8}$, and determine the condition for equality. (2002 Hong Kong Mathematical Olympiad Problem) | Proof: Let $f(a, b, c) = a b^2 + b c^2 + c a^2$, then $f(a, c, b) = a c^2 + c b^2 + b a^2$, because
$$\begin{array}{l}
f(a, b, c) - f(a, c, b) = a b^2 + b c^2 + c a^2 - (a c^2 + c b^2 + b a^2) = \\
a b(a - b) + c^2(a - b) + c(b^2 - a^2) = \\
(a - b)(a b + c^2 - a c - b c) = \\
(a - b)(a - c)(b - c) \geqslant 0
\end{ar... | a = b = \frac{3}{2}, c = 0 | Inequalities | proof | Yes | Yes | inequalities | false | 732,889 |
1. (1) Determine the range of values for the following sum: $S=\frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+$ $\frac{d}{a+c+d}$, where $a, b, c, d$ are any positive real numbers. (16th IMO Problem) | $$\begin{array}{l}
\text { 1. (1) Since } \frac{a}{a+b+d}+\frac{b}{a+b+c}\frac{a}{a+b+c+d}+\frac{b}{a+b}-\frac{b}{b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}=1 \\
\text { Let } a \rightarrow+\infty, b=d=\sqrt{a}, c=1 \text {, the sum } \rightarrow 1, a=c \rightarrow+\infty, b=d=1 \text {, the sum } \rightarrow 2,
\end{a... | (1,2) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,893 |
62. Let $n$ be a positive integer, and real numbers $a_{1}, a_{2}, \cdots, a_{n}$ and $r_{1}, r_{2}, \cdots, r_{n}$ satisfy: $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$ and $0 \leqslant r_{1} \leqslant r_{2} \leqslant \cdots \leqslant r_{n}$. Prove that: $\sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} a_{j} \min \lef... | 62. Construct the $n \times n$ number table
$$A_{1}=\left(\begin{array}{ccccc}
a_{1} a_{1} r_{1} & a_{1} a_{2} r_{1} & a_{1} a_{3} r_{1} & \cdots & a_{1} a_{n} r_{1} \\
a_{2} a_{1} r_{1} & a_{2} a_{2} r_{2} & a_{2} a_{3} r_{2} & \cdots & a_{2} a_{n} r_{2} \\
a_{3} a_{1} r_{1} & a_{3} a_{2} r_{2} & a_{3} a_{3} r_{3} & \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,894 |
(2) Let $a_{1}>1, a_{2}>1, a_{3}>1, a_{1}+a_{2}+a_{3}=S$, and it is known that for $i=1,2,3$, we have $\frac{a_{i}^{2}}{a_{i}-1}>S$. Prove: $\frac{1}{a_{1}+a_{2}}+\frac{1}{a_{2}+a_{3}}+\frac{1}{a_{3}+a_{1}}>1$. (31st Russian Mathematical Olympiad Problem) | (2) It is easy to see that
$$\begin{array}{l}
\frac{a_{1}^{2}}{a_{1}-1}>S \Leftrightarrow a_{1}^{2}>\left(a_{1}+a_{2}+a_{3}\right)\left(a_{1}-F\right) \Leftrightarrow \\
a_{1}+a_{2}+a_{3}>a_{1}\left(a_{2}+a_{3}\right) \Leftrightarrow \\
\frac{1}{a_{2}+a_{3}}>\frac{a_{1}}{a_{1}+a_{2}+a_{3}}
\end{array}$$
Similarly, $\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,895 |
2. If $a, b, c$ are positive numbers, prove that: $\frac{a^{3}}{a^{2}+a b+b^{2}}+\frac{b^{3}}{b^{2}+b c+c^{2}}+\frac{c^{3}}{c^{2}+c a+a^{2}} \geqslant$ $\frac{a+b+c}{3}$. (2003 Beijing Mathematical Competition Problem) | 2. Since $a^{2}+a b+b^{2} \geqslant 3 a b$, we have
$$\begin{aligned}
\frac{a^{3}}{a^{2}+a b+b^{2}}= & \frac{a^{3}+a^{2} b+a b^{2}-\left(a^{2} b+a b^{2}\right)}{a^{2}+a b+b^{2}}=a-\frac{a b(a+b)}{a^{2}+a b+b^{2}} \geqslant \\
& a-\frac{a+b}{3}
\end{aligned}$$
Similarly,
$$\begin{array}{l}
\frac{b^{3}}{b^{2}+b c+c^{2}}... | \frac{a+b+c}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 732,896 |
4. Given positive real numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{n+1}$ satisfy $a_{2}-a_{1}=a_{3}-a_{2}=\cdots=a_{n+1}-a_{n}>$ $0(n \geqslant 2)$, prove that: $\frac{1}{a_{2}^{2}}+\frac{1}{a_{3}^{2}}+\cdots+\frac{1}{a_{n}^{2}} \leqslant \frac{n-1}{2} \cdot \frac{a_{1} a_{n}+a_{2} a_{n+1}}{a_{1} a_{2} a_{n} a_{n+1}}$. (2... | 4. Let $d=a_{i}-a_{i-1}>0$, then we have $a_{k}^{2} \geqslant a_{k}^{2}-d^{2}=\left(a_{k}+d\right)\left(a_{k}-d\right)=$ $a_{k+1} a_{k-1}(k \geqslant 2)$, so
$$\begin{aligned}
\frac{1}{a_{k}^{2}} \leqslant & \frac{1}{a_{k+1} a_{k-1}}=\frac{1}{a_{k+1}-a_{k-1}}\left(\frac{1}{a_{k-1}}-\frac{1}{a_{k+1}}\right)=\frac{1}{2 d... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,898 |
5. Given that $a, b, c, d$ are any positive numbers, prove: $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \geqslant 2$.
(1989 Sichuan Province Mathematics Competition) | 5.
$$\begin{array}{l}
\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}= \\
\frac{a(d+a)+c(b+c)}{(b+c)(d+a)}+\frac{b(c+d)+b(a+b)}{(c+d)(a+b)}
\end{array}$$
By the AM-GM inequality, we get
That is,
Similarly,
Therefore, the right-hand side of (1) $\geqslant \frac{4\left(a^{2}+b^{2}+c^{2}+d^{2}+a d+b c+a b+c d\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,899 |
6. Let $a=\frac{m^{m+1}+n^{n+1}}{m^{m}+n^{n}}$, where $m, n$ are positive integers, prove: $a^{m}+a^{n} \geqslant m^{m}+n^{n}$. (USAMO problem) | 6. Suppose $m \geqslant n$, then
$$\begin{array}{l}
a=\frac{m^{m+1}+n^{n+1}}{m^{m}+n^{n}} \leqslant \frac{m^{m+1}+m n^{n}}{m^{m}+n^{n}}=m \\
a=\frac{m^{m+1}+n^{n+1}}{m^{m}+n^{n}} \geqslant \frac{n m^{m}+n^{n+1}}{m^{m}+n^{n}}=n
\end{array}$$
Thus, $n \leqslant a \leqslant m$, and we have
$$\begin{array}{c}
m^{m}-a^{m}=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,900 |
7. Given that $a, b, c, d$ are real numbers, and $a^{2}+b^{2}+c^{2}+d^{2} \leqslant 1$, prove:
$$(a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4}+(b+d)^{4}+(c+d)^{4} \leqslant 6$$
(28th IMO Preliminary Problem) | $$\text { 7. } \begin{array}{l}
(a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4}+(b+d)^{4}+(c+d)^{4} \leqslant \\
(a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4}+(b+d)^{4}+(c+d)^{4}+ \\
(a-b)^{4}+(a-c)^{4}+(a-d)^{4}+(b-c)^{4}+(b-d)^{4}+(c-d)^{4}= \\
6\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2} \leqslant 6
\end{array}$$
The maximum value ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,901 |
9. Let the sequence $\left\{a_{n}\right\}$ have the sum of its first $n$ terms denoted as $S_{n}$. It is known that $a_{1}=1, a_{2}=6, a_{3}=11$, and $(5 n-8) S_{n+1}-(5 n+2) S_{n}=A n+B, n=1,2,3, \cdots$, where $A, B$ are constants.
(1) Find the values of $A$ and $B$;
(2) Prove that the sequence $\left\{a_{n}\right\}$... | $$\begin{array}{l}
\sqrt{5(5 m n-4)}-\sqrt{(5 m-4)(5 n-4)}= \\
\sqrt{5(5 m n-4)}-\sqrt{(25 m n-20(m+n)+16)}= \\
\sqrt{5(5 m n-4)}-\sqrt{25 m n-40 \sqrt{m n}+16}= \\
\sqrt{5(5 m n-4)}-\sqrt{(5 \sqrt{m n}-4)^{2}}= \\
\sqrt{5(5 m n-4)}-5 \sqrt{m n}+4= \\
4-(5 \sqrt{m n}-\sqrt{5(5 m n-4)})= \\
4-\frac{20}{5 \sqrt{m n}+\sqr... | 4-\frac{20}{5+\sqrt{5(5-4)}}=\sqrt{5}-1 | Algebra | proof | Yes | Yes | inequalities | false | 732,903 |
10. The sequences $a_{0}, a_{1}, \cdots$ and $b_{0}, b_{1}, \cdots$ are defined as follows: $a_{0}=\frac{\sqrt{2}}{2}, a_{n+1}=$ $\frac{\sqrt{2}}{2} \sqrt{1-\sqrt{1-a_{n}^{2}}}, n=0,1,2, \cdots ; b_{0}=1, b_{n+1}=\frac{\sqrt{1+b_{n}^{2}}-1}{b_{n}}, n=0,1,2, \cdots$ Prove that for every $n=0,1,2, \cdots$, the inequality... | 10. $a_{0}=\frac{\sqrt{2}}{2}=\sin \frac{\pi}{2^{2}}$, let $a_{n}=\sin \frac{\pi}{2^{n+2}}$, then
$$a_{n+1}=\frac{\sqrt{2}}{2} \sqrt{1-\cos \frac{\pi}{2^{n+2}}}=\sin \frac{\pi}{2^{n+3}}$$
Similarly, we can use mathematical induction to prove $b_{n}=\tan \frac{\pi}{2^{n+2}}$.
Since for $x \in\left(0, \frac{\pi}{2}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,904 |
Example 8 Let $x, y, z$ be positive real numbers, prove that: $(x y+y z+z x)\left[\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\right.$ $\left.\frac{1}{(z+x)^{2}}\right] \geqslant \frac{9}{4}$ (1996 Iran Mathematical Olympiad) | Proof: Without loss of generality, let $x \geqslant y \geqslant z > 0$, then
$$\begin{array}{l}
(x y + y z + z x) \frac{1}{(x+y)^{2}} + \frac{1}{(y+z)^{2}} + \frac{1}{(z+x)^{2}} - \frac{9}{4} = \\
\frac{x y + z(-x + y)}{(x+y)^{-2}} + \frac{y z + x(y + z)}{(y+z)^{2}} + \frac{z x + y(z + x)}{(z+x)^{2}} - \frac{9}{4} = \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,905 |
Example 1 (1) Let $x, y, z$ be three positive real numbers, and $x y z=1$, prove: $x^{2}+y^{2}+z^{2}+x y+y z+$ $z x \geqslant 2(\sqrt{x}+\sqrt{y}+\sqrt{z}) ;$ | Proof (1) By the binary mean inequality, we have
$$\begin{array}{l}
x^{2}+y z \geqslant 2 \sqrt{x^{2} y z}=2 \sqrt{x x y z}=2 \sqrt{x} \\
y^{2}+z x \geqslant 2 \sqrt{y^{2} z x}=2 \sqrt{y x y z}=2 \sqrt{y} \\
z^{2}+x y \geqslant 2 \sqrt{z^{2} x y}=2 \sqrt{z x y z}=2 \sqrt{z}
\end{array}$$
Adding the above three inequal... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,906 |
11. Real numbers $t_{1}, t_{2}, \cdots, t_{n}$ are $n$ real numbers, satisfying $0<t_{1} \leqslant t_{2} \leqslant \cdots \leqslant t_{n}<1$, prove: $\left(1-t_{n}\right)^{2}\left[\frac{t_{1}}{\left(1-t_{1}^{2}\right)^{2}}+\frac{t_{2}^{2}}{\left(1-t_{2}^{3}\right)^{2}}+\cdots+\frac{t_{n}^{n}}{\left(1-t_{n}^{n+1}\right)... | 11.
$$\begin{aligned}
\text { Left } \leqslant & \sum \frac{t_{k}^{k}}{\left(1+t_{k}+t_{k}^{2}+\cdots+t_{k}^{k}\right)^{2}} \leqslant \\
& \left.\sum \frac{1}{\left(1+t+t_{k}^{2}+\cdots+t_{k}^{k-5}\right)^{2}}-\frac{1}{\left(1+t_{k}+t_{k}^{2}+\cdots+t_{k}^{k}\right)^{2}}\right) \leqslant \\
& \sum\left(\frac{1}{\left(1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,907 |
13. Let $x_{n}=\sqrt{2+\sqrt[3]{3+\cdots+\sqrt[n]{n}}}$, prove: $x_{n+1}-x_{n}<\frac{1}{n!}(n=2,3, \cdots)$. 26th IMO Shortlist Problem. | 13. When $n=2$, it is easy to prove $\sqrt{2}+\sqrt[3]{3}2$, construct the sequences $\left\{a_{i}\right\},\left\{b_{i}\right\},\left\{c_{i}\right\}$ as follows:
$$\begin{array}{l}
b_{i}=\sqrt[i]{i+\sqrt[i+\mathrm{t}]{(i+1)+\cdots+\sqrt[n]{n}}}, i=2,3, \cdots, n, b_{n=\mathrm{F}}=0 \\
c_{i}=a_{i}^{i-1}+a_{i}^{i-2} b_{i... | proof | Algebra | proof | Yes | Yes | inequalities | false | 732,909 |
14. Prove: For any $n \in \mathbf{N}^{*}$, we have $(2 n+1)^{n} \geqslant(2 n)^{n}+(2 n-1)^{n}$. (1987 Soviet Mathematical Competition Problem) | 14. By the binomial theorem, we have
$$(1 \pm x)^{n}=1 \pm n x+a_{2} x^{2} \pm a_{3} x^{3}+a_{4} x^{4} \pm a_{5} x^{5}+\cdots \pm(-\mathrm{k})^{n+1} a_{n} x^{n}$$
where $a_{i}=\mathrm{C}_{n}^{i}(i=1,2, \cdots, n)$, thus
$$(1+x)^{n}-(1-x)^{n}-2 n x=2 a_{3} x^{3}+2 a_{5} x^{5} + \cdots$$
When $x=\frac{1}{2 n}$, the rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,910 |
16. Given non-negative real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}+x+2 y+3 z=\frac{13}{4}$.
(1) Find the maximum value of $x+y+z$; | 16. (1) From the given equation, we have $\left(x+\frac{1}{2}\right)^{2}+(y+1)^{2}+\left(z+\frac{3}{2}\right)^{2}=\frac{27}{4}$. By the Cauchy-Schwarz inequality, we get
$$\begin{array}{l}
{\left[\left(x+\frac{1}{2}\right)+(y+F)+\left(z+\frac{3}{2}\right)\right]^{2} \leqslant} \\
3\left[\left(x+\frac{1}{2}\right)^{2}+(... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,912 |
18. (1) When $x_{1}=\frac{1}{3}, x_{n+1}=x_{n}^{2}+x_{n}$, in which interval of two consecutive integers does the expression $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{2001}}+$ $\frac{1}{1+x_{2002}}$ lie? (2002 2003 Finnish Mathematical Olympiad Problem) | 18. (1) Since $x_{n}^{2}+x_{n}=x_{n}\left(\overline{x_{n}}+1\right)$, and
$$\frac{1}{x_{n}\left(x_{n}+1\right)}=\frac{1}{x_{n}}-\frac{1}{x_{n}+1}, \frac{1}{x_{n}+1}=\frac{1}{x_{n}}-\frac{F}{x_{n+1}}$$
Therefore,
$$\sum_{n=1}^{2002} \frac{1}{x_{n}+1}=\frac{1}{x_{1}}-\frac{1}{x_{2003}}=3-\frac{1}{x_{2003}}$$
By specifi... | 2 < 3 - \frac{1}{x_{2003}} < 3 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,915 |
(2) The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=2, a_{n+1}=a_{n}^{2}-a_{n}+1, n=1,2, \cdots$, prove: $1-$ $\frac{1}{2003^{2003}}<\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2003}}<1$. (2003 Girls' Mathematical Olympiad Problem) | (2) From the given, we have $a_{n+1}-1=a_{n}\left(a_{n}-1\right)$, so
$$\begin{array}{l}
\frac{1}{a_{n+1}-1}=\frac{1}{a_{n}-1}-\frac{1}{a_{n}} \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2003}}=\left(\frac{1}{a_{1}-1}-\frac{1}{a_{2}-1}\right)+ \\
\left(\frac{1}{a_{2}-1}-\frac{1}{a_{3}-1}\right)+\cdots+\left(\... | 1-\frac{1}{2003^{2003}}<\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2003}}<1 | Algebra | proof | Yes | Yes | inequalities | false | 732,916 |
(2) Let $a, b, c$ be three positive real numbers, and $abc=1$, prove that: $\frac{1}{a+b+1}+\frac{1}{b+c+1}+$ $\frac{1}{c+a+1} \leqslant 1 .$(2000 Macau Mathematical Olympiad Problem) | (2) Let $x, y, z$ be three positive real numbers such that $x^{3}=a, y^{3}=b, z^{3}=c$, then $x y z=\sqrt[3]{a b c}=1, a+b+1=x^{3}+y^{3}+x y z=(x+y)\left(x^{2}+y^{2}-x y\right)+x y z \geqslant(x+y)(2 x y- x y)+x y z=x y(x+y+z)=\frac{x+y+z}{z}$. Therefore,
$$\frac{1}{a+b+1} \leqslant \frac{z}{x+y+z}$$
Similarly, we hav... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,917 |
19. (1) Prove: $\frac{1}{1999}<\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \cdots \frac{1997}{1998}<\frac{1}{44}$. (1997 Canadian Mathematical Olympiad) | 19. (1) Let $p=\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \ldots \cdot \frac{1997}{1998}$.
Since $2\frac{1}{3}, \frac{3}{4}>\frac{3}{5}, \frac{5}{6}>\frac{5}{7}, \cdots, \frac{1997}{1998}>\frac{1997}{1999}$, we have
$$p>\frac{1}{3} \cdot \frac{3}{5} \cdot \frac{5}{7} \cdot \cdots \cdot \frac{1997}{1999}=\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,918 |
(2) Prove: $\frac{1}{3^{3}}+\frac{1}{4^{3}}+\frac{1}{5^{3}}+\cdots+\frac{1}{n^{3}}<\frac{1}{12}$. (1990 Irish Mathematical Olympiad) | (2)
$$\begin{aligned}
\frac{1}{k^{3}}= & \frac{\mathrm{F}}{k \cdot k^{2}}=\frac{1}{k \cdot\left(k^{2}-\mathrm{E}\right)}=\frac{1}{(k-1) \cdot k \cdot(k+1)}= \\
& \frac{1}{2}\left[\frac{1}{(k-1) \cdot k}-\frac{1}{k \cdot(k+1)}\right], k=3,4, \cdots, n
\end{aligned}$$
Adding the above inequalities yields
$$\frac{1}{3^{3... | \frac{1}{12}-\frac{1}{2 n \cdot(n+1)}<\frac{1}{12} | Inequalities | proof | Yes | Yes | inequalities | false | 732,919 |
(3) Prove: $2^{\frac{1}{2}} 4^{\frac{1}{4}} 8^{\frac{1}{8}} \cdots\left(2^{n}\right)^{\frac{1}{2^{n}}}<4$. (1996 Irish Mathematical Olympiad Problem) | (3)
$$2^{\frac{1}{2}} 4^{\frac{1}{4}} 8^{\frac{1}{8}} \cdots\left(2^{n}\right)^{\frac{1}{2^{n}}}=2^{\frac{1}{2}+2\left(\frac{1}{2}\right)^{2}+3\left(\frac{1}{2}\right)^{2}+\cdots+n\left(\frac{1}{2}\right)^{n}}$$
Let
$$S=\frac{1}{2}+2\left(\frac{1}{2}\right)^{2}+3\left(\frac{1}{2}\right)^{3}+\cdots+n\left(\frac{1}{2}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,920 |
25. Given a real number $c > -2, x_{1}, x_{2}, \cdots, x_{n}\left(x_{n+1}=x_{1}\right)$ are $n$ positive real numbers, prove the equation
$$\sum_{i=1}^{n} \sqrt{x_{i}^{2} + c x_{i} x_{i+1} + x_{i+1}^{2}} = \sqrt{c+2} \sum_{i=1}^{n} x_{i}$$
holds if and only if $c=2$ or $x_{1}=x_{2}=\cdots=x_{n}$. (2005 Polish Mathemat... | 25. The formula is
$$\begin{array}{l}
\sum_{i=1}^{n} \sqrt{x_{i}^{2}+c x_{i} x_{i+1}+x_{i+1}^{2}}= \\
\sum_{i=1}^{n} \sqrt{\frac{\left(x_{i}+x_{i+1}\right)^{2}+\left(x_{i}-x_{i+1}\right)^{2}}{2}+c \cdot \frac{\left(x_{i}+x_{i+1}\right)^{2}-\left(x_{i}-x_{i+1}\right)^{2}}{4}}= \\
\sum_{i=1}^{n} \sqrt{\frac{2+c}{4}\left(... | proof | Algebra | proof | Yes | Yes | inequalities | false | 732,926 |
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