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Example 2.9 (Han Jingjun) $a, b, c \geqslant 0$, not two of them are 0 at the same time, prove that
$$\sum_{\rightsquigarrow x} \frac{a^{3}}{a^{2}-a b+b^{2}} \leqslant\left(1+\frac{\sqrt{3}}{3}-\frac{4}{\sqrt{3}(\sqrt{3}+3+\sqrt{2 \sqrt{3}})}\right)(a+b+c)$$ | To facilitate the proof, we set
$$\begin{array}{c}
f(a, b, c)=\sum_{\text {cc }} \frac{a^{3}}{a^{2}-a b+b^{2}} \\
\lambda=1+\frac{\sqrt{3}}{3}-\frac{4}{\sqrt{3}(\sqrt{3}+3+\sqrt{2 \sqrt{3}})}
\end{array}$$
We first prove that when $a=\max \{a, b, c\}$, we have
$$\begin{array}{l}
\frac{f(a, b, c)}{a+b+c} \leqslant \fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,070 |
Lemma $2.1\left(x_{1}^{1}, x_{2}^{1}, \cdots, x_{n}^{1}\right)$ is a sequence of real numbers, and we define the $t$-th adjustment as follows.
(1) Select $i, j \in\{1,2, \cdots, n\}$ such that
$$M_{t}=x_{i}^{t}=\max \left\{x_{1}^{t}, x_{2}^{t}, \cdots, x_{n}^{t}\right\}, m_{t}=x_{j}^{t}=\min \left\{x_{1}^{t}, x_{2}^{t}... | To prove that due to the adjustments being infinitely many, there must exist infinitely many adjustments satisfying
$$\frac{M_{t}-x_{0}^{t+1}}{M_{t}-m_{t}} \leqslant \varepsilon \text { or } \frac{x_{0}^{t+1}-m_{t}}{M_{t}-m_{t}} \leqslant \varepsilon$$
For convenience, we only prove the case where each adjustment sati... | proof | Algebra | proof | Yes | Yes | inequalities | false | 732,071 |
Example 1.6 (Romanian Mathematical Olympiad 2007) For positive numbers $a, b, c$ satisfying
$$\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1} \geqslant 1$$
Prove that
$$a+b+c \geqslant a b+b c+c a$$ | Prove by Cauchy inequality
$$\left(a+b+c^{2}\right)(a+b+1) \geqslant(a+b+c)^{2}$$
i.e.,
$$\frac{a+b+c^{2}}{(a+b+c)^{2}} \geqslant \frac{1}{a+b+1}$$
Thus,
$$\sum \frac{a+b+c^{2}}{(a+b+c)^{2}} \geqslant \sum \frac{1}{a+b+1} \geqslant 1$$
It follows that
$$\sum\left(a+b+c^{2}\right) \geqslant(a+b+c)^{2}$$
Expanding an... | a+b+c \geqslant a b+b c+c a | Inequalities | proof | Yes | Yes | inequalities | false | 732,073 |
Example $2.10$ For $a, b, c>0$ satisfying $a+b+c=3$, prove that
$$\frac{a b}{7+2 c^{2}}+\frac{b c}{7+2 a^{2}}+\frac{c a}{7+2 b^{2}} \leqslant \frac{1}{3}$$ | Proof: Without loss of generality, let $a=\min \{a, b, c\}$, and set $f(a, b, c)=\frac{1}{3}-\sum_{c c c} \frac{b c}{7+2 a^{2}}$. Then,
$$\begin{array}{l}
f(a, b, c)-f\left(a, \frac{b+c}{2}, \frac{b+c}{2}\right)= \\
(b-c)^{2}\left(\frac{1}{4\left(7+2 a^{2}\right)}-\frac{a(b+c)\left(7+2\left(b^{2}+3 b c+c^{2}\right)\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,076 |
Example $2.11 a, b, c, d>0, a+b+c+d=4$, prove that
$$\frac{1}{5-a b c}+\frac{1}{5-b c d}+\frac{1}{5-c d a}+\frac{1}{5-d a b} \leqslant 1$$ | Proof: Let $w=a b c, x=b c d, y=c d a, z=d a b$, without loss of generality, assume $w \leqslant x \leqslant y \leqslant z$. Let
$$g(w, x, y, z)=\frac{1}{5-x}-\frac{1}{5-y}+\frac{1}{5-z}+\frac{1}{5-w}$$
We first find the maximum value of $g$ when $0 \leqslant w \leqslant x \leqslant y \leqslant z \leqslant 4$, then
$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,077 |
Example 2.12 Non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1} a_{2} \cdots a_{n}=1$, for $n \geqslant 4$, prove that $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}+\frac{3 n}{a_{1}+a_{2}+\cdots+a_{n}} \geqslant n+3$ | To prove, without loss of generality, assume \(a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}\), and let
$$f\left(a_{1}, a_{2}, \cdots, a_{n}\right)=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}+\frac{3 n}{a_{1}+a_{2}+\cdots+a_{n}}$$
By the corollary, we only need to prove
$$\begin{array}{l}
f\left(a_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,078 |
Example 2.13 Positive numbers $x, y, z, t$ satisfy $x+y+z+t=4$, prove
$$(1+3 x)(1+3 y)(1+3 z)(1+3 t) \leqslant 125+131 x y z t$$ | Consider the function
$$f(x, y, z, t)=(1+3 x)(1+3 y)(1+3 z)(1+3 t)+131 x y z t$$
Without loss of generality, assume $x \geqslant y \geqslant z \geqslant t$, at this point $y+t \leqslant \frac{x+y+z+t}{2}=2$, and we have
$$\begin{array}{l}
f(x, y, z, t)-f\left(\frac{x+z}{2}, y, \frac{x+z}{2}, t\right)= \\
9(1+3 y)(1+3 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,079 |
Example 3.2 (Huang Chendi) For all $a, b, c \in [0,1]$, prove
$$(a+b+c)\left(\frac{1}{bc+1}+\frac{1}{ca+1}+\frac{1}{ab+1}\right) \leqslant 5$$ | $$\begin{array}{l}
(a+b+c)\left(\frac{1}{bc+1}+\frac{1}{ca+1}+\frac{1}{ab+1}\right)= \\
\frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}+\frac{b+c}{bc+1}+\frac{c+a}{ca+1}+\frac{a+b}{ab+1} \leqslant \\
\frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}+1+1+1 \leqslant \\
\frac{a}{bc+1}+\frac{b}{ca+b}+\frac{c}{ab+c}+3= \\
\frac{a}... | 5 | Inequalities | proof | Yes | Yes | inequalities | false | 732,082 |
Example 3.3 (2006 China National Training Team Test Question) $a, b, c, d \in \mathbf{R}^{+}, abcd=1$, prove that
$$\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geqslant 1$$ | Notice that when $xy=1$, we have
$$\frac{1}{1+x}+\frac{1}{1+y}=1$$
Thus, we have
$$1=\frac{1}{1+ab}+\frac{1}{1+cd}$$
Therefore, we consider proving the local inequality
$$\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}} \geqslant \frac{1}{1+ab}$$
The above inequality holds because
$$\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}-\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,083 |
Example $1.7$ a, b, c are positive numbers, prove
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{16(ab+bc+ca)}{a^{2}+b^{2}+c^{2}} \geqslant 8$$ | Proof: Let $a+b+c=1, ab+bc+ca=x$, then the original inequality is equivalent to
$$\frac{3abc+1-2x}{x-abc}+\frac{16x}{1-2x} \geqslant 8$$
In fact, we have
$$\begin{array}{r}
\frac{3abc+1-2x}{x-abc}+\frac{16x}{1-2x} \geqslant \frac{1-2x}{x}+\frac{16x}{1-2x}= \\
\frac{(6x-1)^2}{x(1-2x)}+8 \geqslant 8
\end{array}$$
There... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,084 |
Example $3.4 a, b, c, d>0, a+b+c+d=1$, prove:
$(1-\sqrt{a})(1-\sqrt{b})(1-\sqrt{c})(1-\sqrt{d}) \geqslant \sqrt{a b c d}$ | Proof Inspired by the previous problem, we consider grouping two of them together, and we attempt to prove
$$(1-\sqrt{a})(1-\sqrt{b}) \geqslant \sqrt{c d}$$
In fact
$$\begin{aligned}
2(1-\sqrt{a})(1-\sqrt{b})= & 2+2 \sqrt{a b}-2 \sqrt{a}-2 \sqrt{b}= \\
& (\sqrt{a}+\sqrt{b}-1)^{2}+1-a-b \geqslant \\
& c+d \geqslant 2 \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,085 |
Example 3.5 Let $n(n \geqslant 3)$ be an integer. Prove that for positive real numbers $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$, the inequality
$$\frac{x_{n} x_{1}}{x_{2}}+\frac{x_{1} x_{2}}{x_{3}}+\cdots+\frac{x_{n-1} x_{n}}{x_{1}} \geqslant x_{1}+x_{2}+\cdots+x_{n}$$
holds. | Prove a lemma first: If $0<x \leqslant y, 0<a<1$, then
$$x+y \leqslant a x+\frac{y}{a}$$
In fact, from $a x \leqslant x \leqslant y$ we get $(1-a)(y-a x) \geqslant 0$, i.e., $a^{2} x+y \geqslant a x+a y$, thus
$$x+y \leqslant a x+\frac{y}{a}$$
Now let $(x, y, a)=\left(x_{i}, x_{n-1} \cdot \frac{x_{i+1}}{x_{2}}, \frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,086 |
Example $3.60<A, B, C<\frac{\pi}{2}, A+B+C=\pi$, prove that
$$\sqrt{1-\sin A \sin B}+\sqrt{1-\sin B \sin C}+\sqrt{1-\sin C \sin A} \geqslant \frac{3}{2}$$ | To prove that each term in this problem contains a square root, direct proof is not easy. Consider tackling the square roots locally first. Let $A=\frac{\pi}{2}-\frac{A_{1}}{2}$ and similarly for the other two. Thus, the problem is transformed into proving in any triangle $\Delta A_{1} B_{1} C_{1}$ that
$$\sqrt{1-\sin ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,087 |
Example 3.7 (Yang Xuezhi) $x, y, z, w>0, \alpha, \beta, \gamma, \theta$ satisfy $\alpha+\beta+\gamma+\theta=(2 k+$ 1) $\pi(k \in \mathbf{Z})$, then we have
$$x \sin \alpha+y \sin \beta+z \sin \gamma+w \sin \theta \leqslant \sqrt{\frac{(x y+z w)(x z+y w)(x w+y z)}{x y z w}}$$ | Proof: Let \( u = x \sin \alpha + y \sin \beta, v = z \sin \gamma + w \sin \theta \), then
\[
\begin{aligned}
u^{2} = & (x \sin \alpha + y \sin \beta)^{2} \leqslant \\
& (x \sin \alpha + y \sin \beta)^{2} + (x \cos \alpha - y \cos \beta)^{2} = \\
& x^{2} + y^{2} - 2 x y \cos (\alpha + \beta)
\end{aligned}
\]
Thus,
\[
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,088 |
Example 3.9 Given $a, b, c > 0, a + b + c = 3$, prove that
$$\sqrt{3 - bc} + \sqrt{3 - ca} + \sqrt{3 - ab} \geqslant 3 \sqrt{2}$$ | To prove that it is sufficient to consider $a, b, c$ as non-negative. The original inequality, after being homogenized, becomes
$$\begin{array}{l}
\sqrt{(a+b+c)^{2}-3 b c}+\sqrt{(a+b+c)^{2}-3 c a}+ \\
\sqrt{(a+b+c)^{2}-3 a b} \geqslant \sqrt{6}(a+b+c)
\end{array}$$
Squaring both sides, it is equivalent to
$$2 \sum \s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,090 |
Example $3.10 a, b, c \geqslant 0$, prove that
$$\sum \sqrt{a^{2}+b c} \leqslant \frac{3}{2}(a+b+c)$$ | Assume $a \geqslant b \geqslant c$, then
$$2 a+c \geqslant 2 \sqrt{a^{2}+b c}$$
Thus, it suffices to prove
$$a+3 b+2 c \geqslant 2\left(\sqrt{b^{2}+a c}+\sqrt{c^{2}+a b}\right)$$
Squaring both sides and simplifying, this is equivalent to
$$a^{2}+5 b^{2}+2 a b+12 b c \geqslant 8 \sqrt{\left(b^{2}+a c\right)\left(c^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,091 |
For example, $3.11 x, y, z \geqslant 0$ satisfy $x+y+z=1$, prove that
$$\sqrt{x+y^{2}}+\sqrt{y+z^{2}}+\sqrt{z+x^{2}} \geqslant 2$$ | Proof. First, we prove the following conclusion.
If $a, b, c, d \geqslant 0$ and $a+b, c+d, (a-b)^{2} \leqslant (c-d)^{2}$, then we have
$$\sqrt{a}+\sqrt{b} \geqslant \sqrt{c}+\sqrt{d}$$
Squaring both sides and simplifying, it is equivalent to
$$\sqrt{a b} \geqslant \sqrt{c d} \Leftrightarrow a b \geqslant c d \Leftri... | 2 | Inequalities | proof | Yes | Yes | inequalities | false | 732,092 |
Example 3.12 (2006 China National Team Training Problem) For $x, y, z$ not all positive, satisfying
$$k\left(x^{2}-x+1\right)\left(y^{2}-y+1\right)\left(z^{2}-z+1\right) \geqslant(x y z)^{2}-x y z+1$$
Find the minimum value of the real number $k$. | For a 3-variable 6th-degree symmetric inequality, the equality often holds when two numbers are equal or one of them is 0. After testing, it is found that when \( x = y = \frac{1}{2}, z = 0 \), \( k \) is minimized, at which point \( k = \frac{16}{9} \). We now prove that the inequality holds when \( k = \frac{16}{9} \... | \frac{16}{9} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,093 |
Example $3.13 a, b, c \geqslant 0$, prove that
$$2\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geqslant\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)(1+a b c)$$ | Prove the local inequality
$$\begin{array}{l}
2\left(a^{3}+1\right)^{3}-\left(1+a^{2}\right)^{3}\left(1+a^{3}\right)= \\
(a-1)^{2}\left(1+a^{3}\right)\left(a^{4}+2 a^{3}+2 a+1\right) \geqslant 0
\end{array}$$
Similarly, we have
$$\begin{aligned}
2\left(b^{3}+1\right)^{3}-\left(1+b^{2}\right)^{3}\left(1+b^{3}\right) & ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,094 |
Example $1.8 a, b, c \geqslant 0$, prove that
$$\sqrt{2 a^{2}+5 a b+2 b^{2}}+\sqrt{2 a^{2}+5 a c+2 c^{2}}+\sqrt{2 b^{2}+5 b c+2 c^{2}} \leqslant 3(a+b+c)$$ | Prove that by AM-GM inequality,
$$2 a^{2}+5 a b+2 b^{2}=(2 a+b)(2 b+a) \leqslant \frac{1}{4}(2 a+b+2 b+a)^{2}=\frac{9}{4}(a+b)^{2}$$
Therefore,
$$\sum \sqrt{2 a^{2}+5 a b+2 b^{2}} \leqslant \frac{3}{2} \sum(a+b)=3(a+b+c)$$
Equality holds if and only if \(a=b=c\). | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,095 |
Example 3.14 (2002 Taiwan Mathematical Olympiad) $a, b, c, d \in\left(0, \frac{1}{2}\right]$, prove that
$$\frac{a^{4}+b^{4}+c^{4}+d^{4}}{a b c d} \geqslant \frac{(1-a)^{4}+(1-b)^{4}+(1-c)^{4}+(1-d)^{4}}{(1-a)(1-b)(1-c)(1-d)}$$ | Prove that
$$\frac{x^{4}+y^{4}+z^{4}+w^{4}}{x y z w}=\sum_{c y c} \frac{(x-y)^{4}}{x y z w}+4 \sum_{c y c} \frac{(x y)^{2}}{z w}+\frac{2\left(x^{2}+z^{2}\right)\left(y^{2}+w^{2}\right)}{x y z w}$$
Let $x=a, y=b, z=c, w=d$ and $x=1-a, y=1-b, z=1-c, w=1-d$, which are the left and right sides of the inequality, respectiv... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,096 |
Example 3.15 (2007 Girls' Mathematical Olympiad) 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
$$\frac{a_{1}}{a_{2}^{2}+1}+\frac{a_{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n}}{a_{1}^{2}+1}.$$ | This problem first appeared in Mathematical Reflections, as a generalization of a 2003 Bulgarian competition problem (the original problem was for 3 variables), and was later selected as a 2007 Girls' Mathematical Olympiad problem. It is said that few people solved this problem that year, indicating its considerable di... | \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,097 |
Example 3.17 (2005 China National Training Team Problem) $a, b, c>0, ab+bc+ca=\frac{1}{3}$, prove that
$$\frac{1}{a^{2}-bc+1}+\frac{1}{b^{2}-ca+1}+\frac{1}{c^{2}-ab+1} \leqslant 3$$ | Prove (Han Jingjun) the original inequality is equivalent to
$$\frac{1}{a(a+b+c)+\frac{2}{3}}+\frac{1}{b(a+b+c)+\frac{2}{3}}+\frac{1}{c(a+b+c)+\frac{2}{3}} \leqslant 3$$
By Cauchy's inequality, we have
$$\begin{array}{c}
\frac{1}{a(a+b+c)+\frac{2}{3}} \leqslant \frac{a(a+b+c)+\frac{3}{2}(b+c)^{2}(a+b+c)^{2}}{(a+b+c)^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,099 |
Example 3.18 (46th IMO) $x, y, z > 0, xyz \geqslant 1$ Prove
$$\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-z^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}-y^{2}}{z^{5}+x^{2}+y^{2}} \geqslant 0$$ | Prove that the original inequality is equivalent to
$$\sum \frac{x^{2}+y^{2}+z^{2}}{x^{5}+y^{2}+z^{2}} \leqslant 3$$
By the Cauchy inequality, we have
$$\left(x^{5}+y^{2}+z^{2}\right)\left(y z+y^{2}+z^{2}\right) \geqslant\left(\sqrt{x^{5} y z}+y^{2}+z^{2}\right)^{2} \geqslant\left(x^{2}+y^{2}+z^{2}\right)^{2}$$
Thus,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,100 |
Example 3.19 (42nd IMO) $a, b, c>0$, prove that
$$\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1$$ | To prove that the denominators of each fraction in this problem are different, and it is difficult to successfully apply the important inequality to each term directly. Since it involves square roots, we cannot apply the previously introduced decomposition method. Therefore, we use the method of undetermined coefficien... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,101 |
Example 3.20 Let $r_{a}, r_{b}, r_{c}$ be the radii of the excircles opposite to the sides $a, b, c$ of $\triangle A B C$, respectively. Prove that
$$\frac{a^{2}}{r_{b}^{2}+r_{c}^{2}}+\frac{b^{2}}{r_{a}^{2}+r_{c}^{2}}+\frac{c^{2}}{r_{a}^{2}+r_{b}^{2}} \geqslant 2$$ | Prove that by making the algebraic substitution $x=-a+b+c, y=-b+a+c, z=-c+a+b$, then $x, y, z>0$, note that
$$\begin{array}{c}
S_{\triangle A B C}=\frac{1}{4} \sqrt{(x+y+z) x y z} \\
r_{a}=\frac{2 S_{\triangle A B C}}{b+c-a}=\frac{1}{2 x} \sqrt{(x+y+z) x y z}
\end{array}$$
and so on, through calculation, the original ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,102 |
Let $3.21 a, b, c, d$ be non-negative real numbers. Prove
$$\begin{array}{c}
\sqrt{1+\frac{7 a}{b+c+d}}+\sqrt{1+\frac{7 b}{c+d+a}}+\sqrt{1+\frac{7 c}{d+a+b}}+ \\
\sqrt{1+\frac{7 d}{a+b+c}} \geqslant 4 \sqrt{\frac{10}{3}}
\end{array}$$ | By the homogeneity of the inequality, without loss of generality, assume $a+b+c+d=4$. Let $f(x)=\sqrt{1+\frac{7 x}{4-x}}$, then we have
$$f(x) \geqslant \frac{1}{3} \sqrt{\frac{2}{15}}(8+7 x)$$
The above inequality holds because
$$\frac{14(x-1)^{2}(2+7 x)}{135(4-x)} \geqslant 0$$
Substituting $a, b, c, d$ into $f(x)$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,103 |
Example $3.22 a, b, c>0$, prove that
$$\frac{a^{2}}{\sqrt{a^{2}+\frac{1}{4} a b+b^{2}}}+\frac{b^{2}}{\sqrt{b^{2}+\frac{1}{4} b c+c^{2}}}+\frac{c^{2}}{\sqrt{c^{2}+\frac{1}{4} c a+a^{2}}} \geqslant \frac{2}{3}(a+b+c)$$ | Notice that equality holds if and only if $a=b=c$. We let
$$\frac{x^{2}}{\sqrt{x^{2}+\frac{1}{4} x+1}} \geqslant A x+B$$
where $A, B \in \mathbf{R}$ and $A+B=\frac{2}{3}, x=\frac{a}{b}$.
Then it is not difficult to solve $A=1, B=-\frac{1}{3}$, and at this point the inequality holds, i.e.,
$$\frac{a^{2}}{\sqrt{a^{2}+\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,104 |
Example 3.23 If $x, y, z \geqslant 0, x+y+z=1$, prove that
$$\frac{5}{2} \leqslant \frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{1}{1+z^{2}} \leqslant \frac{27}{10}$$ | $$\begin{array}{l}
\frac{1}{1+x^{2}} \leqslant-\frac{27}{50}(x-2) \Leftrightarrow \\
f(x)=27 x^{3}-54 x^{2}+27 x-4 \geqslant 0 \\
f^{\prime}(x)=27(3 x-1)(x-1)
\end{array}$$
This indicates
$$\max _{0 \leqslant x \leqslant 1} f(x)=\max \{f(1 / 3), f(1)\}=0$$
Thus,
$$\sum \frac{1}{1+x^{2}} \leqslant-\frac{27}{50} \sum x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,105 |
Example 1.9 For positive numbers $a, b, c, d$, prove
$$\sum \frac{a}{3 a^{2}+2 b^{2}+c^{2}} \leqslant \frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$$ | Prove that by only using $AM-GM$ and Cauchy inequality, we have
$$\frac{18 a}{3 a^{2}+2 b^{2}+c^{2}}=\frac{18 a}{2\left(a^{2}+b^{2}\right)+a^{2}+c^{2}} \leqslant \frac{9}{2 b+c} \leqslant \frac{2}{b}+\frac{1}{c}$$
Similarly, we can obtain the other 3 inequalities. Adding these 4 inequalities together will complete the... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,106 |
Example 3.24 (2004 China Western Mathematical Competition) Let $a, b, c$ be real numbers, satisfying $a+b+c=3$, prove
$$\frac{1}{5 a^{2}-4 a+11}+\frac{1}{5 b^{2}-4 b+11}+\frac{1}{5 c^{2}-4 c+11} \leqslant \frac{1}{4}$$ | Prove that, without loss of generality, let $a=\max (a, b, c)$. We first prove that when $x \leqslant \frac{9}{5}$, we have
$$\frac{1}{5 x^{2}-4 x+11} \leqslant \frac{1}{24}(3-x) \Leftrightarrow(9-5 x)(x-1)^{2} \geqslant 0$$
We will discuss this in cases.
(1) If $a \leqslant \frac{9}{5}$, then
$$\sum \frac{1}{5 a^{2}-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,107 |
Example 3.25 (Crux) $a, b, c>0, a^{2}+b^{2}+c^{2}=1$, prove that
$$\frac{1}{1-a b}+\frac{1}{1-b c}+\frac{1}{1-c a} \leqslant \frac{9}{2}$$ | Prove that the tangent line method fails in this problem, and we attempt to set $g(x)$ as a quadratic function.
Assume $a \geqslant b \geqslant c$, it is clear that $2 a b \leqslant a^{2}+b^{2}+c^{2}=1$, thus $\max (a b, b c, c a) \leqslant \frac{1}{2}$. Considering $a b$ as $x$, we set
$$\begin{array}{c}
\frac{1}{1-x}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,108 |
For example, $4.1 a, b, c>0$, prove that
$$\sum \frac{a^{3}}{a^{2}+2 b^{2}} \geqslant \sum \frac{a^{3}}{2 a^{2}+b^{2}}$$ | Prove that the original inequality is equivalent to
$$\sum_{\text {cyc }} \frac{a^{3}\left(a^{2}-b^{2}\right)}{\left(a^{2}+2 b^{2}\right)\left(2 a^{2}+b^{2}\right)} \geqslant 0$$
Using the tangent line method (or the method of undetermined coefficients), we have
$$\begin{array}{l}
\sum_{c c}\left(\frac{a^{3}(a+b)(a-b)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,111 |
Example 4.2 (2005 IMO) $x, y, z$ are real numbers and $x y z \geqslant 1$, prove:
$$\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}-z^{2}}{z^{5}+y^{2}+x^{2}} \geqslant 0$$ | Proof: First, we homogenize it:
$$\begin{array}{r}
\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}} \geqslant \frac{x^{5}-x^{2} \cdot x y z}{x^{5}+\left(y^{2}+z^{2}\right) x y z} \geqslant \frac{x^{4}-x^{2} y z}{x^{4}+y z\left(y^{2}+z^{2}\right)} \\
\frac{x^{4}-x^{2} y z}{x^{4}+y z\left(y^{2}+z^{2}\right)} \geqslant \frac{2 x^{4}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,112 |
Example 4.3 (1999 Macedonian Mathematical Olympiad) $a, b, c>0, a^{2}+b^{2}+c^{2}=1$, prove that
$$a+b+c+\frac{1}{a b c} \geqslant 4 \sqrt{3}$$ | Prove that after homogenization, it is equivalent to
$$a+b+c+\frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a b c} \geqslant 4 \sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}$$
Notice that
$$\begin{array}{l}
\sqrt{3\left(a^{2}+b^{2}+c^{2}\right)} \backsim a+b+c, \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a b c} \backsim 3(a+b+c) \Le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,113 |
Example 4.4 (2004 Mosp) $a, b, c \geqslant 0$, prove that
$$a^{3}+b^{3}+c^{3}+3 a b c \geqslant a b \sqrt{2 a^{2}+2 b^{2}}+b c \sqrt{2 b^{2}+2 c^{2}}+c a \sqrt{2 c^{2}+2 a^{2}}$$ | To prove this is another form of the strengthened 3rd Schur's Inequality.
We know
$$\sqrt{2 a^{2}+2 b^{2}} \backsim(a+b), \sqrt{2 b^{2}+2 c^{2}} \backsim(b+c), \sqrt{2 c^{2}+2 a^{2}} \backsim(c+a)$$
To complete the square, subtract $\sum_{y, m} a^{2} b$ from both sides, which is equivalent to
$$\sum_{c y c}\left(a^{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,114 |
Example $4.5$ Given $a, b, c>0$, and $a b c=1$, prove that
$$\frac{1}{(1+a)^{3}}+\frac{1}{(1+b)^{3}}+\frac{1}{(1+c)^{3}}+\frac{5}{(1+a)(1+b)(1+c)} \geqslant 1$$ | Prove that multiplying both sides by $(1+a)(1+b)(1+c)$ gives
$$\begin{array}{l}
\sum \frac{(1+b)(1+c)}{(1+a)^{2}}+5 \geqslant(1+a)(1+b)(1+c)=2+\sum a+\sum a b \Leftrightarrow \\
\sum \frac{(1+b)(1+c)}{(1+a)^{2}}+3 \geqslant \sum a \sum a b \Leftrightarrow \\
\sum\left(\frac{(1+b)(1+c)}{(1+a)^{2}}+1-a-b c\right) \geqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,115 |
Example $4.6 a, b, c \geqslant 1, a+b+c=9$, prove
$$\sqrt{a b+b c+c a} \leqslant \sqrt{a}+\sqrt{b}+\sqrt{c}$$ | Prove the inequality after squaring both sides $\Leftrightarrow$
$$a b+b c+c a \leqslant 9+2(\sqrt{a b}+\sqrt{b c}+\sqrt{c a})$$
Below we implement the completion of the square $\Leftrightarrow$
$$\begin{array}{l}
2(a+b+c)-2(\sqrt{a b}+\sqrt{b c}+\sqrt{c a}) \leqslant 27-(a b+b c+c a) \Leftrightarrow \\
\sum(\sqrt{a}-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,116 |
Example 1.10 (2009 Serbia Mathematical Olympiad) Positive real numbers $x, y, z$ satisfy $x+y+z=$ $xy+yz+zx$, prove
$$\frac{1}{x^{2}+y+1}+\frac{1}{y^{2}+z+1}+\frac{1}{z^{2}+x+1} \leqslant 1$$ | Prove that by Cauchy's inequality,
$$\begin{array}{c}
\left(x^{2}+y+1\right)\left(1+y+z^{2}\right) \geqslant(x+y+z)^{2} \Rightarrow \\
\frac{1}{x^{2}+y+1} \leqslant \frac{1+y+z^{2}}{(x+y+z)^{2}}
\end{array}$$
Thus,
$$\text { LHS } \leqslant \frac{3+x+y+z+x^{2}+y^{2}+z^{2}}{(x+y+z)^{2}}$$
It suffices to prove that
$$(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,117 |
Example 4.7 (1996 Iran Mathematical Olympiad) For all $a, b, c>0$, prove
$$\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}} \geqslant \frac{9}{4(a b+b c+c a)}$$ | Proof (Yang Xuezhi) We prove a stronger proposition.
For $x, y, z > 0, u, v, w > 0$, we have
$$\begin{array}{l}
\sum \frac{1}{(y+z)(v+w)} \geqslant \frac{9}{2 \sum x(v+w)} \Leftrightarrow \\
\sum \frac{\sum x(v+w)}{(y+z)(v+w)} \geqslant \frac{9}{2} \Leftrightarrow \\
\left(\sum \frac{x}{y+z}-\frac{3}{2}\right)+\left(\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,118 |
Example 4.8 (Vasile Cirtoaje) $a, b, c>0, a b+b c+c a=1$, prove that
$$\sum \frac{1+a^{2} b^{2}}{(a+b)^{2}} \geqslant \frac{5}{2}$$ | Prove the completion of the formula.
$$\begin{aligned}
\sum \frac{1+a^{2} b^{2}}{(a+b)^{2}}-\frac{5}{2}= & \sum \frac{(1-a b)^{2}}{(a+b)^{2}}+\sum \frac{2 a b}{(a+b)^{2}}-\frac{5}{2}= \\
& \sum \frac{c^{2}(a+b)^{2}}{(a+b)^{2}}-\sum \frac{(a-b)^{2}}{2(a+b)^{2}}-1= \\
& \sum a^{2}-\sum a b-\sum \frac{(a-b)^{2}}{2(a+b)^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,119 |
Example 4.9 (Han Jingjun) Non-negative real numbers $a, b, c$ satisfy $a+b+c=1$, prove that
$$\frac{\sqrt{a}}{b+c a}+\frac{\sqrt{b}}{c+a b}+\frac{\sqrt{c}}{a+b c} \geqslant \frac{9 \sqrt{3}}{4}$$ | Prove first a lemma: If $a, b, c \geqslant 0, a+b+c=1$, then
$$64(a b+b c+c a) \geqslant 243(a+b)^{2}(b+c)^{2}(c+a)^{2}$$
Let $p=a+b+c=1, q=a b+b c+c a, r=a b c$, then (*) is equivalent to $64 q \geqslant 243(q-r)^{2}$. We discuss in two cases.
(1) If $q \geqslant \frac{3}{25}$, by the 3rd degree Schur's inequality, w... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,120 |
Example $4.10$ a, b, c > 0, prove that
$$a^{2}+b^{2}+c^{2} \geqslant \frac{9 a b^{3}}{5 a^{2}+4 b^{2}}+\frac{9 b c^{2}}{5 b^{2}+4 c^{2}}+\frac{9 c a^{3}}{5 c^{2}+4 a^{2}}$$ | Prove the inequality is equivalent to
$$\begin{array}{l}
\sum\left(b^{2}-\frac{9 a b^{3}}{5 a^{2}+4 b^{2}}\right) \geqslant 0 \Leftrightarrow \\
\sum \frac{b^{2}(5 a-4 b)(a-b)}{5 a^{2}+4 b^{2}} \geqslant 0 \Leftrightarrow \\
\sum\left(\frac{18 b^{2}(5 a-4 b)(a-b)}{5 a^{2}+4 b^{2}}-\left(a^{2}-b^{2}\right)\right) \geqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,121 |
Example 4.11 (Romanian Mathematical Olympiad 2007) \(a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n} \in\) R, satisfying
$$\sum_{i=1}^{n} a_{i}^{2}=\sum_{i=1}^{n} b_{i}^{2}=1, \sum_{i=1}^{n} a_{i} b_{i}=0$$
Prove that
$$\left(\sum_{i=1}^{n} a_{i}\right)^{2}+\left(\sum_{i=1}^{n} b_{i}\right)^{2} \leqslant n$$ | Proof: Let $A=\sum_{i=1}^{n} a_{i}, B=\sum_{i=1}^{n} b_{i}$, then
$$\begin{array}{l}
0 \leqslant \sum_{i=1}^{n}\left(1-A a_{i}-B b_{i}\right)^{2}= \\
\quad \sum_{i=1}^{n}\left(1+A^{2} a_{i}^{2}+B^{2} b_{i}^{2}-2 A a_{i}-2 B b_{i}+2 A B a_{i} b_{i}\right)= \\
\quad \sum_{i=1}^{n} 1+A^{2} \sum_{i=1}^{n} a_{i}^{2}+B^{2} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,122 |
Example 4.12 (Crux1998; Mohammed Aassila, Komal) $a, b, c>0$, prove that
$$\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)} \geqslant \frac{3}{1+a b c}$$ | Proof Notice that
$$\begin{array}{l}
\sum \frac{1+a b c}{a(1+b)}-3=\sum \frac{1-a b+(b c-1) a}{a+a b}= \\
\sum\left(\frac{1-a b}{a+a b}+\frac{a b-1}{1+a}\right)=\sum \frac{(a b-1)^{2}}{(a+a b)(1+a)}
\end{array}$$
Thus, the proposition is proved!
Note In fact, we have a stronger inequality:
$$\frac{1}{a(b+1)}+\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,123 |
Example $4.14$ a, b, c $\in \mathbf{R}$, prove that
$$\begin{array}{c}
\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right) \geqslant \\
3\left(a^{2} b+b^{2} c+c^{2} a\right)\left(a b^{2}+b c^{2}+c a^{2}\right)
\end{array}$$ | Proof
$$\text { LHS - RHS }=(a-b)^{2}(b-c)^{2}(c-a)^{2}$$
The above is obvious.
Note This problem can also be proven as follows.
Proof By Cauchy's inequality we have
$$\begin{array}{l}
\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)= \\
\frac{1}{16}\left(3(a+b)^{2}+(a-b)^{2}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,125 |
Example $4.15 a, b, c \in \mathbf{R}, k \geqslant 0$, prove that
$$\sum \frac{(a-k b)(a-k c)}{(b-c)^{2}} \geqslant \frac{8+8 k-k^{2}}{4}$$ | Notice that we have
$$\frac{(a-k b)(a-k c)}{(b-c)^{2}}+\frac{k^{2}}{4}=\frac{(2 a-k b-k c)^{2}}{4(b-c)^{2}}$$
It suffices to prove
$$\sum \frac{(2 a-k b-k c)^{2}}{(b-c)^{2}} \geqslant 8(k+1)$$
It is easy to see that
$$\sum \frac{2 a-k b-k c}{b-c} \cdot \frac{2 b-k c-k a}{c-a}=-4(k+1)$$
Thus, the original inequality ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,126 |
For example, $4.16 x, y, z \geqslant 0$, prove that
$$\frac{x}{1+x+x y}+\frac{y}{1+y+y z}+\frac{z}{1+z+z x} \leqslant 1$$ | $$\begin{array}{l}
\frac{1}{y z+1+y}-\frac{1}{\frac{1}{x}+1+y}+\frac{1}{1+\frac{1}{y z}+\frac{1}{z}}-\frac{1}{x+1+\frac{1}{z}} \geqslant 0 \Leftrightarrow \\
\frac{1-x y z}{(y z+1+y)(1+x+x y)}+\frac{(x y z-1) z}{(y z+1+y)(x z+z+1)} \geqslant 0 \Leftrightarrow \\
\frac{(1-x y z)^{2}}{(y z+1+y)(1+x+x y)(x z+z+1)} \geqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,127 |
Example 1 Given that $a, b, c$ are positive real numbers, prove that for any real numbers $x, y, z$, we have
$$\begin{array}{l}
x^{2}+y^{2}+z^{2} \geqslant 2 \sqrt{\frac{a b c}{(a+b)(b+c)(c+a)}}\left(\sqrt{\frac{a+b}{c}} x y+\right. \\
\left.\sqrt{\frac{b+c}{a}} y z+\sqrt{\frac{c+a}{b}} z x\right)
\end{array}$$
and de... | Prove that the left side - right side $=$
$$\begin{array}{l}
=\left(\frac{b}{b+c} x^{2}+\frac{a}{c+a} y^{2}-2 \sqrt{\frac{a b}{(b+c)(c+a)}} x y\right)+ \\
\left(\frac{c}{c+a} y^{2}+\frac{b}{a+b} z^{2}-2 \sqrt{(c+a)(a+b)} y z\right)+ \\
\left(\frac{c}{b+c} x^{2}+\frac{a}{a+z^{2}}-2 \sqrt{(b+c)(a+b)} x z\right)= \\
\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,128 |
Example 10 - Given $n$ real numbers $a_{\mathrm{E}} \leqslant \sigma_{2} \leqslant \cdots \leqslant a_{n-1} \leqslant a_{n}$, let $x=\frac{1}{n}\left(a_{1}+a_{2}+\cdots+\right.$ $\left.a_{n}\right), y=\frac{1}{n}\left(a_{1}^{2} \pm a_{2}^{2-}+\cdots+a_{n}^{2}\right)$, prove: $2 \sqrt{y-x^{2}} \leqslant a_{n}-a_{1} \leq... | Given that $n x=a_{1}+a_{2}+\cdots+a_{n}, n y=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}$, then
$$\begin{array}{l}
n^{2}\left(y-x^{2}\right)=n\left(-a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}= \\
(n-F) \sum_{i=1}^{n} a_{i}^{2}-2_{1 \leqslant i<j \leqslant n}^{-} a_{i} a_{j}= \\
(n-F... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,130 |
7. If $x, y, z$ are three positive numbers, prove: $\sqrt{3}\left(\frac{y z}{x}+\frac{z x}{y}+\frac{x y}{z}\right) \geqslant(y z+z x+x y)$ $\sqrt{\frac{x+y+z}{x y z}}$, and the equality holds if and only if $x=y=z$. (Mathematical Bulletin Problem 444) | 7. By the basic inequality, we have
$$3\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+b+c)^{2}, a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$$
Therefore,
$$\begin{array}{c}
3\left[(y z)^{2}+(z x)^{2}+(x y)^{2}\right] \geqslant(y z+z x+x y)^{2} \\
(y z)^{2}+(z x)^{2}+(x y)^{2} \geqslant y z \cdot z x+z x \cdot x y+x y \cdot y z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,131 |
9. Prove that for any positive numbers $a, b, c, d$, we have
$$\sqrt{\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}} \geqslant \sqrt[3]{\frac{a b c+a b d+a c d+b c d}{4}}$$ | $$\begin{array}{l}
\text { 9. } \frac{a b c+a b d+a c d+b c d}{4}=\frac{1}{2}\left(a b \cdot \frac{c+d}{2}+c d \cdot \frac{a+b}{2}\right) \leqslant \\
\frac{1}{2}\left[\left(\frac{a+b}{2}\right)^{2} \cdot \frac{c+d}{2}+\left(\frac{c+d}{2}\right)^{2} \cdot \frac{a+b}{2}\right]= \\
\frac{1}{2}\left(\frac{a+b}{2}\right)\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,133 |
10. Given three positive real numbers $x, y, z$ satisfying $x+y+z+\frac{1}{2} \sqrt{x y z}=16$, prove:
(1) $\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{1}{8} \sqrt{x y z} \leqslant 7$; (2) $\sqrt{x}+\sqrt{y}+\sqrt{z} \geqslant 4+\frac{1}{4} \sqrt{x y z}$.
(2004 Henan Province Mathematics Competition Problem) | 10. ( (1) Since $x+4 \geqslant 4 \sqrt{x}, y+4 \geqslant 4 \sqrt{y}, z+4 \geqslant 4 \sqrt{z}$, therefore, $x+y+z \mp 12 \geqslant$ $4(\sqrt{x}+\sqrt{y}+\sqrt{z})$, combining with $x+y+z+\frac{1}{2} \sqrt{x y z}=16$, we get $\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{1}{8} \sqrt{x y z} \leqslant 7$.
(2) From $x+y+\frac{1}{2} \sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,134 |
11. (1) Let $a, b, c$ be positive numbers, and 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)
(2) Let $a, b, c$ be positive numbers, and their sum equals 2. Prove: $\frac{a}{1-a} \cdot \frac{b}{1-b}... | 11. (1) From the inequality $\frac{1}{x}+\frac{1}{y} \geqslant \frac{4}{x+y}$, where $x, y>0$, we can obtain $\frac{1}{a+b}+\frac{1}{b+c} \geqslant$ $\frac{4}{a+2 b+c}, \frac{1}{b+c}+\frac{1}{c+a} \geqslant \frac{4}{b+2 c+a}, \frac{1}{c+a}+\frac{1}{a+b} \geqslant \frac{4}{2 a+b+c}$. Adding these three inequalities, we ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,135 |
13. Let $a, b, c$ be positive numbers, and $a+b+c=3$, prove: $\sqrt{a}+\sqrt{b}+\sqrt{c} \geqslant a b+b c+c a$. (28th Russian Mathematical Olympiad problem) | 13. Since $(a+b+c)^{2}=9$, then
$$a b+b c+c a=\frac{9-\left(a^{2}+b^{2}+c^{2}\right)}{2}$$
It is sufficient to prove that $2 \sqrt{a}+2 \sqrt{b}+2 \sqrt{c}+a^{2}+b^{2}+c^{2} \geqslant 9$. For this, we first prove that $2 \sqrt{a}+a^{2} \geqslant 3 a$. In fact, $2 \sqrt{a}+a^{2}=\sqrt{a}+\sqrt{a}+a^{2} \geqslant 3 \cdo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,137 |
14. Let $a, b, c$ be the lengths of the sides of a triangle, prove that:
$$\sqrt{\frac{a}{b+c-a}}+\sqrt{\frac{b}{c+a-b}}+\sqrt{\frac{c}{a+b-c}} \geqslant 3$$ | 14. By the binary mean inequality, we have
$$\sqrt{\frac{a}{b+c-a}}=\frac{a}{\sqrt{a(b+c-a)}} \geqslant \frac{2 a}{a+(b+c-a)}=\frac{2 a}{b+c}$$
Similarly, we get
$$\sqrt{\frac{b}{c+a-b}} \geqslant \frac{2 b}{c+a}, \sqrt{\frac{c}{a+b-c}} \geqslant \frac{2 c}{a+b}$$
Adding these three inequalities, we obtain
$$\sqrt{\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,138 |
15. Let $a, b, c$ be positive numbers, prove that:
$$\sqrt{a b(a+b)}+\sqrt{b c(b+c)}+\sqrt{c a(c+a)} \leqslant \frac{3}{2} \sqrt{(a+b)(b+c)(c+a)}$$ | 15. The original inequality is equivalent to
$$\sqrt{\frac{a b}{(c+a)(b+c)}}+\sqrt{\frac{b c}{(a+b)(c+a)}}+\sqrt{\frac{c a}{(b+c)(a+b)}} \leqslant \frac{3}{2}$$
By the AM-GM inequality, we have
$$\sqrt{\frac{a b}{(c+a)(b+c)}}=\sqrt{\frac{a}{c+a}} \sqrt{\frac{b}{b+c}} \leqslant \frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,139 |
16. Let $a, b, c$ be positive numbers. Prove that: $\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}} \geqslant 3 \sqrt{2}$. | 16. By the binary mean inequality,
$$\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}} \geqslant \sqrt{2}\left(\sqrt[4]{\frac{b c}{a^{2}}}+\sqrt[4]{\frac{a c}{b^{2}}}+\sqrt[4]{\frac{a b}{c^{2}}}\right)$$
By the ternary mean inequality,
$$\sqrt[4]{\frac{b c}{a^{2}}}+\sqrt[4]{\frac{a c}{b^{2}}}+\sqrt[4]{\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,140 |
17. Let $x, y, z$ be positive numbers, and $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $S=\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}$.
(1988 Soviet Union Mathematical Olympiad Problem) | 17. Since $S$ is a positive number, $S$ and $S^{2}$ attain their minimum values at the same point. Given that $x^{2}+y^{2}+z^{2}=1$, by the AM-GM inequality, we have
$$\begin{aligned}
S^{2}= & \left(\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}\right)^{2}=\frac{x^{2} y^{2}}{z^{2}}+\frac{y^{2} z^{2}}{x^{2}}+\frac{z^{2} x^{2... | \sqrt{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,142 |
18. If $x, y, z$ are all positive real numbers, find the maximum value of $\frac{x y z}{(1+5 x)(-4 x+3 y)(5 y+6 z)(z+18)}$ and prove your conclusion. (2003 Singapore Mathematical Olympiad) | 18. Given a fixed $y$
$$\begin{array}{l}
\frac{x}{(1+5 x)(4 x+3 y)}=\frac{x}{20 x^{2}+(15 y+4) x+3 y}= \\
\frac{1}{20 x+\frac{3 y}{x}+(15 y+4)} \leqslant \\
\frac{1}{2 \sqrt{20 x \cdot \frac{3 y}{x}}+(-15 y+4)}= \\
\frac{1}{(\sqrt{15 y}+2)^{2}}
\end{array}$$
Equality holds if and only if $x=\sqrt{\frac{3 y}{20}}$. Sim... | \frac{1}{5120} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,143 |
20. Let $a, b, c$ be positive numbers, and satisfy $a^{2}+b^{2}+c^{2}=3$, prove: $\frac{1}{1+2 a b}+\frac{1}{1+2 b c}+$ $\frac{1}{1+2 c a} \geqslant 1$ (2004 Estonian Olympiad Problem) | 20. From the arithmetic mean being greater than or equal to the geometric mean and the arithmetic mean being greater than or equal to the harmonic mean, we get
$$\frac{1}{1+2 a b}+\frac{1}{1+2 b c}+\frac{1}{1+2 c a} \geqslant \frac{1}{1+a^{2}+b^{2}}+\frac{1}{1+b^{2}+c^{2}}+\frac{1}{1+c^{2}+a^{2}} \geqslant$$
$$3 \cdot ... | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 732,145 |
23. Let $a, b, c$ be positive numbers, and $a^{2}+b^{2} \leq c^{2}=1$, prove:
$$\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\frac{c}{1-c^{2}} \geqslant \frac{3 \sqrt{3}}{2}$$
(30th IMO Canadian Training Problem) | 23. $\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\frac{c}{1-c^{2}} \geqslant \frac{3 \sqrt{3}}{2}$ is equivalent to
$$\frac{a^{2}}{a\left(1-a^{2}\right)}+\frac{b^{2}}{b\left(1-b^{2}\right)}+\frac{c^{2}}{c\left(1-c^{2}\right)} \geqslant \frac{3 \sqrt{3}}{2}$$
Given that $a^{2}+b^{2}+c^{2}=1$. If we can prove that $x\left(1-x^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,149 |
24. Given that $x, y, z$ are positive numbers, and satisfy $x y z(x+y+z)=1$, find the minimum value of the expression $(x+y)(x+z)$. (1989 All-Soviet Union Mathematical Olympiad Problem) | 24. Since $(x+y)(x+z)=y z+x(x+y+z)$, by the AM-GM inequality we have
$$\begin{array}{l}
(x+y)(x+z)=y z+x(x+y+z) \geqslant 2 \sqrt{y z \cdot x(x+y+z)}= \\
2 \sqrt{x y z \cdot(x+y+z)}=2
\end{array}$$
Furthermore, when $x=\sqrt{2}-1, y=z=1$, the above expression achieves equality, hence the minimum value of $(x+y)(x+z)$ ... | 2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,150 |
Example 12 Let $x, y, z$ be real numbers, $k_{1}, k_{2}, k_{3} \in\left(0, \frac{1}{2}\right)$, and $k_{1}+k_{2}+k_{3}=1$, prove: $k_{1} k_{2} k_{3}(x+y+z)^{2} \geqslant x y k_{3}\left(1-2 k_{3}\right)+y z k_{1}\left(1-2 k_{1}\right)+z x k_{2}\left(1-2 k_{2}\right)$. (1999 China National Training Team Test Question) | Proof: First, we prove a lemma: In $\triangle ABC$, for any real numbers $x, y, z$, we have
$$x^{2}+y^{2}+z^{2} \geqslant 2 x y \cos C + 2 y z \cos A + 2 z x \cos B$$
In fact, since $\cos A = -\cos (B+C)$, we have
$$\begin{array}{l}
x^{2}+y^{2}+z^{2}-(2 x y \cos C + 2 y z \cos A + 2 z x \cos B) = \\
(x - y \cos C - z ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,152 |
26. Let $x_{1}, x_{2} \in \mathbf{R}$, and $x_{1}^{2}+x_{2}^{2} \leqslant 1$, prove that for any $y_{1}, y_{2} \in \mathbf{R}$, we have $\left(x_{1} y_{1}+ \right.$ $\left.x_{2} y_{2}-1\right)^{2} \geqslant\left(x_{1}^{2}+x_{2}^{2}-1\right)\left(y_{1}^{2}+y_{2}^{2}-1\right)$. (2000 Poland-Austria Mathematical Olympiad ... | 26. If $y_{1}^{2}+y_{2}^{2}-\mathrm{F} \geqslant 0$, then the inequality obviously holds - if $y_{1}^{2}+y_{2}^{2}-1<0$, then by the mean inequality we get $x_{1} y_{1} \leqslant \frac{x_{1}^{2}+y_{1}^{2}}{2}, x_{2} y_{2} \leqslant \frac{x_{2}^{2}+y_{2}^{2}}{2}$, then
Thus,
$$\begin{array}{c}
x_{1} y_{1}+x_{2} y_{2} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,153 |
28. Let $a, b, c$ be positive numbers, prove:
$$\begin{array}{l}
\text { (1) } 4\left(a^{3}+b^{3}\right) \geqslant(a+b)^{3} \\
\text { (2) } 9\left(a^{3}+b^{3}+c^{3}\right) \geqslant(a+b+c)^{3}
\end{array}$$
(1996 British Mathematical Olympiad) | 28. $(1)-$
$$\begin{array}{l}
a^{3}+\left(-\frac{a+b}{2}\right)^{3}+\left(\frac{a+b}{2}\right)^{3} \geqslant 3 a\left(\frac{a+b}{2}\right)^{2} \\
b^{3}+\left(\frac{a+b}{2}\right)^{3}+\left(\frac{a+b}{2}\right)^{3} \geqslant 3 b\left(\frac{a+b}{2}\right)^{2}
\end{array}$$
Adding the two equations and simplifying yields... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,155 |
30. If positive real numbers $a, b, c$ satisfy $a+b+c=1$, prove that: $a \sqrt[3]{1+b-c}+b \sqrt[3]{1+c-a}+$ $c \sqrt[3]{1+a-b} \leqslant 1$. (2005 Japan Mathematical Olympiad Problem) | 30=Notice that $1 \leftarrow b-c=a+b \pm c+b-c=a+2 b>0$, so by the AM-GM inequality we have
$$\sqrt[3]{1+b-c} \leqslant \frac{1+(1+b-c)}{3}=1+\frac{b-c}{3}$$
Multiplying both sides by $a$ gives
$$a \sqrt[3]{1+b-c} \leqslant a+\frac{a b-a c}{3}$$
Similarly,
$$b \sqrt[3]{1+c}-a \leqslant b+\frac{b c-a b}{3}, c \sqrt[3]... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,159 |
33. Let $a, b, c \in(0,1), x, y, z \in(0,+\infty), a^{x}=b c, b^{y}=c a, c^{z}=a b$, prove that: $\frac{1}{2+x}+\frac{1}{2+y}+\frac{1}{2+z} \leqslant \frac{3}{4}$. (2006 Romanian Mathematical Olympiad Problem) | 33. From the given, we have $x=\frac{\ln (b c)}{\ln a}, y=\frac{\ln (c a)}{\ln b}, z=\frac{\ln (a b)}{\ln c}$. Let $p=-\ln a, q=-\ln b, r=-\ln c$, then $p, q, r$ are all positive numbers, and $x=\frac{q+r}{p}, y=\frac{p+r}{q}, z=\frac{p+q}{r}$.
$$\begin{array}{l}
\frac{1}{2+x}+\frac{1}{2+y}+\frac{1}{2+z}=\frac{p}{2 p+q... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 732,160 |
34. Let $a, b, c$ be positive numbers, find the minimum value of $\frac{a^{2}+b^{2}}{c^{2}+a b}+\frac{b^{2}+c^{2}}{a^{2}+b c}+\frac{c^{2}+a^{2}}{b^{2}+c a}$. (2004 Indian Mathematical Olympiad) | 34. Let $x=a^{2}+b^{2}, y=b^{2}+c^{2}, z=c^{2}+a^{2}$, then
$$\begin{array}{l}
\frac{a^{2}+b^{2}}{c^{2}+a b}+\frac{b^{2}+c^{2}}{a^{2}+b c}+\frac{c^{2}+a^{2}}{b^{2}+c a} \geqslant \frac{a^{2}+b^{2}}{c^{2}+\frac{a^{2}+b^{2}}{2}}+\frac{b^{2}+c^{2}}{a^{2}+\frac{b^{2}+c^{2}}{2}}+\frac{c^{2}+a^{2}}{b^{2}+\frac{c^{2}+a^{2}}{2... | 3 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,161 |
35. (1) Let $x, y, u, v \in \mathbf{R}^{+}$, prove the inequality: $\frac{u}{x}+\frac{v}{y} \geqslant \frac{4(u y+v x)}{(x+y)^{2}}$.
(2) Let $a, b, c, d \in \mathbf{R}^{+}$, prove the inequality: $\frac{a}{b+2 c+d}+\frac{b}{c+2 d+a}+\frac{c}{d+2 a+b}+$ $\frac{d}{a+2 b+c} \geqslant 1$. (2005 Romanian Mathematical Olympi... | 35. (1) Since $(x+y)^{2} \geqslant 4 x y$, therefore, $\frac{1}{x y} \geqslant \frac{4}{(x+y)^{2}}$. Multiplying both sides by $u y+v x$ gives $\frac{u}{x}+\frac{v}{y} \geqslant \frac{4(u y+v x)}{(x+y)^{2}}$.
(2) Using the conclusion from (1): Let $u=a, v=c, x=b+2 c+d, y=d+2 a+b$, so
$$\begin{aligned}
\frac{a}{b+2 c+d}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,162 |
36. Let $a, b, c \in \mathbf{R}^{+}$, prove the inequality: $a b c \geqslant(-a+b+c)(a-b+c)(a+b-$ c). (1983 Swiss Mathematical Olympiad) | 36. $-a+b+c, a-b+c, a+b-c$ can have at most one negative number, otherwise, since the sum of any two of them is positive, a contradiction can be derived. If $-a+b+c, a-b+c, a+b-c$ has one negative number, the inequality obviously holds. If all three are positive, then
$$\begin{array}{l}
\sqrt{(-a+b+c)(a-b+c)} \leqslant... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,164 |
38. Given that $a, b, c, d$ are positive real numbers, and $a+b+c+d=4$. Prove:
$$\frac{a}{1+b^{2}}+\frac{b}{1+c^{2}}+\frac{c}{1+d^{2}}+\frac{d}{1+a^{2}} \geqslant 2$$
(2006 Russian Winter Camp Problem) | 38. $\frac{a}{1+b^{2}}=\frac{a b^{2}+a-a b^{2}}{1+b^{2}}=a-b \cdot \frac{a b}{1+b^{2}} \geqslant a-b \cdot \frac{a b}{2 b}=a-\frac{a b}{2}$
Similarly,
$$\frac{b}{1+c^{2}} \geqslant b-\frac{b c}{2}, \frac{c}{1+d^{2}} \geqslant c-\frac{c d}{2}, \frac{d}{1+a^{2}} \geqslant d-\frac{d a}{2}$$
Thus,
$$\begin{array}{r}
\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,166 |
39. Given that $a, b, c$ are positive real numbers, prove: $\frac{2\left(a^{3}+b^{3}+c^{3}\right)}{a b c}+\frac{9(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}} \geqslant 33$. Canada Curx problem | 39.
$$\begin{array}{l}
\frac{2\left(a^{3}+b^{3}+c^{3}\right)}{a b c}+\frac{9(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}-33= \\
\frac{2\left(a^{3}+b^{3}+c^{3}-6 a b c\right)}{a b c}+\frac{9\left[(a+b+c)^{2}-3\left(a^{2}+b^{2}+c^{2}\right)\right]}{a^{2}+b^{2}+c^{2}}= \\
\frac{2(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)}{a b c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,167 |
41. Let $a, b, c, d, e, f \in \mathbf{R}$, and $a+b+c+d+e+f=0$, prove: $a b+b c+c d+$ $d e+e f+f a \leqslant \frac{1}{2}\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}\right)$. (2003 French Mathematical Olympiad Problem) | 41. Since $(a+c+e)(b+d+f)=-(a+c+e)^{2} \leqslant 0$, we have
$$(a+c+e)(b+d+f)=(a b+b c+c d+d e+e f+f a)+(a d+b e+c f) \leqslant 0$$
That is,
$$\begin{array}{l}
a b+b c+c d+d e+e f+f a \leqslant-(a d+b e+c f) \leqslant \\
\frac{1}{2}\left(a^{2}+d^{2}\right)+\frac{1}{2}\left(b^{2}+e^{2}\right)+\frac{1}{2}\left(c^{2}+f^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,169 |
42. Let $a, b, c \in \mathbf{R}^{+}$, and $a b c=1$, prove the inequality: $\frac{1+a b}{1+a}+\frac{1+b c}{1+b}+\frac{1+c a}{1+c} \geqslant 3$. (2006 Morocco Mathematical Olympiad Problem) | 42. Since $a b c=1$, by the AM-GM inequality we have
$$\begin{array}{l}
\frac{1+a b}{1+a}+\frac{1+b c}{1+b}+\frac{1+c a}{1+c} \geqslant 3 \sqrt[3]{\frac{(1+a b)(1+b c)(1+c a)}{(1+a)(1+b)(1+c)}}= \\
\sqrt[3]{\frac{\left(1+\frac{1}{c}\right)\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)}{(1+a)(1+b)(1+c)}}=3 \sqrt[3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,170 |
44. Let $x, y, z$ be positive real numbers, prove that: $\left(\frac{x+y}{2}\right)^{2}\left(\frac{y+z}{2}\right)^{2}\left(\frac{z+x}{2}\right)^{2} \geqslant x y z\left(\frac{x+y+z}{3}\right)^{3}$. (Mathematical Communications, Issue 12, 1986, Problem Curx2108) | 44. Since $x, y, z$ are positive, we have
$$\begin{array}{c}
x^{2} y+y z^{2} \geqslant 2 x y z, y^{2} z+x^{2} z \geqslant 2 x y z, z^{2} x+x y^{2} \geqslant 2 x y z \\
\frac{1}{2}\left(x^{2} y^{2}+y^{2} z^{2}\right) \geqslant x y^{2} z, \frac{1}{2}\left(y^{2} z^{2}+z^{2} x^{2}\right) \geqslant x y z^{2}, \frac{1}{2}\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,172 |
45. Given that $x, y, z$ are all positive numbers.
(1) Prove: $\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y} \geqslant \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$;
(2) If $x+y+z \geqslant x y z$, find the minimum value of $u=\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y}$. (2006 Shaanxi Provincial Mathematics Competition) | 45. (1) Since $x, y, z$ are all positive numbers, we have $\frac{x}{y z}+\frac{y}{z x}=\frac{1}{z}\left(\frac{x}{y}+\frac{y}{x}\right) \geqslant \frac{2}{z}$. Similarly, $\frac{y}{z x}+\frac{z}{x y} \geqslant \frac{2}{x}, \frac{x}{y z}+\frac{z}{x y} \geqslant \frac{2}{y}$. The equality holds if and only if $x=y=z$. Add... | \sqrt{3} | Inequalities | proof | Yes | Yes | inequalities | false | 732,173 |
Example 14 Let $x_{1}, x_{2}, \cdots, x_{n} \in(0,1)$, and $x_{1}+x_{2}+\cdots+x_{n}=1$, prove that: $(n-$ 1) $\left(\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+\cdots+\frac{1}{1-x_{n}}\right) \geqslant(n+1)\left(\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n}}\right)$. (2004 Romanian Mathematical Olympiad) | $$\begin{array}{l}
(n-1)\left(\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+\cdots+\frac{1}{1-x_{n}}\right)- \\
(n+1)\left(\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n}}\right)= \\
\sum_{k=1}^{n}\left[(n-1) \frac{1}{1-x_{k}}-(n+1) \frac{1}{1+x_{k}}\right]=2 \sum_{k=1}^{n} \frac{n x_{k}-1}{1-x_{k}^{2}}= \\
2 \sum_{k... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,174 |
46. Given that $x, y, z$ are positive numbers, and $x+y+z=1$, prove:
$$\left(\frac{1}{x}-x\right)\left(\frac{1}{y}-y\right)\left(\frac{1}{z}-z\right) \geqslant\left(\frac{8}{3}\right)^{3}$$ | 46. By the AM-GM inequality, we have $\frac{x y}{z}+\frac{y z}{x} \geqslant 2 y, \frac{y z}{x}+\frac{z x}{y} \geqslant 2 z, \frac{x y}{z}+\frac{z x}{y} \geqslant 2 x$. Adding these three inequalities, we get $\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y} \geqslant x+y+z=1$. Also, by $1=x+y+z \geqslant 3 \sqrt[3]{x y z}$, w... | \left(\frac{8}{3}\right)^{3} | Inequalities | proof | Yes | Yes | inequalities | false | 732,175 |
47. Let three positive real numbers $x, y, z$. Try to find the maximum value of the algebraic expression $\frac{16 x+9 \sqrt{2 x y}+9 \sqrt[3]{3 x y z}}{x+2 y+z}$. (2006 Henan Province Mathematics Competition Problem) | 47. Since
$$\begin{array}{l}
\frac{16 x+9 \sqrt{2 x y}+9 \sqrt[3]{3 x y z}}{x+2 y+z}= \\
\frac{16 x+\frac{9 \sqrt{x \cdot 18 y}}{3}+\frac{3 \sqrt[3]{x \cdot 18 y \cdot 36 z}}{3}}{x+2 y+z} \leqslant \\
\frac{16 x+\frac{3(x+18 y)}{2}+\frac{x+18 y+36 z}{2}}{x+2 y+z}=18
\end{array}$$
Therefore, the equality holds if and o... | 18 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,176 |
49. Let $a, b, c$ be positive real numbers, and $a+b+c=3$, prove that: $\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+$ $\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}}+\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \leqslant 5$. (2006 Northern China Mathematical Olympiad) | 49. Since $a+b+c=3$, we have
$$\begin{array}{l}
\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}}+\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}}= \\
\frac{a^{2}+(a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+(a+b+c)^{2}}{2 b^{2}+(c+a)^{2}}+\frac{c^{2}+(a+b+c)^{2}}{2 c^{2}+(a+b)^{2}}= \\
\frac{2 a^{2}+(b+c)^{2}+2 a(b+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,177 |
48. Given that $a, b, c$ are positive numbers, and $a+b+c=1$, prove:
$$\sqrt{\frac{1-a}{a}} \cdot \sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-b}{b}} \cdot \sqrt{\frac{1-c}{c}}+\sqrt{\frac{1-a}{a}} \cdot \sqrt{\frac{1-c}{c}} \geqslant 6$$
(2006 Albanian Mathematical Olympiad Problem) | 48. Since $a+b+c=1$, we have
$$\begin{array}{l}
\sqrt{\frac{1-a}{a}} \cdot \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-b}{b}} \cdot \sqrt{\frac{1-c}{c}} + \sqrt{\frac{1-a}{a}} \cdot \sqrt{\frac{1-c}{c}} = \\
\sqrt{\frac{b+c}{a}} \cdot \sqrt{\frac{c+a}{b}} + \sqrt{\frac{c+a}{b}} \cdot \sqrt{\frac{a+b}{c}} + \sqrt{\frac{b+c}{a}... | 6 | Inequalities | proof | Yes | Yes | inequalities | false | 732,178 |
54. Let $a, b, c, d$ be positive real numbers, prove: $\frac{a+b+c+d}{a b c d} \leqslant \frac{F}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}+\frac{1}{d^{3}}$. (2005 Austrian Mathematical Olympiad) | $\begin{array}{l}\text { 54. } \frac{a+b+c+d}{a b c d}=\frac{1}{b c d}+\frac{1}{a c d}+\frac{1}{a b d}+\frac{1}{a b c} \leqslant \frac{1}{3}\left(\frac{1}{b^{3}}+\frac{1}{c^{3}}+\frac{1}{d^{3}}\right)+\frac{1}{3}\left(\frac{1}{a^{3}}+\right. \\ \left.\frac{1}{c^{3}}+\frac{1}{d^{3}}\right)+\frac{1}{3}\left(\frac{1}{a^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,183 |
55. Let $a, b, c$ be positive real numbers, prove:
$$\frac{a^{2}}{a^{2}+2 b c}+\frac{b^{2}}{b^{2}+2 c a}+\frac{c^{2}}{c^{2}+2 a b} \geqslant 1 \geqslant \frac{b c}{a^{2}+2 b c}+\frac{c a}{b^{2}+2 c a}+\frac{a b}{c^{2}+2 a b}$$
(1997 Romanian Mathematical Olympiad) | 55. $\quad \frac{a^{2}}{a^{2}+2 b c}+\frac{b^{2}}{b^{2}+2 c a}+\frac{c^{2}}{c^{2}+2 a b} \geqslant$
$$\begin{array}{c}
\frac{a^{2}}{a^{2}+b^{2}+c^{2}}+\frac{b^{2}}{b^{2}+c^{2}+a^{2}}+\frac{c^{2}}{c^{2}+a^{2}+b^{2}}=1 \\
\frac{a^{2}}{a^{2}+2 b c}+\frac{b^{2}}{b^{2}+2 c a}+\frac{c^{2}}{c^{2}+2 a b}+2\left(\frac{b c}{a^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,184 |
56. Given that $x, y, z$ are positive numbers, and $x+y+z=\sqrt{x y z}$, prove: $x y+y z+z x \geqslant 9(x+y+z)$. (1996 Belarus Mathematical Olympiad Problem) | $\begin{array}{l}\text { 56. } x y+y z+z x=x y z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=(x+y+z)^{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=(x+ \\ y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+y+z) \geqslant 9(x+y+z) .\end{array}$
The translation is as follows:
$\begin{array}{l}\text { 56. }... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,186 |
57. Let $\theta<a, b-c<1$, prove that $: \sqrt{a b c}+\sqrt{(1-a)(1-b)(1-c)}<1$. (2002 Romanian Mathematical Olympiad) | $\begin{array}{l}\text { 57. Since } 0<a, b, c<1 \text {, then } \sqrt{a b c}+\sqrt{(1-a)(1-b)(1-c)}<\sqrt[3]{a b c}+ \\ \sqrt[3]{(1-a)(1-b)(1-c)} \leqslant \frac{a+b+c}{3}+\frac{(1-a)+(1-b)+(1-c)}{3}=1 .\end{array}$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,187 |
58. Given $a, b, c \geqslant F$, prove: $\sqrt{a-\mathrm{F}}+\sqrt{b-1}+\sqrt{c-1} \leqslant \sqrt{(a b+1) c}$.
(-1989 Yugoslav Mathematical Olympiad, 1998 Hong Kong Mathematical Olympiad) | 58. If $x, y \geqslant 1$, then $\sqrt{x-1}+\sqrt{y-1} \leqslant \sqrt{x y}$. In fact,
$$\begin{aligned}
\sqrt{x-1}+\sqrt{y-1} \leqslant \sqrt{x y} \Leftrightarrow & x+y-2+2 \sqrt{(x-1)(y-1)} \leqslant x y \Leftrightarrow \\
& 2 \sqrt{(x-1)(y-1)} \leqslant(x-1)(y-1)+1
\end{aligned}$$
The last inequality is obtained by... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,188 |
60. Let $a, b, c, d>0$, prove the inequality:
$$\sqrt[3]{a b}+\sqrt[3]{c d} \leqslant \sqrt[3]{(a+b+c)(b+c+d)}$$ | 60. According to the AM-GM inequality, we have $\sqrt[3]{x y z} \leqslant \frac{x+y+z}{3}$. Let $x=\frac{a}{a+b+c}, y=\frac{b+c}{b+c+d}, z=\frac{b}{b+c}$, then
$$\sqrt[3]{\frac{a b}{(a+b+c)(b+c+d)}} \leqslant \frac{1}{3}\left(\frac{a}{a+b+c}+\frac{b+c}{b+c+d}+\frac{b}{b+c}\right)$$
Let $x=\frac{b+c}{a+b+c}, y=\frac{d}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,190 |
64. Let $a>b>0, f(-x)=\frac{2(a+b) x+2 a b}{4 x+a+b}$, prove: there exists a unique positive number $x$, such that $f(x)=\frac{a^{\frac{5}{3}}+6^{\frac{1}{3}}}{2}$. (2006 China Southeast Mathematical Olympiad) | 64. Let $t=\frac{a^{\frac{1}{2}}+b^{\frac{1}{3}}}{2}$, from $t=\frac{2(a+b) x+2 a b}{4 x+a+b}$, we get
$$[2(a+b)-4 t] x=t(a+b)-2 a b$$
To prove (1) has a unique positive solution $x$, it suffices to prove $2(a+b)=-4 t>0=$ and $t(a+b)-2 a b>0$, which means
$$\frac{2 a b}{a+b}v$, which means to prove
$$\frac{2 u^{3} v^{... | proof | Algebra | proof | Yes | Yes | inequalities | false | 732,193 |
69. Let $a, b, c$ be positive numbers, and $a+b+c=1$, prove: $\frac{a^{7}+b^{7}}{a^{5}+b^{5}}+\frac{b^{7}+c^{7}}{b^{5}+c^{5}}+\frac{c^{7}+a^{7}}{c^{5}+a^{5}} \geqslant \frac{1}{3}$. (2000 Kazakhstan Mathematical Olympiad Problem) | 69. Since
$$\begin{array}{l}
2\left(a^{7}+b^{7}\right)-\left(a^{5}+b^{5}\right)\left(a^{2}+b^{2}\right)=\left(a^{5}-b^{5}\right)\left(a^{2}-b^{2}\right)= \\
(a-b)^{2}(a+b)\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right) \geqslant 0
\end{array}$$
Therefore, $\frac{a^{7}+b^{7}}{a^{5}+b^{5}} \geqslant \frac{a^{2}+b^{... | \frac{1}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 732,199 |
72. Let $a, b, c$ be positive numbers, and $a+b+c=1$, prove: $a \sqrt{b}+b \sqrt{c}+c \sqrt{a} \leqslant \frac{1}{\sqrt{3}}$. (2005 Bosnia and Herzegovina Mathematical Olympiad Problem) | 72.
$$\begin{array}{l}
a \sqrt{b} \frac{1}{\sqrt{3}}+b \sqrt{c} \frac{1}{\sqrt{3}}+c \sqrt{a} \frac{1}{\sqrt{3}} \leqslant a \cdot \frac{b+\frac{1}{3}}{2}+b \cdot \frac{c+\frac{1}{3}}{2}+c \cdot \frac{a+\frac{1}{3}}{2}= \\
\frac{a b+b c+c a}{2}+\frac{a+b+c}{6} \leqslant \frac{(a+b+c)^{2}}{6}+\frac{a+b+c}{6}=\frac{1}{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,203 |
73. Let $a, b$ be positive numbers, prove: $\frac{1}{2}(a+b)^{2}+\frac{1}{4}(a+b) \geqslant a \sqrt{b}+b \sqrt{a} \cdot$ (1987 Yugoslav Mathematical Olympiad Problem) | $\begin{array}{l}\text { 73. } \frac{1}{2}(a+b)^{2}+\frac{1}{4}(a+b)=\frac{1}{4}(a+b)\left(2 a+\frac{1}{2}+2 b+\frac{1}{2}\right) \geqslant \\ \frac{1}{4}(a+b)\left(2 \sqrt{2 a \cdot \frac{1}{2}}+2 \sqrt{2 b \cdot \frac{1}{2}}\right)=\frac{1}{2}(a+b)(\sqrt{a}+\sqrt{b}) \geqslant \\ \sqrt{a b}(\sqrt{a}+\sqrt{b})=a \sqrt... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,204 |
74. Let $a, b, c$ be positive numbers, prove:
$$\frac{1}{3}(a+b+c) \leqslant \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}} \leqslant \frac{\mathrm{F}}{3}\left(\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b}\right)$$
(2007 Irish Mathematical Olympiad) | 74.
$$\begin{aligned}
a^{2}+\left(\frac{a+b+c}{3}\right)^{2} & \geqslant \frac{2 a(a+b+c)}{3} \\
b^{2}+\left(\frac{a+b+c}{3}\right)^{2} & \geqslant \frac{2 b(a+b+c)}{3} \\
c^{2}+\left(\frac{a+b+c}{3}\right)^{2} & \geqslant \frac{2 c(a+b+c)}{3}
\end{aligned}$$
Adding them up, we get $a^{2}+b^{2}+c^{2} \geqslant \frac{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,205 |
Example 17 Given $\sum_{i=1}^{n} \frac{F}{1+w_{i}}=1$, prove: $\sum_{i=k}^{n} \sqrt{w_{i}} \geqslant(n-1)=\sum_{i=1}^{n} \frac{1}{\sqrt{w_{i}}}$. (1993 Austrian-Polish Mathematical Team Competition Problem) | Proof: Let $x_{i}=\frac{1}{1+w_{i}}(i=1,2, \ldots, n)$; then $\bar{w}_{i}=\frac{1}{x_{i}}-1$.
The original inequality condition is transformed into $\sum_{r=1}^{n} x_{i}=1$, and the conclusion is transformed into
$$\sum_{i=1}^{n} \sqrt{\frac{1-x_{i}}{x_{i}}} \geqslant(-n-1) \sum_{i=1}^{n} \sqrt{\frac{x_{i}}{-1-x_{i}}}$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,207 |
76. Let $u, v, w$ be positive real numbers, satisfying the condition $u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v}=1$, find the minimum value of $u+v+$ $w$ ( (3rd China Girls Mathematical Olympiad problem) | 76. By the AM-GM inequality and the conditions given in the problem, we have
$$u \cdot \frac{v+w}{2}+v \cdot \frac{w+u}{2}+w \cdot \frac{u+v}{2} \geqslant u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v}=1$$
Thus, \(uv + vw + wu \geqslant 1\). Therefore,
$$\begin{array}{l}
(u+v+w)^{2}=u^{2}+v^{2}+w^{2}+2uv+2vw+2wu= \\
\frac{u^{... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,208 |
77. Let $a, b, c$ be positive real numbers, prove that: $\left(a^{5}-a^{2}+3\right)\left(b^{5}-b^{2}+3\right)\left(c^{5}-c^{2}+3\right) \geqslant$ $(a+b+c)^{3}$ (2004 USA Mathematical Olympiad Problem) | 77. Notice that, when $a>0$, we have
$$\left(a^{5}-a^{2}+3\right)-\left(a^{3}+2\right)=a^{5}-a^{3}-a^{2}+1=\left(a^{3}-1\right)\left(a^{2}-1\right) \geqslant 0$$
Therefore, $a^{5}-a^{2}+3 \geqslant a^{3}+2$.
Next, we prove
$$\left(a^{3}+2\right)\left(b^{3}+2\right)\left(c^{3}+2\right) \geqslant(-a+b+c)^{3}$$
Expandin... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,209 |
79. Let $0<\alpha, \beta, \gamma<\frac{\pi}{2}$, and $\sin ^{3} \alpha+\sin ^{3} \beta+\sin ^{3} \gamma=1$, prove that $\tan ^{2} \alpha+\tan ^{2} \beta+$ $\tan ^{2} \gamma \geqslant \frac{3 \sqrt{3}}{2}$. (2005 China Southeast Mathematical Olympiad Problem) | 79. Let $a=\sin \alpha, b=\sin \beta, c=\sin \gamma$, then $a^{3}+b^{3}+c^{3}=1$, then
$$a-a^{3}=\frac{1}{\sqrt{2}} \sqrt{2 a^{2}\left(1-a^{2}\right)^{2}} \leq \frac{1}{\sqrt{2}} \sqrt{\left(\frac{2 a^{2}+\left(1-a^{2}\right)+\left(1-a^{2}\right)}{3}\right) 3}=\frac{2}{3 \sqrt{3}}$$
Similarly, $b-b^{3} \leqslant \frac... | \tan ^{2} \alpha+\tan ^{2} \beta+\tan ^{2} \gamma \geqslant \frac{3 \sqrt{3}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,211 |
80. Let $x, y, z$ be non-negative real numbers, and satisfy $x+y+z=1$, prove: $x^{2} y+y^{2} z+z^{2} x \leqslant \frac{4}{27}$
(1999 Canadian Mathematical Olympiad) | 80. Let $x=\frac{a^{2}}{a^{2}+1}, y=\frac{b^{2}}{b^{2}+1}, z=\frac{c^{2}}{c^{2}+1}$, then $0<x, y, z<1$, and $x+y+z=1$, $a^{2}=\frac{x}{1-x}=\frac{x}{y+z}, b^{2}=\frac{y}{1-y}=\frac{y}{z+x}, c^{2}=\frac{z}{1-z}=\frac{z}{x+y}$, to prove $a b c \leqslant \frac{\sqrt{2}}{4}$, it is sufficient to prove $a^{2} b^{2} c^{2} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,212 |
84. Let \(x ; y, z\) be real numbers greater than -1, prove:
$$\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \geqslant 2$$ | 84. Since $x, y, z > -1$, the denominators on the left side of the inequality are all positive, so
$$\begin{array}{l}
\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \geqslant \\
\frac{1+x^{2}}{1+\frac{1+y^{2}}{2}+z^{2}}+\frac{1+y^{2}}{1+\frac{1+z^{2}}{2}+x^{2}}+\frac{1+z^{2}}{1+\frac{1+x^... | 2 | Inequalities | proof | Yes | Yes | inequalities | false | 732,216 |
85. Let $a, b, c$ be positive real numbers, prove that
$$\frac{a^{3}-2 a+2}{b+c}+\frac{b^{3}}{c+a b+2}+\frac{c^{3}-2 c+2}{a+b} \geqslant \frac{3}{2}$$ | 85. By the AM-GM inequality, $a^{3}+2=a^{3}+1+1 \geqslant 3 a$, so $a^{3}-2 a+2 \geqslant a$. Similarly, $b^{3}-2 b+2 \geqslant b$, $c^{3}-2 c+2 \geqslant c$. It suffices to prove $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}$. This has been proven in Section 1. | \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,217 |
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