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Example 2.2.10 Let $a, b, c > 0$, and satisfy $3 \max \left(a^{2}, b^{2}, c^{2}\right) \leq 2\left(a^{2}+b^{2}+c^{2}\right)$, prove
$$\frac{a}{\sqrt{2 b^{2}+2 c^{2}-a^{2}}}+\frac{b}{\sqrt{2 c^{2}+2 a^{2}-b^{2}}}+\frac{c}{\sqrt{2 a^{2}+2 b^{2}-c^{2}}} \geq \sqrt{3}$$ | Proof: By Holder's inequality, we have
$$\left(\sum_{c y c} \frac{a}{\sqrt{2 b^{2}+2 c^{2}-a^{2}}}\right)\left(\sum_{c y c} \frac{a}{\sqrt{2 b^{2}+2 c^{2}-a^{2}}}\right)\left(\sum_{c y c} a\left(2 b^{2}+2 c^{2}-a^{2}\right)\right) \geq(a+b+c)^{3}$$
Thus, it suffices to prove $\square$
$$(a+b+c)^{3} \geq 3 \sum_{c y c}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,691 |
Example 1.1.5 Let $a, b, c$ be the side lengths of a triangle with perimeter 3. Prove that
\[
\frac{1}{\sqrt{a+b-c}}+\frac{1}{\sqrt{b+c-a}}+\frac{1}{\sqrt{c+a-b}} \geq \frac{9}{a b+b c+c a} \quad \text{(Pham Kim Hung)}
\] | Proof: Let $x=\sqrt{b+c-a}, y=\sqrt{c+a-b}, z=\sqrt{a+b-c}$, we have $x^{2}+y^{2}+z^{2}=3$. The inequality becomes
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{36}{9+x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}$$
Let $m=x y, n=y z, p=z x$, then the above inequality is equivalent to
$$(m+n+p)\left(m^{2}+n^{2}+p^{2}+9\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,692 |
Theorem 4 (Chebyshev's Inequality) Let $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ and $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ be two increasing sequences of real numbers, then
$$a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n} \geq \frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(b_{1}+b_{2}+\cdots+b_{n}\right)$$ | Prove: By directly expanding the expression, we have
$$n\left(a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(b_{1}+b_{2}+\cdots+b_{n}\right)=\sum_{i, j=1}^{n}\left(a_{i}-a_{j}\right)\left(b_{i}-b_{j}\right) \geq 0$$
Note: Using the same method, we can obtain that if the se... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,693 |
Example 3.1.1 Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=n$, prove that
$$a_{1}^{n+1}+a_{2}^{n+1}+\cdots+a_{n}^{n+1} \geq a_{1}^{n}+a_{2}^{n}+\cdots+a_{n}^{n}$$ | To prove: In order to solve this problem using the AM-GM inequality, we must go through two steps: first, prove $n \sum_{i=1}^{n} a_{i}^{n+1}+n \geq(n+1) \sum_{i=1}^{n} a_{i}^{n}$, and then prove $\sum_{i=1}^{n} a_{i}^{n} \geq n$. To solve this problem using the Cauchy-Schwarz inequality, we must use an inductive metho... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,694 |
Example 3.1.2 Let $a, b, c, d>0$, and $a^{2}+b^{2}+c^{2}+d^{2}=4$, prove:
$$\frac{a^{2}}{b+c+d}+\frac{b^{2}}{c+d+a}+\frac{c^{2}}{d+a+b}+\frac{d^{2}}{a+b+c} \geq \frac{4}{3}$$ | Proof: Note that, if $(a, b, c, d)$ is in increasing order, then
$$\frac{1}{b+c+d} \geq \frac{1}{c+d+a} \geq \frac{1}{d+a+b} \geq \frac{1}{a+b+c}$$
Therefore, by Chebyshev's inequality, we have
$$4 L H S \geq\left(\sum_{\text {cyc }} a^{2}\right)\left(\sum_{c y c} \frac{1}{b+c+d}\right) \geq \frac{16\left(a^{2}+b^{2}+... | \frac{a^{2}}{b+c+d}+\frac{b^{2}}{c+d+a}+\frac{c^{2}}{d+a+b}+\frac{d^{2}}{a+b+c} \geq \frac{4}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 731,695 |
Example 3.1.3 Let $a, b, c>1$, and satisfy the condition $\frac{1}{a^{2}-1}+\frac{1}{b^{2}-1}+\frac{1}{c^{2}-1}=1$, prove that $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} \leq 1 \quad($ Poru Loh,Crux $)$ | Proof: Note that, if $a \geq b \geq c$, then we have
$$\frac{a-2}{a+1} \geq \frac{b-2}{b+1} \geq \frac{c-2}{c+1} ; \quad \frac{a+2}{a-1} \leq \frac{b+2}{b-1} \leq \frac{c+2}{c-1}$$
By Chebyshev's inequality, we have
$$3\left(\sum_{c y c} \frac{a^{2}-4}{a^{2}-1}\right) \leq\left(\sum_{c y c} \frac{a-2}{a+1}\right)\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,696 |
Example 3.1.4 Let $a, b, c, d, e \geq 0$, and $\frac{1}{4+a}+\frac{1}{4+b}+\frac{1}{4+c}+\frac{1}{4+d}+\frac{1}{4+e}=1$, prove: $\frac{a}{4+a^{2}}+\frac{b}{4+b^{2}}+\frac{c}{4+c^{2}}+\frac{d}{4+d^{2}}+\frac{e}{4+e^{2}} \leq 1$ | Proof: Given the condition, we have $\sum_{c y c} \frac{1-a}{4+a}=0$, we need to prove
$$\sum_{c y c} \frac{1}{4+a} \geq \sum_{c y c} \frac{a}{4+a} \Leftrightarrow \sum_{c y c} \frac{1-a}{4+a} \cdot \frac{1}{4+a^{2}} \geq 0$$
Assume $a \geq b \geq c \geq d \geq e$, then
$$\begin{array}{l}
\frac{1-a}{4+a} \leq \frac{1-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,697 |
Example 3.1.5 Let $a, b, c, d>0$, and satisfy $a+b+c+d=4$, prove: $\frac{1}{11+a^{2}}+\frac{1}{11+b^{2}}+\frac{1}{11+c^{2}}+\frac{1}{11+d^{2}} \leq \frac{1}{3} \quad$ (Pham Kim Hung) | Prove: The inequality can be transformed as follows
$$\sum_{c y c}\left(\frac{1}{11+a^{2}}-\frac{1}{12}\right) \geq 0$$
or equivalently,
$$\sum_{c y c}(1-a) \cdot \frac{a+1}{a^{2}+11} \geq 0$$
Notice that, if $(a, b, c, d)$ is in increasing order, then
$$\frac{a+1}{a^{2}+11} \geq \frac{b+1}{b^{2}+11} \geq \frac{c+1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,698 |
Example 3.2.1 Let $a, b, c, d>0$, and satisfy $a+b+c+d=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$, prove the inequality $2(a+b+c+d) \geq \sqrt{a^{2}+3}+\sqrt{b^{2}+3}+\sqrt{c^{2}+3}+\sqrt{d^{2}+3}$ (Pham Kim Hung) | Proof: A cursory look at this inequality might make you hesitate. The relationship between variables $a, b, c, d$ is difficult to manipulate; moreover, the inequality contains square roots, which adds to the complexity. Surprisingly, a simple method is to use Chebyshev's inequality to unravel this mystery. Let's explor... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,700 |
Example 3.2.2 Let $a, b, c>0$, and satisfy $a+b+c=3$, prove: $\frac{1}{c^{2}+a+b}+\frac{1}{a^{2}+b+c}+\frac{1}{c^{2}+c+a} \leq 1$ | Prove: The inequality is equivalent to
$$\sum_{c y c}\left(\frac{1}{c^{2}-c+3}-\frac{1}{3}\right) \geq 0 \Leftrightarrow \sum_{c y c} \frac{a(a-1)}{a^{2}-a+3} \geq 0$$
or
$$\sum_{c y c} \frac{a-1}{a-1+\frac{3}{a}} \geq 0$$
According to Chebyshev's inequality and the condition $a+b+c=3$, it suffices to prove that if $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,701 |
Example 3.2.3 Let $a, b, c>0, 0 \leq k<2$, prove:
$$\frac{a^{2}-b c}{b^{2}+c^{2}+k a^{2}}+\frac{b^{2}-c a}{c^{2}+a^{2}+k b^{2}}+\frac{c^{2}-a b}{a^{2}+b^{2}+k c^{2}} \geq 0$$ | Prove: Although this problem can be solved using the same method as Example 2.1.1, here we provide a simpler solution using Chebyshev's inequality. Notice that, if $a \geq b$, then for any positive real number $c$, we have $\left(a^{2}-b c\right)(b+c) \geq\left(b^{2}-c a\right)(c+a)$ and
$$\left(b^{2}+c^{2}+k a^{2}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,702 |
Example 1.1.6 Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, and satisfy $a_{i} \in[0, i] i \in\{1,2, \cdots, n\}$, prove:
$2^{n} a_{1}\left(a_{1}+a_{2}\right) \cdots\left(a_{1}+a_{2}+\cdots+a_{n}\right) \geq(n+1) a_{1}^{2} a_{2}^{2} \cdots a_{n}^{2}$
【Phan Thanh Nam】 | Prove: According to the AM-GM inequality, we have
$$\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{k}=1 \cdot\left(\frac{a_{1}}{1}\right)+2 \cdot\left(\frac{a_{2}}{2}\right)+\cdots+k \cdot\left(\frac{a_{k}}{k}\right) \\
\geq \frac{k(k+1)}{2}\left(\frac{a_{1}}{1}\right)^{\frac{2}{k(k+1)}} \cdot\left(\frac{a_{2}}{2}\right)^{\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,703 |
Example 3.2.4 Let $a, b, c>0$, prove:
$$\sqrt{a^{2}+8 b c}+\sqrt{b^{2}+8 c a}+\sqrt{c^{2}+8 a b} \leq 3(a+b+c)$$ | Prove: The inequality is equivalent to
$$\sum_{\text {cyc }}\left(3 a-\sqrt{a^{2}+8 b c}\right) \geq 0 \Leftrightarrow \sum_{c y c} \frac{a^{2}-b c}{3 a+\sqrt{a^{2}+8 b c}} \geq 0$$
or
$$\sum_{c y c} \frac{\left(a^{2}-b c\right)(b+c)}{\left(3 a+\sqrt{a^{2}+8 b c}\right)(b+c)} \geq 0$$
According to Chebyshev's inequal... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,704 |
Example 3.2.5 Let $a, b, c, d>0$, and $a^{2}+b^{2}+c^{2}+d^{2}=4$, prove: $\frac{1}{5-a}+\frac{1}{5-b}+\frac{1}{5-c}+\frac{1}{5-d} \leq 1 \quad$ (Pham Kim Hung) | Prove: The inequality is equivalent to
$$\sum_{c y c}\left(\frac{1}{5-a}-\frac{1}{4}\right) \leq 0 \Leftrightarrow \sum_{c y c} \frac{a-1}{5-a} \leq 0 \Leftrightarrow \sum_{c y c} \frac{(a-1)(a+1)}{(5-a)(a+1)} \leq 0 \Leftrightarrow \sum_{c y c} \frac{a^{2}-1}{4 a-a^{2}+5} \leq 0$$
Notice that $\sum_{c y c}\left(a^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,705 |
Example 3.2.6 Let $a_{1}, a_{2}, \cdots, a_{n}>0$, and satisfy $a_{1}+a_{2}+\cdots+a_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}$, prove the following inequality
$$\frac{1}{n^{2}+a_{1}^{2}-1}+\frac{1}{n^{2}+a_{2}^{2}-1}+\cdots+\frac{1}{n^{2}+a_{n}^{2}-1} \geq \frac{1}{a_{1}+a_{2}+\cdots+a_{n}}$$
(Pham Ki... | Proof: Without loss of generality, we assume $a_{1} \geq a_{2} \geq \cdots \geq a_{n}$. The given condition is equivalent to
$$\frac{1-a_{1}^{2}}{a_{1}}+\frac{1-a_{2}^{2}}{a_{2}}+\cdots+\frac{1-a_{n}^{2}}{a_{n}}=0$$
Let $S=\sum_{i=1}^{n} a_{i}, k=n^{2}-1$, according to $(*)$, the inequality can be rewritten as
$$\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,706 |
Example 3.2.7 Let $a, b, c>0$, and $a+b+c=3$, prove: $\frac{1}{9-ab}+\frac{1}{9-bc}+\frac{1}{9-ca} \leq \frac{3}{8}$ | Proof: Let $x=b c, y=c a, z=a b$, then the inequality becomes
$$\sum_{c y c} \frac{1}{9-x} \leq \frac{3}{8} \Leftrightarrow \sum_{c y c} \frac{1-x}{9-x} \geq 0$$
Assume $a_{x}, a_{y}, a_{z}$ are the coefficients we are looking for. We transform the inequality into
$$\sum_{c y c} a_{x}(1-x) \cdot \frac{1}{a_{x}(9-x)} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,707 |
Example 3.2.8 Let $a, b, c>0$, and $a^{4}+b^{4}+c^{4}=3$, prove: $\frac{1}{4-ab}+\frac{1}{4-bc}+\frac{1}{4-ca} \leq 1$ | Proof: Let $x=ab, y=ca, z=bc$, then the inequality is equivalent to
$$\frac{1-x}{4-x}+\frac{1-y}{4-y}+\frac{1-z}{4-z} \geq 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}} \geq 0$$
Notice that $a^{4}+b^{4}+c^{4}=3$, thus $x^{2}+y^{2}+z^{2} \leq 3$. Therefore, if $x ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,708 |
Example 3.2.9 Let $a_{1}, a_{2}, \cdots, a_{n}>0$, and satisfy $a_{1}+a_{2}+\cdots+a_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}$, prove: $\frac{1}{n-1+a_{1}^{2}}+\frac{1}{n-1+a_{2}^{2}}+\cdots+\frac{1}{n-1+a_{n}^{2}} \leq 1 \quad$ (Pham Kim Hung) | Prove: The inequality can be rewritten as follows
$$\sum_{i=1}^{n}\left(\frac{1}{n-1+a_{i}^{2}}-\frac{1}{n}\right) \leq 0$$
or equivalently,
$$\sum_{i=1}^{n} \frac{a_{i}^{2}-1}{n-1+a_{i}^{2}} \geq 0$$
Assuming \(a_{1} \geq a_{2} \geq \cdots \geq a_{n}\), according to the problem, we have
$$\sum_{i=1}^{n} \frac{a_{i}^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,709 |
Theorem 5, If $f(x)$ is a real function defined on $[a, b] \subset R$, and for any $x \in [a, b], f^{\prime \prime}(x) \geq 0$, then $f(x)$ is a convex function on $[a, b]$. | Proof: We will prove that for any $x, y \in [a, b]$ and $0 \leq t \leq 1$, we always have
$$t f(x) + (1-t) f(y) \geq f(t x + (1-t) y)$$
In fact, assuming $t$ and $y$ are constants, define
$$g(x) = t f(x) + (1-t) f(y) - f(t x + (1-t) y)$$
Taking the derivative with respect to $x$,
$$g'(x) = t f'(x) - t f'(t x + (1-t) ... | proof | Calculus | proof | Yes | Yes | inequalities | false | 731,710 |
Example 1.1.7 Let $a, b, c$ be positive real numbers, prove: $\frac{1}{a^{3}+b^{3}+a b c}+\frac{1}{b^{3}+c^{3}+a b c}+\frac{1}{c^{3}+a^{3}+a b c} \leq \frac{1}{a b c}$ | Proof: Notice that $a^{3}+b^{3} \geq a b(a+b)$, so
$$\frac{a b c}{a^{3}+b^{3}+a b c} \leq \frac{a b c}{a b(a+b)+a b c}=\frac{c}{a+b+c}$$
Similarly, we can obtain the other two inequalities, and by adding them, we have
$$\frac{a b c}{a^{3}+b^{3}+a b c}+\frac{a b c}{b^{3}+c^{3}+a b c}+\frac{a b c}{c^{3}+a^{3}+a b c} \le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,714 |
Example 4.1.2 Let real numbers $a_{1}, a_{2}, \cdots, a_{n} \in\left(\frac{1}{2}, 1\right]$, prove:
$$\frac{a_{1} a_{2} \cdots a_{n}}{\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{n}} \geq \frac{\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right)}{\left(n-a_{1}-a_{2} \cdots-a_{n}\right)^{n}}$$ | Prove that the inequality is equivalent to
$$\sum_{i=1}^{n}\left[\ln a_{i}-\ln \left(1-a_{i}\right)\right] \geq n \ln \left(\sum_{i=1}^{n} a_{i}\right)-n \ln \left(n-\sum_{i=1}^{n} a_{i}\right)$$
Notice the function $f(x)=\ln x-\ln (1-x)$, its second derivative is
$$f^{\prime \prime}(x)=\frac{-1}{x^{2}}+\frac{1}{(1-x)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,719 |
Example 4.1.3 Let $a, b, c>0$, prove: $\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geq 1$ | Proof: Although this problem has been solved using Hölder's inequality, the proof given by Jensen's inequality is still very elegant. Without loss of generality, we assume $a+b+c=1$. Since $f(x)=\frac{1}{\sqrt{x}}$ is a convex function, by Jensen's inequality we get:
$$a f\left(a^{2}+8 b c\right)+b f\left(b^{2}+8 c a\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,720 |
Example 4.1.4 Let $a, b, c, d>0$, and $a+b+c+d=4$, prove:
$$\frac{a}{b^{2}+b}+\frac{b}{c^{2}+c}+\frac{c}{d^{2}+d}+\frac{d}{a^{2}+a} \geq \frac{8}{(a+c)(b+d)}$$ | Prove: Let $f(x)=\frac{1}{x(x+1)}$, then $f(x)(x>0)$ is a convex function. According to Jensen's inequality, we have
$$\frac{a}{4} \cdot f(b)+\frac{b}{4} \cdot f(c)+\frac{c}{4} \cdot f(d)+\frac{d}{4} \cdot f(a) \geq f\left(\frac{a b+b c+c d+d a}{4}\right)$$
Thus, the inequality can be rewritten as follows:
$$\sum_{c y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,721 |
Example 4.1.5 Let $a, b, c>0$, prove: $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} \leq \frac{3}{\sqrt{2}}$ | Proof: Note that $f(x)=\sqrt{x}$ is a convex function. According to Jensen's inequality, we have
$$\begin{array}{l}
\sum_{c y c} \sqrt{\frac{a}{a+b}}=\sum_{c y c} \frac{a+c}{2(a+b+c)} \cdot \sqrt{\frac{4 a(a+b+c)^{2}}{(a+b)(a+c)^{2}}} \\
\leq \sqrt{\sum_{c y c} \frac{a+c}{2(a+b+c)} \cdot \frac{4 a(a+b+c)^{2}}{(a+b)(a+c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,722 |
Example 4.1.6 Let $a, b, c \geq 0$, prove: $\frac{a}{\sqrt{4 b^{2}+b c+4 c^{2}}}+\frac{b}{\sqrt{4 c^{2}+c a+4 a^{2}}}+\frac{c}{\sqrt{4 a^{2}+a b+4 b^{2}}} \geq 1$ | Proof: We can assume $a+b+c=1$. Since $f(x)=\frac{1}{\sqrt{x}}$ is a convex function, according to Jensen's inequality, we have
$$a f\left(4 b^{2}+b c+4 c^{2}\right)+b f\left(4 c^{2}+c a+4 a^{2}\right)+c f\left(4 a^{2}+a b+4 b^{2}\right) \geq f(M)$$
Here
$$M=a\left(4 b^{2}+b c+4 c^{2}\right)+b\left(4 c^{2}+c a+4 a^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,723 |
Example 4.1.7 Let $a, b, c>0$, prove: $\sqrt{\frac{a}{4 a+4 b+c}}+\sqrt{\frac{b}{4 b+4 c+a}}+\sqrt{\frac{c}{4 c+4 a+b}} \leq 1$ | Proof: Note that the function $f(x)=\sqrt{x}$ is a convex function, so by Jensen's inequality, we have
$$\begin{array}{l}
\sum_{c y c} \sqrt{\frac{a}{4 a+4 b+c}}=\sum_{c y c} \frac{(4 a+4 b+c)}{9(a+b+c)} \cdot \sqrt{\frac{81 a(a+b+c)^{2}}{(4 a+4 b+c)(4 a+4 c+b)^{2}}} \\
\leq \sqrt{\sum_{c y c} \frac{(4 a+4 b+c)}{9(a+b+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,724 |
Example 1.1.8 If $x_{1}, x_{2}, \cdots, x_{n}>0$, and satisfy $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n}}=1$, prove:
$$x_{1} x_{2} \cdots x_{n} \geq(n-1)^{n}$$ | Prove: The given condition can be transformed into $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n-1}}=\frac{x_{n}}{1+x_{n}}$. Applying the AM-GM inequality, we have
$$\frac{x_{n}}{1+x_{n}} \geq \frac{n-1}{\sqrt[n-1]{\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n-1}\right)}}$$
Similarly, we ca... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,725 |
Example 4.1.8 Let $a, b, c > 0$ and satisfy $a^{2} + b^{2} + c^{2} = 3$, prove:
$\sqrt{\frac{a}{a^{2} + b^{2} + 1}} + \sqrt{\frac{b}{b^{2} + c^{2} + 1}} + \sqrt{\frac{c}{c^{2} + a^{2} + 1}} \leq \sqrt{3}$ (Pham Kim Hung) | Prove: Applying Jensen's inequality to the convex function $f(x)=\sqrt{x}$, we have
$$\begin{array}{l}
\sum_{c y c} \sqrt{\frac{a}{a^{2}+b^{2}+1}}=\sum_{c y c} \frac{c^{2}+a^{2}+1}{3\left(a^{2}+b^{2}+c^{2}\right)} \cdot \sqrt{\frac{9 a\left(a^{2}+b^{2}+c^{2}\right)^{2}}{\left(a^{2}+b^{2}+1\right)\left(c^{2}+a^{2}+1\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,726 |
Example 4.2.1 Let $a, b, c>0, a, b \in[1,2]$, prove: $a^{3}+b^{3}+c^{3} \leq 5 a b c$ | Proof: First, we provide a basic solution to this simple problem. Since $a, b, c \in [1,2]$, if $a \geq b \geq c$: then $\square$
$$\begin{array}{l}
a^{3}+2 \leq 5 a \Leftrightarrow(a-2)\left(a^{2}+2 a-1\right) \leq 0 \\
5 a+b^{3} \leq 5 a b+1 \Leftrightarrow(b-1)\left(b^{2}+b+1-5 a\right) \leq 0 \\
5 a b+c^{3} \leq 5 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,727 |
Example 4.2.2 Let $a, b, c>0, a, b, c \in[1,2]$, prove: $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leq 10$
(Olympiad 30-4, Vietnam) | Prove: The inequality is transformed into
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c} \leq 7$$
Without loss of generality, we can assume $a \geq b \geq c$, then
$$(a-b)(a-c) \geq 0 \Rightarrow\left\{\begin{array}{l}
\frac{a}{c}+1 \geq \frac{a}{b}+\frac{b}{c} \\
\frac{c}{a}+1 \geq \frac{c}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,728 |
Example 4.2.3 Let real numbers $x_{1}, x_{2}, \cdots, x_{2005} \in[-1,1]$, find the minimum value of the following expression
$$P=x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{2004} x_{2005}+x_{2005} x_{1}$$ | Solution: Since this inequality is cyclic rather than symmetric, we have specified the order of the variables. If we rely on the relation $\left(x_{i}-1\right)\left(x_{i}+1\right) \leq 0$, we will not succeed.
Intuitively, we believe that the expression reaches its maximum value when the sequence $x_{1}, x_{2}, \cdots... | -2003 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,729 |
Lemma 4, Let $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ be a real function defined on $[a, b] \times[a, b] \times \cdots \times[a, b] \subset R^{n}(a<b)$, satisfying that for all $k \in\{1,2, \cdots, n\}$, if we fix $n-1$ variables $x_{j}(j \neq k)$, then $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=f\left(x_{k}\right)$... | Proof: In fact, we only need to prove that if $f(x)$ is a convex function defined on $[a, b]$, then for all $x \in [a, b]$, we have
$$f(x) \leq \max \{f(a), f(b)\}$$
Substantially, because $\{t a + (1-t) b \mid t \in [0,1]\} = [a, b]$, for all $x \in [a, b]$, there exists a number $t \in [0,1]$ such that $x = t a + (1... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,730 |
Example 4.2.4 Given positive real numbers $x_{1}, x_{2}, \cdots, x_{n} \in[a, b]$, find the maximum value of the following expression
$$\left(x_{1}-x_{2}\right)^{2}+\left(x_{1}-x_{3}\right)^{2}+\cdots+\left(x_{1}-x_{n}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\cdots+\left(x_{n-1}-x_{n}\right)^{2}$$ | Proof: Let the expression above be denoted as $F$. Note that $F$, when expressed as a function of $x_{1}$ (with other variables fixed), is given by
$$f\left(x_{1}\right)=(n-1) x_{1}^{2}-2\left(\sum_{i=2}^{n} x_{i}\right) x_{1}+c$$
Here, $c$ is a constant. It is clear that $f$ is a convex function (since $\left.f^{\pri... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,731 |
Example 4.2.5 Let $n \in N$, find the minimum value of the following expression
$$f(x)=|1+x|+|2+x|+\cdots+|n+x|, \quad(x \in R)$$ | Let $I_{1}=[-1,+\infty), I_{n+1}=(-\infty,-n], I_{k}=[-k,-k+1], k \in\{2,3, \cdots, n\}$. If $x \in I_{1}$, then $f(x)=\sum_{i=1}^{n}(1+x) \geq \sum_{i=1}^{n}(i-1)=\frac{n(n-1)}{2}=f(-1)$. If $x \in I_{n}$, then
$$f(x)=\sum_{i=1}^{n}(-1-x) \geq \sum_{i=1}^{n}(-i+n)=\frac{n(n-1)}{2}=f(-n)$$
Assume $x \in I_{k}(1<k<n+1)... | \left\{\begin{array}{cc}
m(m+1) & n=2 m(m \in N) \\
(m+1)^{2} & n=2 m+1(m \in N)
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,732 |
Lemma 5, Assume $f(x)$ is a real convex function defined on $[a, b] \in R$, real numbers $x_{1}, x_{2}, \cdots, x_{n} \in[a, b]$ and satisfy $x_{1}+x_{2}+\cdots+x_{n}=s=$ constant $(n a \leq s \leq n b)$, consider the following expression
$$F=f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{n}\right)$$ | Notice that this lemma follows directly from the case $n=2$. In fact, it is only necessary to prove: if $x, y \in [a, b]$ and $2a \leq x+y \leq 2b$, then
$$f(x)+f(y) \leq\left\{\begin{array}{l}
f(a)+f(s-a), s \leq a+b \\
f(b)+f(s-b), s \geq a+b
\end{array}\right.$$
Indeed, assume $s \leq a+b$, then $s-a \leq b$. Since... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,733 |
Example 4.2.6 Let $a_{1}, a_{2}, \cdots, a_{n}>0, a_{1}, a_{2}, \cdots, a_{n} \in[0,2]$, and satisfy $a_{1}+a_{2}+\cdots+a_{n}=n$, find the maximum value of the following expression.
$$S=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}$$ | Solution: Applying the lemma above to the convex function $f(x)=x^{2}$, we get that $S$ reaches its maximum value if and only if $k$ numbers are equal to $2$, and $n-k-1$ numbers are equal to $0$. In this case, we have $S=4 k+(n-2 k)^{2}$. Since $a_{1}, a_{2}, \cdots, a_{n} \in[0,2]$, we must have $0 \leq n-2 k \leq 2$... | 2n+1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,734 |
Example 4.2.7 Let $a, b, c \in [0,2]$, and satisfy $a+b+c=5$, prove: $a^{2}+b^{2}+c^{2} \leq 9$ | Proof: Assuming $a \leq b \leq c$, according to Lemma 5, we infer that $a^{2}+b^{2}+c^{2}$ reaches its maximum value if and only if $a=0$ or $b=c=2$. In the first case, $a=0$ is impossible, because $4 \geq b+c=5$, leading to a contradiction. In the second case, we have $a=1$, thus $\max \left\{a^{2}+b^{2}+c^{2}\right\}... | 9 | Inequalities | proof | Yes | Yes | inequalities | false | 731,735 |
Example 1.1 Given $x, y>0, x^{2}+y^{3} \geqslant x^{3}+y^{4}$, prove that $x^{2}+y^{2} \leqslant 2$ | Prove that because $x^{2}-x^{3} \geqslant y^{4}-y^{3} \geqslant y^{3}-y^{2}(\mathrm{AM}-\mathrm{GM}$ inequality), so
$$x^{2}+y^{2} \geqslant x^{3}+y^{3}$$
and because
$$x^{2}+y^{2} \leqslant \frac{2 x^{3}+1}{3}+\frac{2 y^{3}+1}{3}(\mathrm{AM}-\mathrm{GM} \text { inequality })$$
combining the above two inequalities, w... | x^{2}+y^{2} \leqslant 2 | Inequalities | proof | Yes | Yes | inequalities | false | 731,736 |
Example 1.2 Positive numbers $a, b, c$ satisfy $a+b+c=1$, prove
$$\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b} \geqslant 2$$ | $$\begin{array}{l}
\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}+1= \\
\left(\frac{a^{2}}{b+c}+a\right)+\left(\frac{b^{2}}{c+a}+b\right)+\left(\frac{c^{2}}{a+b}+c\right)= \\
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}
\end{array}$$
Therefore, the original inequality is equivalent to
$$\frac{a+b}{b+c}+\frac{b+c}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,737 |
Example 1.11 (Han Jingjun) Real numbers $a_{i}, b_{i}(i=1,2, \cdots, n)$ satisfy $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}=1$, prove that
$$\sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}+\sum_{1 \leqslant i<j \leqslant n} b_{i} b_{j}+\sum_{i=1}^{n} a_{i} b_{i} \leqslant 1$$ | $$\begin{aligned}
\text { LHS } \leqslant & \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}+\sum_{1 \leqslant i<j<n} b_{i} b_{j}+\frac{1}{2} \sum_{i=1}^{n}\left(a_{i}^{2}+b_{i}^{2}\right)= \\
& \frac{1}{2}\left[\left(\sum_{i=1}^{n} a_{i}\right)^{2}+\left(\sum_{i=1}^{n} b_{i}\right)^{2}\right]=1
\end{aligned}$$ | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 731,738 |
Example 4.17 (Fujian Middle School Mathematics, Chen Ji) $p, q, r \geqslant 0, a, b, c \in \mathbf{R}$, prove that
$$[(q+r) a+(r+p) b+(p+q) c]^{2} \geqslant 4(p+q+r)(p b c+q c a+r a b)$$ | Proof Without loss of generality, let $a=\max \{a, b, c\}$, then
$$\begin{array}{l}
{[(q+r) a+(r+p) b+(p+q) c]^{2}-4(p+q+r)(p b c+q c a+r a b)=} \\
{[(q-r) a+(r+p) b-(p+q) c]^{2}+4 q r(a-b)(a-c) \geqslant 0}
\end{array}$$
Note This problem has a geometric background.
Let $P$ be an interior point of $\triangle A B C$, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,739 |
Example $4.18$ a, b, c $\in \mathbf{R}$, prove that
$$3\left(a^{4}+b^{4}+c^{4}\right)+4 a^{3} b+4 b^{3} c+4 c^{3} a \geqslant 0$$ | Prove that there is an identity:
$$\begin{array}{l}
3\left(a^{4}+b^{4}+c^{4}\right)+4 a^{3} b+4 b^{3} c+4 c^{3} a= \\
\frac{1}{7} \sum_{y c}\left(4 a^{2}-2 b^{2}-c^{2}+2 a c+4 a b\right)^{2} \geqslant 0
\end{array}$$
Equality holds if and only if $a=b=c=0$.
Note: The identity in this problem, how did it come about? Us... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,740 |
Example 4.19 (Crux, Vasile Cirtoaje) $a, b, c \in \mathbf{R}$, prove that
$$\left(a^{2}+b^{2}+c^{2}\right)^{2} \geqslant 3\left(a^{3} b+b^{3} c+c^{3} a\right)$$ | $$\begin{array}{l}
4\left(\left(a^{2}+b^{2}+c^{2}\right)-(b c+c a+a b)\right)\left(\left(a^{2}+b^{2}+c^{2}\right)^{2}-3\left(a^{3} b+b^{3} c+c^{3} a\right)\right)= \\
\left(\left(a^{3}+b^{3}+c^{3}\right)-5\left(a^{2} b+b^{2} c+c^{2} a\right)+4\left(b^{2} a+c^{2} b+a^{2} c\right)\right)^{2}+ \\
3\left(\left(a^{3}+b^{3}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,741 |
Example $4.20$ Given $a, b, c > 0$, prove that
$$\sqrt{\frac{a^{4}+2 b^{2} c^{2}}{a^{2}+2 b c}}+\sqrt{\frac{b^{4}+2 c^{2} a^{2}}{b^{2}+2 c a}}+\sqrt{\frac{c^{4}+2 a^{2} b^{2}}{c^{2}+2 a b}} \geqslant a+b+c$$ | Prove using the Cauchy inequality:
$$\left(a^{4}+b^{2} c^{2}+b^{2} c^{2}\right)(a+b+c)(a+c+b) \geqslant\left(a^{2}+2 b c\right)^{3}$$
Similarly, there are other two inequalities, thus
$$\sum \sqrt{\frac{a^{4}+2 b^{2} c^{2}}{a^{2}+2 b c}} \geqslant \sum \frac{a^{2}+2 b c}{a+b+c}=a+b+c$$
Adding them up proves the origi... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,742 |
Example $4.22 a, b, c>0$, prove that
$$\sum \sqrt{\frac{a(b+c)}{a^{2}+b c}} \geqslant 2$$ | Let's set \( A = a(b+c), B = a^2 + bc \), then
\[
\sum \sqrt{\frac{a(b+c)}{a^2 + bc}} = \sum \frac{a(b+c)}{\sqrt{(ab + ac)(a^2 + bc)}} \geqslant \sum \frac{2a(b+c)}{(a+b)(a+c)} = 2
\]
The desired inequality is proved. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,744 |
Example $4.23 a, b, c \geqslant 0$ and satisfy $a+b+c=1$, prove that
$$\sqrt{a+b^{2}}+\sqrt{b+c^{2}}+\sqrt{c+a^{2}} \geqslant 2$$ | Prove (Vo Quoc Ba Can) by subtracting $\sum a$ from both sides, then $\Leftrightarrow$
$$\sum\left(\sqrt{a+b^{2}}-b\right) \geqslant 1 \Leftrightarrow \sum \frac{a}{b+\sqrt{a+b^{2}}} \geqslant 1$$
By AM - GM inequality, we have
$$\begin{aligned}
\frac{a}{b+\sqrt{a+b^{2}}}= & \frac{a(a+b)}{b(a+b)+(a+b) \sqrt{a+b^{2}}} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,745 |
Example $4.24 a, b, c>0, a+b+c \geqslant 3$, prove that
$$\frac{1}{a^{2}+b+c}+\frac{1}{b^{2}+c+a}+\frac{1}{c^{2}+a+b} \leqslant 1$$ | Notice that $1+b+c \approx a^{2}+b+c$, we can consider multiplying both the numerator and the denominator by $1+b+c$, and then perform the bounding
$$\sum \frac{1+b+c}{\left(a^{2}+b+c\right)(1+b+c)} \leqslant \sum \frac{1+b+c}{(a+b+c)^{2}}=\frac{3+2 \sum a}{\left(\sum a\right)^{2}} \leqslant 1$$
Proved. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,746 |
Example $4.25 a, b, c \geqslant 0$, and at most one of them is 0. Prove that
$$\sum \frac{\sqrt{a b+4 b c+4 a c}}{a+b} \geqslant \frac{9}{2}$$ | To prove this problem, which is quite challenging, direct bounding is not effective. Therefore, we consider squaring both sides of the inequality. The inequality is then equivalent to:
$$\sum \frac{b c+4 a c+4 a b}{(b+c)^{2}}+2 \sum \frac{\sqrt{(4 a b+c(4 a+b))(4 a b+c(a+4 b))}}{(a+c)(b+c)} \geqslant \frac{81}{4}$$
Us... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,747 |
Example $4.26 a, b, c \geqslant 0, a+b+c=3$, prove that
$$\sqrt{3-a b}+\sqrt{3-b c}+\sqrt{3-c a} \geqslant 3 \sqrt{2}$$ | Proof (Liu Yuzhan) From $3-a b \approx 3-\frac{(a+b)^{2}}{4}, \frac{(3+c)^{2}}{8} \approx 3-\frac{(a+b)^{2}}{4}$, and $\sum \sqrt{\frac{(3+c)^{2}}{8}}=3 \sqrt{2}$, we think of proving
$$\begin{array}{l}
{\left[\sqrt{3-a b}-\sqrt{3-\frac{(a+b)^{2}}{4}}\right] \geqslant \sum\left[\sqrt{\frac{(3+c)^{2}}{8}}-\sqrt{3-\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,748 |
For example, $4.27$ a, b, c$ are non-negative real numbers, satisfying $a+b+c=1$, prove that
$$\sum a \sqrt{8 b^{2}+c^{2}} \leqslant 1$$ | Prove that noting $\sqrt{8 b^{2}+c^{2}} \approx \frac{8 b}{3}+\frac{c}{3}$, then
$$\sqrt{8 b^{2}+c^{2}}-\left(\frac{8 b}{3}+\frac{c}{3}\right)=\frac{\frac{8}{9}(b-c)^{2}}{\sqrt{8 b^{2}+c^{2}}+\frac{8 b}{3}+\frac{c}{3}} \leqslant \frac{2(b-c)^{2}}{3(2 b+c)}$$
i.e. $\square$
$$3 b+c-\frac{3 b c}{2 b+c} \geqslant \sqrt{8... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,750 |
Example $4.28 a, b, c \geqslant 0$, prove that
$$3 \sum a \geqslant 2 \sum \sqrt{a^{2}+b c}$$ | Proof Note that the equality condition of the inequality is $\backsim(1,1,0)$. To maintain the non-negativity of the original square roots after differentiation, consider subtracting $2 a+2 b+2 c$ from both sides, then $\Leftrightarrow$
$$\sum \frac{b c}{\sqrt{a^{2}+b c}+a} \leqslant \frac{\sum a}{2}$$
We examine $\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,751 |
Example 4.29 (Sun Shibao) Let $x, y, z \geqslant 0$, prove
$$1 \leqslant \sum \frac{x^{2}}{\sqrt{\left(x^{2}+y^{2}+x y\right)\left(x^{2}+z^{2}+z x\right)}} \leqslant \frac{2 \sqrt{3}}{3}$$ | To prove the right side of the inequality, by the Cauchy inequality, we have
$$\begin{array}{l}
\left(x^{2}+x y+y^{2}\right)\left(x^{2}+x z+z^{2}\right)= \\
{\left[\left(y+\frac{x}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2} x\right)^{2}\right]\left[\left(z+\frac{x}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2} x\right)^{2}\right]... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,752 |
Example 5.1 (Guha Inequality) If $p>0, a, b, c, \alpha, \beta, \gamma \geqslant 0$, and $a^{\frac{1}{p}}+$ $c^{\frac{1}{p}} \leqslant b^{\frac{1}{p}}, \alpha^{\frac{1}{p+1}}+\gamma^{\frac{1}{p+1}} \geqslant \beta^{\frac{1}{p+1}}$, then $\alpha b c-\beta a c+\gamma a b \geqslant 0$ | Prove that from the generalization of the Cauchy inequality, we have
$$\left[a^{\frac{1}{p+1}}(\alpha c)^{\frac{1}{p+1}}+c^{\frac{1}{p+1}}(\gamma a)^{\frac{1}{1+p}}\right]^{p+1} \leqslant(\alpha c+\gamma a)\left(a^{\frac{1}{p}}+c^{\frac{1}{p}}\right)^{p}$$
That is,
$$a c\left(\alpha^{\frac{1}{p+1}}+\gamma^{\frac{1}{1+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,753 |
Example $5.2 a, b, c>0$, satisfying $a+b+c=3$, prove that
$$\sum_{c y c} \frac{a}{3 a^{2}+a b c+27} \leqslant \frac{3}{31}$$ | Prove that from the 3rd Schur inequality, we have
$$3 a b c \geqslant 4(a b+b c+c a)-9$$
Let \( s = \frac{4(a b + b c + c a) + 72}{9} \), we only need to prove
$$\begin{array}{l}
\sum \frac{3 a}{9 a^{2} + 4(a b + b c + c a) + 72} \leqslant \frac{3}{31} \Leftrightarrow \\
\sum\left(1 - \frac{31 a(a + b + c)}{9 a^{2} + ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,754 |
Example 5.3 (2008 Turkey) For positive real numbers $a, b, c$, if $a+b+c=1$, prove that
$$\frac{a^{2} b^{2}}{c^{3}\left(a^{2}-a b+b^{2}\right)}+\frac{b^{2} c^{2}}{a^{3}\left(b^{2}-b c+c^{2}\right)}+\frac{c^{2} a^{2}}{b^{3}\left(c^{2}-c a+a^{2}\right)} \geqslant \frac{3}{a b+b c+c a}$$ | Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$, we only need to prove
$$\sum \frac{z^{3}}{x^{2}-x y+y^{2}} \geqslant \frac{3(x y+y z+z x)}{x+y+z}$$
In fact, we have a stronger inequality:
$$\sum \frac{z^{3}}{x^{2}-x y+y^{2}} \geqslant x+y+z \geqslant \frac{3(x y+y z+z x)}{x+y+z}$$
We will directly prove the strong... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,755 |
Example 5.5 (2006 IMO Shortlist) $a, b, c$ are the lengths of the sides of a triangle, prove that
$$\sum \frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}} \leqslant 3$$ | Given that $a, b, c$ are the lengths of the sides of a triangle, there exist $x, y, z > 0$ such that $\sqrt{b} + \sqrt{c} - \sqrt{a} = x$, $\sqrt{c} + \sqrt{a} - \sqrt{b} = y$, and $\sqrt{a} + \sqrt{b} - \sqrt{c} = z$. The original inequality is equivalent to
$$\begin{array}{l}
\sum \frac{\sqrt{x^{2} + x y + x z - y z}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,757 |
Example 5.6 (Han Jingjun) Prove: For non-negative real numbers $a, b, c$, we have
$$a \sqrt{a^{2}+b c}+b \sqrt{b^{2}+c a}+c \sqrt{c^{2}+a b} \geqslant \sqrt{2}(a b+b c+c a)$$ | Proof (Zheng Fan (Gold Medalist of the 50th IMO)) from the 4th Schur inequality we have
$$\begin{array}{l}
a \sqrt{a^{2}+b c}+b \sqrt{b^{2}+c a}+c \sqrt{c^{2}+a b} \geqslant \\
\frac{1}{\sqrt{2}}(a(a+\sqrt{b c})+b(b+\sqrt{c a})+c(c+\sqrt{a b})) \geqslant \\
\frac{1}{\sqrt{2}}(\sqrt{a b}(a+b)+\sqrt{b c}(b+c)+\sqrt{c a}(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,758 |
Example 5.7 (1996 Iran Mathematical Olympiad) For all $x, y, z \geqslant 0$, with at most one of them being 0, prove that
$$\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}} \geqslant \frac{9}{4(x y+y z+z x)}$$ | Prove that
$$\begin{array}{l}
(x y+y z+z x)\left(\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}\right)= \\
(x y+y z+x z) \frac{(x+y)^{2}(y+z)^{2}+(y+z)^{2}(z+x)^{2}+(z+x)^{2}(x+y)^{2}}{(x+y)^{2}(y+z)^{2}(z+x)^{2}}
\end{array}$$
But we have
$$\begin{array}{l}
(x y+y z+z x)\left((x+y)^{2}(y+z)^{2}+(y+z)^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,759 |
Example 1.13 (Han Jingjun) $a, b, c>0, a b c=1$. Prove that
$$\sum \frac{a^{2}+b^{2}}{a^{2}+b^{2}+1} \geqslant \sum \frac{a+b}{a^{2}+b^{2}+1}$$ | Prove that the original inequality is equivalent to
$$3 \geqslant \sum \frac{a+b+1}{a^{2}+b^{2}+1} \Leftrightarrow 3 \geqslant \sum \frac{a^{3}+b^{3}+1}{a^{6}+b^{6}+1}$$
Notice that
$$\begin{array}{l}
\sum \frac{a^{3}+b^{3}+1}{a^{6}+b^{6}+1} \leqslant \sum \frac{3}{a^{3}+b^{3}+1}=\sum \frac{3}{a^{3}+b^{3}+a b c} \leqs... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,760 |
Example $5.8$ Given $a, b, c \geqslant 0, a^{3}+b^{3}+c^{3}=3$, prove that
$$a^{4} b^{4}+b^{4} c^{4}+c^{4} a^{4} \leqslant 3$$ | Prove that if we directly homogenize the degree, it would be too high and difficult to handle. Consider using the conditions to gradually "increase the degree".
$$\begin{aligned}
9 \sum b^{4} c^{4}= & 9 \sum b^{3} c^{3} \cdot b c \leqslant 3 \sum b^{3} c^{3}\left(b^{3}+c^{3}+1\right)= \\
& 3 \sum b^{3} c^{3}\left(b^{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,761 |
For example, $5.9 x, y, z>0$, prove
$$\frac{y^{3}}{x^{3}}+\frac{z^{3}}{y^{3}}+\frac{x^{3}}{z^{3}} \geqslant \sum \sqrt{\frac{y\left(z^{2}+x y\right)}{2 x^{2} z}}$$ | Prove (Han Jingjun) Let $\frac{y}{x}=a, \frac{z}{y}=b, \frac{x}{z}=c$, then $a b c=1$, so the inequality is equivalent to
$$\sum a^{3} \geqslant \sum a \sqrt{\frac{b+c}{2}}$$
Using the 3rd degree Schur's inequality, we have
$$\sum a^{3} \geqslant \sum a^{2}(b+c)-3 a b c=\sum a^{2}(b+c)-3$$
On the other hand
$$\begin{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,762 |
Example $5.10$ a, b, c $\geqslant 0$, prove that
$$\sum a(a-b)(a-c)(a-2 b)(a-2 c) \geqslant 0$$ | Prove that since
$$\begin{array}{c}
(a-2 b)(a-2 c)=(a-b-c)^{2}-(b-c)^{2} \\
\sum a(b-c)^{2}(a-b)(a-c)=0
\end{array}$$
the original inequality is equivalent to
$$\sum a(b+c-a)^{2}(a-b)(a-c) \geqslant 0$$
Assume without loss of generality that $a \geqslant b \geqslant c \geqslant 0$, then we have
$$\begin{array}{l}
a(b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,763 |
Example 5.11 Given that $a, b, c$ are positive numbers, prove that
$$\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}+\frac{3(a b+b c+c a)}{(a+b+c)^{2}} \geqslant 4$$ | Without loss of generality, let $c=\min (a, b, c)$, we have
$$\begin{array}{l}
\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}-3= \frac{1}{(a+c)(b+c)}(a-b)^{2}+ \\
\frac{1}{(a+b)(b+c)}(a-c)(b-c) \\
\frac{3(a b+b c+c a)}{(a+b+c)^{2}}-1=-\frac{1}{(a+b+c)^{2}}(a-b)^{2}-\frac{1}{(a+b+c)^{2}}(a-c)(b-c)
\end{array}$$
There... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,764 |
Theorem 5.1 $a, b, c, x, y, z$ are non-negative real numbers and satisfy $a \geqslant b \geqslant c, a x \geqslant b y$ or $c z \geqslant b y$, then
$$x(a-b)(a-c)+y(b-c)(b-a)+z(c-a)(c-b) \geqslant 0$$ | First, let's consider the case $a x \geqslant b y$.
It is easy to prove that $a x - b y \geqslant c(x - y)$. In fact, if $x - y \leqslant 0$, then by $a x - b y \geqslant 0$ and $c(x - y) \leqslant 0$, we can get $a x - b y \geqslant c(x - y)$. If $x - y \geqslant 0$, we have $b(x - y) \geqslant c(x - y)$, so
$$a x - b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,765 |
Example 5.13 Let $a, b, c>1$ and satisfy $a+b+c=9$. Prove
$$\sqrt{a b+b c+c a} \leqslant \sqrt{a}+\sqrt{b}+\sqrt{c}$$ | $$\begin{array}{l}
\sqrt{a}=x, \sqrt{b}=y, \sqrt{c}=z \text {, we have } \\
x^{2}+y^{2}+z^{2}=9, 9 \min \left\{x^{2}, y^{2}, z^{2}\right\} \geqslant x^{2}+y^{2}+z^{2}
\end{array}$$
Then the original inequality is equivalent to
$$\begin{array}{l}
x+y+z \geqslant \sqrt{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}} \Leftrightarro... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,767 |
Example 5.14 For positive real numbers $a, b, c$, prove
(1) $\frac{a^{2}+2 b c}{(b+c)^{2}}+\frac{b^{2}+2 a c}{(c+a)^{2}}+\frac{c^{2}+2 a b}{(a+b)^{2}} \geqslant \frac{9}{4}$.
(2) $\frac{a^{2}+b c}{b^{2}+b c+c^{2}}+\frac{b^{2}+a c}{c^{2}+c a+a^{2}}+\frac{c^{2}+a b}{a^{2}+a b+b^{2}} \geqslant 2$. | To prove (1), by Theorem 5.1, we can obtain
$$\frac{(a-b)(a-c)}{(b+c)^{2}}+\frac{(b-c)(b-a)}{(c+a)^{2}}+\frac{(c-a)(c-b)}{(a+b)^{2}} \geqslant 0$$
This is because, by the symmetry of the inequality, we can assume $a \geqslant b \geqslant c$, and in this case, it is clear that $\frac{1}{(b+c)^{2}} \geqslant \frac{1}{(c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,768 |
Example 5.15 (Tiks) $a, b, c>0$, prove
$$\frac{a^{2}}{(2 a+b)(2 a+c)}+\frac{b^{2}}{(2 b+c)(2 b+a)}+\frac{c^{2}}{(2 c+a)(2 c+b)} \leqslant \frac{1}{3}$$ | To prove the desired inequality, we can write it as:
$$\begin{array}{l}
\frac{1}{3}-\sum \frac{a^{2}}{(2 a+b)(2 a+c)}= \\
\sum\left(\frac{a}{3(a+b+c)}-\frac{a^{2}}{(2 a+b)(2 a+c)}\right)= \\
\sum \frac{1}{3(a+b+c)} \cdot \frac{a}{(2 a+b)(2 a+c)}(a-b)(a-c)= \\
\frac{1}{3(a+b+c)} \sum \frac{a}{(2 a+b)(2 a+c)}(a-b)(a-c)
\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,769 |
Example 1.14 For non-negative real numbers $a, b, c$ satisfying $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=3$, prove
$$\sqrt{\frac{a+b c^{2}}{2}}+\sqrt{\frac{b+c a^{2}}{2}}+\sqrt{\frac{c+a b^{2}}{2}} \leqslant \frac{3}{a b c}$$ | $$\begin{aligned}
\sum \sqrt{\frac{a+b c^{2}}{2}}= & \sum \sqrt{\frac{a b+b^{2} c^{2}}{2 b}} \leqslant \frac{1}{2} \sum\left(\frac{1}{b}+\frac{a b+b^{2} c^{2}}{2}\right)= \\
& \frac{1}{4}\left(\sum \frac{2}{a}+\sum a b+\sum a^{2} b^{2}\right) \leqslant \frac{3}{a b c}
\end{aligned}$$
Prove that using the AM-GM inequal... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,771 |
Note (8) is equivalent to: $a, b, c$ are the three sides of a triangle, $x, y, z \in \mathbf{R}$, prove that
$$a^{2}(x-y)(x-z)+b^{2}(y-x)(y-z)+c^{2}(z-y)(z-x) \geqslant 0$$ | Prove that let $t_{1}=x-y, t_{2}=y-z$, then
$$\begin{array}{l}
a^{2}(x-y)(x-z)+b^{2}(y-x)(y-z)+c^{2}(z-y)(z-x)= \\
a^{2} t_{1}\left(t_{1}+t_{2}\right)-b^{2} t_{1} t_{2}+c^{2} t_{2}\left(t_{1}+t_{2}\right)= \\
a^{2} t_{1}^{2}+\left(a^{2}-b^{2}+c^{2}\right) t_{1} t_{2}+c^{2} t_{2}^{2}= \\
t_{2}^{2}\left(a^{2}\left(\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,772 |
Example $5.16$ a, b, c are positive numbers, prove that
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3 \sqrt[3]{a b c}}{2(a+b+c)} \geqslant 2$$ | Proof: Let $a+b+c=1, ab+bc+ca=x$, using the Cauchy inequality, we need to prove
$$\frac{1}{2(ab+bc+ca)}+\frac{3}{2} \sqrt[3]{abc} \geqslant 2$$
In fact, we easily get $ab+bc+ca \leqslant \frac{1}{3}$.
Thus, if $ab+bc+ca \leqslant \frac{1}{4}$, the original inequality is obviously true.
And if $ab+bc+ca \geqslant \frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,773 |
Example 5.17 Given non-negative real numbers $a, b, c$ satisfy $a b + b c + c a + 6 a b c = 9$. Determine the maximum value of $k$ such that the following inequality always holds.
$$a + b + c + k a b c \geqslant k + 3$$ | Let $a=b=3, c=0$, then we have $k \leqslant 3$. Now, let's prove that
$$a+b+c+3 a b c \geqslant 6$$
Let $p=a+b+c, q=a b+b c+c a, r=a b c$, then from the problem, we have
$$9=q+6 r \geqslant 3 r^{\frac{2}{3}}+6 r \Rightarrow r \leqslant 1 \Rightarrow q=9-6 r \geqslant 3$$
And $p^{2} \geqslant 3 q$, so $p \geqslant 3$.... | 3 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,774 |
Example $5.18 a, b, c \geqslant 1, a+b+c=9$, prove
$$\sqrt{a b+b c+c a} \leqslant \sqrt{a}+\sqrt{b}+\sqrt{c}$$ | Proof (Han Jingjun) By making the substitution $a=x+1, b=y+1, c=z+1$, the condition becomes $x+y+z=6, x, y, z \geqslant 0$, and we need to prove
$$\sqrt{\sum(x+1)(y+1)} \leqslant \sum \sqrt{x+1}$$
Squaring both sides, this is equivalent to
$$\begin{array}{l}
\sum x y+2 \sum x+3 \leqslant \sum x+3+2 \sum \sqrt{(x+1)(y+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,775 |
Example 5.19 (Han Jingjun) $a, b, c \geqslant 0$, at most one of them is 0, prove that
$$\sum \frac{a}{2} \geqslant \sum \frac{b c}{\sqrt{(a+b)(a+c)}} \geqslant \frac{\sqrt{3 \sum a b}}{2}$$ | To prove the right inequality, by the Cauchy inequality we have
$$\sum \frac{b c}{\sqrt{(a+b)(a+c)}} \geqslant \frac{\left(\sum b c\right)^{2}}{\sum b c \sqrt{(a+b)(a+c)}}$$
Thus, it suffices to prove
$$\begin{array}{c}
\left(\sum b c\right)^{\frac{3}{2}} \geqslant \frac{\sqrt{3}}{2} \sum b c \sqrt{(a+b)(a+c)} \Leftri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,776 |
For example, $5.20 x, y, z>0$, prove
$$3 \sqrt[4]{\frac{x^{2}+y^{2}+z^{2}}{12}} \geqslant \frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \geqslant \sqrt[4]{\frac{27(y z+z x+x y)}{4}}$$ | To prove the right side of the inequality, using the generalized Cauchy inequality, we have:
$$\begin{array}{l}
\left(\sum \frac{x}{\sqrt{x+y}}\right)\left(\sum \frac{x}{\sqrt{x+y}}\right)\left(\sum x(x+y)\right) \geqslant (x+y+z)^{3} \Leftrightarrow \\
\sum \frac{x}{\sqrt{x+y}} \geqslant \sqrt{\frac{(x+y+z)^{3}}{x^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,777 |
Example $5.21 a, b, c>0$, no two are zero at the same time, prove
$$\frac{a b+a c-b c}{b^{2}+c^{2}}+\frac{b c+b a-c a}{c^{2}+a^{2}}+\frac{c a+c b-a b}{a^{2}+b^{2}} \geqslant \frac{3}{2}$$ | Proof: Let $a+b+c=p=1, ab+bc+ca=q, abc=r$, the original inequality is equivalent to
$$12 q^{3}-11 q^{2}+2 q+2 r(2 q+1)-9 r^{2} \geqslant 0$$
Let $f(r)=2 r(2 q+1)-9 r^{2}+12 q^{3}-11 q^{2}+2 q \Rightarrow f^{\prime}(r)=4 q+2-18 r \geqslant 0$.
(1) $q \leqslant \frac{1}{4}$, noting that $r \geqslant 0$, then $\Rightarro... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,778 |
Example $5.22 a, b, c \geqslant 0, a^{2}+b^{2}+c^{2}=1$, prove that
$$(2-3 a b)(2-3 b c)(2-3 c a) \geqslant 1$$ | Proof: Let $x=a+b+c, y=ab+bc+ca, z=abc$, then $a^{2}+b^{2}+c^{2}=x^{2}-2y=1$, so we only need to prove
$$\begin{array}{l}
(2-3ab)(2-3bc)(2-3ca) \geq 1 \Leftrightarrow \\
7-12y+3 \cdot 3z(2x-3z) \geq 0
\end{array}$$
Let $f(t)=t(u-t)\left(t \leq \frac{u}{2}\right)$, where $t=3z, u=2x$, then
$$f'(t)=u-2t \geq 0$$
And al... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,779 |
Example $5.23 a, b, c \geqslant 0$, no two are zero at the same time, prove that
$$\sum \sqrt{\frac{a^{2}+a b+b^{2}}{c^{2}+a b}} \geqslant \frac{3 \sqrt{6}}{2}$$ | Prove that by Cauchy's generalization:
$$\left(\sum \sqrt{\frac{a^{2}+a b+b^{2}}{c^{2}+a b}}\right)^{2}\left(\sum \frac{(a+b)^{3}\left(c^{2}+a b\right)}{a^{2}+a b+b^{2}}\right) \geqslant 8\left(\sum a\right)^{3}$$
Thus, it suffices to prove:
$$\begin{array}{l}
\frac{16}{27}\left(\sum a\right)^{3} \geqslant \sum \frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,780 |
Example 5.24 (Vo Quoc Ba Can) $a, b, c>0, a+b+c=3$, prove that
$$\frac{a}{2 b+1}+\frac{b}{2 c+1}+\frac{c}{2 a+1} \leqslant \frac{1}{a b c}$$ | Prove that after expansion, it is equivalent to
$$7+4 a b c \sum a b+4 \sum a b \geqslant 2 \sum a^{2} b^{2}+13 a b c+4 a b c \sum_{\text {ec }} a^{2} c$$
From $\sum_{c \tau} a^{2} c+a b c \leqslant 4$ (proof can be found in the chapter on breaking symmetry and case analysis), we have
$$7+4 a b c \sum a b+4 \sum a b \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,781 |
Example 1.15 (Vasile) $a, b, c>0, \sum a \sum \frac{1}{a}=10$. Prove that
$$\frac{b^{2}+c^{2}}{a^{2}}+\frac{b^{2}+a^{2}}{c^{2}}+\frac{a^{2}+c^{2}}{b^{2}} \geqslant \frac{31}{3}$$ | Prove that the condition is equivalent to $\sum\left(\frac{a}{b}+\frac{b}{a}\right)=7$, by Cauchy inequality we have
$$\begin{array}{l}
\sum\left(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}\right)^{2} \geqslant \frac{1}{3}\left(\sum\left(\frac{a}{b}+\frac{b}{a}\right)\right)^{2}=\frac{49}{3} \Leftrightarrow \\
\frac{b^{2}+... | \frac{31}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 731,782 |
Example 5.25 Non-negative real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}=3$, prove
$$\frac{a}{a b+3}+\frac{b}{b c+3}+\frac{c}{c a+3} \leqslant \frac{3}{4}$$ | To prove that the complete expansion of the above expression is equivalent to proving
$$
4 a b c \sum a b+12 \sum_{\text {cyc }} a b^{2}+36 a b c+36 \sum a \leqslant 3 a^{2} b^{2} c^{2}+9 a b c \sum a+27 \sum a b+81
$$
Since $\sum_{\text{cyc}} a^{2} c+a b c \leqslant \frac{4}{27}\left(\sum a\right)^{3}$ (proof can be f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,783 |
Example 5.26 Let \(a, b, c\) be positive real numbers. Prove that
\[
\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{7(a b+b c+c a)}{a^{2}+b^{2}+c^{2}} \geqslant \frac{17}{2}
\] | Prove that we only need to consider the case $a \leqslant c \leqslant b$: This is because
$$\frac{a}{c}+\frac{b}{c}+\frac{c}{a}+\frac{(b-a)(c-a)(b-c)}{a b c}=\frac{a}{c}+\frac{c}{b}+\frac{b}{a}$$
Let $a+b+c=p, a b+b c+c a=q, a b c=r$, then
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{p q-3 r-\sqrt{-4 p^{3} r-4 q^{3}+p^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,784 |
Example 5.27 Let $a, b, c$ be non-negative real numbers, and not two of them are zero at the same time, prove that
$$\frac{a^{3}}{a^{2}+b^{2}}+\frac{b^{3}}{b^{2}+c^{2}}+\frac{c^{3}}{c^{2}+a^{2}} \geqslant \frac{\sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}}{2}$$ | Prove that the inequality can be written in the following form:
$$\begin{array}{l}
\left(\sum \frac{a^{3}+b^{3}}{a^{2}+b^{2}}-\frac{a+b}{2}\right) \geqslant \sqrt{3 \sum a^{2}}-\sum a+\sum \frac{b^{3}-a^{3}}{a^{2}+b^{2}} \Leftrightarrow \\
\sum \frac{(a-b)^{2}(a+b)}{2\left(a^{2}+b^{2}\right)} \geqslant \sum \frac{(a-b)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,785 |
Example $6.1 a, b, c, d \in \mathbf{R}$, prove that
$$\sqrt{\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}} \geqslant \sqrt[3]{\frac{a b c+b c d+b d a+c d a}{4}}$$ | Proof: Let $x=a^{2}+d^{2}, y=b^{2}+c^{2}$.
By the AM - GM inequality, we have
$$\begin{array}{l}
\sqrt[3]{\frac{a b c+b c d+b d a+c d a}{4}}=\sqrt[3]{\frac{(a+d) b c+(b+c) a d}{4}} \leqslant \\
\sqrt[3]{\frac{\sqrt{2\left(a^{2}+d^{2}\right)} \frac{b^{2}+c^{2}}{2}+\sqrt{2\left(b^{2}+c^{2}\right)} \frac{a^{2}+d^{2}}{2}}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,786 |
Example 6.2 Given that $a, b, c$ are positive numbers, prove that
$$\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right) \geqslant(a b+b c+c a)^{3}$$ | Proof: Let $k=\sqrt[3]{\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)}$, then the conclusion is transformed to $k \geqslant a b+b c+c a$. By the AM - GM inequality,
$$\begin{array}{l}
\frac{a^{2}}{a^{2}+a b+b^{2}}+\frac{c^{2}}{b^{2}+b c+c^{2}}+\frac{c a}{c^{2}+c a+a^{2}} \geqslant ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,787 |
Example $6.4 a, b, c \geqslant 0$, prove that
$$\left(a^{2}+b^{2}+c^{2}\right)^{2} \geqslant 4(a-b)(b-c)(c-a)(a+b+c)$$ | Assume without loss of generality that $a \geqslant b, a \geqslant c$.
If $b \geqslant c$, then the inequality is obviously true.
If $b \leqslant c$, then by the $\mathrm{AM}-\mathrm{GM}$ inequality, we have
$$\begin{array}{l}
(a-b)(b-c)(c-a)(a+b+c)= \\
(a-b)(c-b)(a-c)(a+b+c)= \\
\left(a^{2}-b^{2}+a c-b c\right)(c-b)(a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,789 |
Example 6.5 (2008 IMO Shortlist) Let $a, b, c, d$ be positive real numbers, satisfying $a b c d=1$ and
$$a+b+c+d>\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$$
Prove that
$$a+b+c+d<\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}$$ | Proof: By Cauchy's inequality, we know
$$\begin{array}{l}
(a+b+c+d)(a b+b c+c d+d a)> \\
\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)(a b+b c+c d+d a) \geqslant(a+b+c+d)^{2} \Rightarrow \\
a b+b c+c d+d a>a+b+c+d
\end{array}$$
It suffices to prove
$$\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d} \geqs... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,790 |
Example $6.6 a^{2}+b^{2}+c^{2}=1, a, b, c \in \mathbf{R}^{+}$, prove:
$$\sum \frac{1}{1-\left(\frac{a+b}{2}\right)^{2}} \leqslant \frac{9}{2}$$ | Proof (Li Chao, 2009 Chinese National Training Team Member) For conditions that appear twice, a natural idea is to align the numerator and denominator of the left side of the inequality in both directions. By the AM-GM inequality, we have
$$\begin{array}{l}
\sum \frac{1}{1-\left(\frac{a+b}{2}\right)^{2}}-3=\sum \frac{\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,791 |
Example 1.16 (Han Jingjun) $x, y \in \mathbf{R}$, and $x^{2}+y^{2}=1$, prove that
$$\sqrt{1-x}+\sqrt{1-\frac{1}{2} x-\frac{\sqrt{3}}{2} y} \geqslant \frac{\sqrt{2}}{2}$$ | Prove that after completing the square inside the square root, it is equivalent to
$$\frac{\sqrt{2}}{2}\left[\sqrt{(1-x)^{2}+y^{2}}+\sqrt{\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{\sqrt{3}}{2}\right)^{2}}\right] \geqslant \frac{\sqrt{2}}{2}$$
By Minkowski's inequality, we have
$$\sqrt{(1-x)^{2}+y^{2}}+\sqrt{\left(x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,793 |
For example, $6.8 x_{i} \geqslant 0, i=1,2, \cdots, n, n \geqslant 2$, find the smallest $C$ such that
$$C\left(\sum_{i=1}^{n} x_{i}\right)^{4} \geqslant \sum_{1<i<j<n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right)$$ | On one hand, taking $x_{1}=x_{2}=1, x_{3}=x_{4}=\cdots=x_{n}$, we get $C \geqslant \frac{1}{8}$.
On the other hand, by the AM - GM inequality, we have
$$\begin{aligned}
\left(\sum_{i=1}^{n} x_{i}\right)^{4}= & \left(\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\right)^{2} \geqslant 4\left(\s... | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,794 |
Example $6.10 a, b, c \geqslant 0$, prove that
$$\sqrt[3]{\frac{(a+2 b)(b+2 c)(c+2 a)}{27}} \geqslant \sqrt{\frac{a b+a c+b c}{3}}$$ | Prove that cubing both sides of the inequality is equivalent to
$$(a+2 b)(b+2 c)(c+2 a) \geqslant(a b+b c+c a) \sqrt{3(a b+b c+c a)}$$
By the AM - GM inequality, we have
$$\sqrt{3(a b+b c+c a)} \leqslant \frac{1}{2}\left(a+b+c+\frac{3(a b+b c+c a)}{a+b+c}\right)$$
Thus, it suffices to prove
$$\begin{array}{l}
2(a+2 b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,795 |
Example 6.11 (Turkevici) Let $a, b, c, d$ be non-negative real numbers. Prove that
$$a^{4}+b^{4}+c^{4}+d^{4}+2 a b c d \geqslant a^{2} b^{2}+a^{2} c^{2}+a^{2} d^{2}+b^{2} c^{2}+b^{2} d^{2}+c^{2} d^{2}$$ | Assume without loss of generality that $a \geqslant b \geqslant c \geqslant d$, the original inequality is equivalent to
$$\left(a^{2}-c^{2}\right)^{2}+\left(b^{2}-c^{2}\right)^{2}+\left(a^{2}-d^{2}\right)^{2}+\left(b^{2}-d^{2}\right)^{2} \geqslant 2(a b-c d)^{2}$$
By the AM - GM inequality and Cauchy inequality, we h... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,796 |
Example 6.13 Given that $m$ is a positive even number, and real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $\sum_{i=1}^{n} x_{i}^{m}=1$, find $\max \min _{1 \leqslant i<j \leqslant n}\left|x_{i}-x_{j}\right|$. | Proof (Sun Shibao) First, we prove a lemma:
Lemma If $m$ is a positive even number, then for any real numbers $a, b$, we have
$$\frac{a^{m}+b^{m}}{2} \geqslant\left(\frac{a-b}{2}\right)^{m}$$
Proof of the lemma: Let $x, y \geqslant 0$, by the power mean inequality, we have $\frac{x^{m}+y^{m}}{2} \geqslant\left(\frac{x... | \frac{2}{\sqrt[m]{\sum_{i=1}^{n}(n+1-2 i)^{m}}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,798 |
Example 6.14 (Carleman Inequality) \(a_{i} \geqslant 0, i=1,2, \cdots, n\), prove that
\[
\sum_{k=1}^{n} \sqrt[k]{\prod_{i=1}^{k} a_{i}} < e \sum_{i=1}^{n} a_{i}
\] | The $e$ on the right side of the inequality easily evokes $\lim _{k \rightarrow+\infty}\left(1+\frac{1}{k}\right)^{k}=e$. Furthermore, $\left(1+\frac{1}{k}\right)^{k}<e$.
We first attempt to use the $\mathrm{AM}-\mathrm{GM}$ inequality to eliminate the square root.
Let $a_{1} b_{1}=a_{2} b_{2}=\cdots=a_{n} b_{n}$, whe... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,799 |
Example 6. $15 x \in R$, find the maximum value of the following function
$$f(x)=|\sin x(a+\cos x)|$$ | By Cauchy inequality and AM-GM inequality, we have
$$\begin{aligned}
f^{2}(x)= & \frac{1}{\lambda^{2}} \sin ^{2} x(a \lambda+\lambda \cos x)^{2} \leqslant \\
& \frac{1}{\lambda^{2}} \sin ^{2} x\left(\lambda^{2}+\cos ^{2} x\right)\left(a^{2}+\lambda^{2}\right) \leqslant \\
& \frac{1}{\lambda^{2}}\left(\frac{\sin ^{2} x+... | \frac{\sqrt{a^{4}+8 a^{2}}-a^{2}+4}{8} \sqrt{\frac{\sqrt{a^{4}+8 a^{2}}+a^{2}+2}{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,800 |
Example 6. Given $16 a, b, c \geqslant 0$, satisfying $a+b+c=3$, prove that
$$\sqrt{a+b c}+\sqrt{b+c a}+\sqrt{c+a b} \geqslant \sqrt{2}(a b+b c+c a)$$ | Prove that by AM-GM inequality we have
$$\sum\left(\sqrt{a+b c}+\sqrt{a+b c}+\frac{(a+b c)^{2}}{2 \sqrt{2}}\right) \geqslant \sum \frac{3(a+b c)}{\sqrt{2}}$$
Therefore, it suffices to prove that
$$\sum\left(\frac{3(a+b c)}{\sqrt{2}}-\frac{(a+b c)^{2}}{2 \sqrt{2}}\right) \geqslant 2 \sqrt{2}(a b+b c+c a)$$
And we have... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,801 |
Example 6.17 (Vo Quoc Ba Can) $a, b, c$ are non-negative real numbers, and $abc=1$, prove that
$$\frac{a}{\sqrt[k]{8+b^{3}}}+\frac{b}{\sqrt[k]{8+c^{3}}}+\frac{c}{\sqrt[k]{8+a^{3}}} \geqslant \sqrt[k]{3^{k-2}}$$
for all $k \geqslant 1, k \in \mathbf{R}$. | Prove that by the weighted AM-GM inequality we have
$$a^{3}+8+9(k-1) \geqslant 9 k\left(\frac{8+a^{3}}{9}\right)^{\frac{1}{4}}$$
which is equivalent to
$$\frac{a^{3}+9 k-1}{9 k} \geqslant \sqrt[n]{\frac{8+a^{3}}{9}}$$
Similarly, we can get
$$\begin{array}{l}
\frac{b^{3}+9 k-1}{9 k} \geqslant\left(\frac{8+b^{3}}{9}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,803 |
Example 1.17 For non-negative real numbers $a, b, c$ satisfying $a^{2}+b^{2}+c^{2}=3$, prove
$$\frac{a^{3}}{a^{5}+a^{2} b+c^{3}}+\frac{b^{3}}{b^{5}+b^{2} c+a^{3}}+\frac{c^{3}}{c^{5}+c^{2} a+b^{3}} \leqslant 1$$ | Prove that
$$\frac{a^{3}}{a^{5}+a^{2} b+c^{3}}=\frac{a}{a^{3}+b+\frac{c^{3}}{a^{2}}}$$
By Cauchy's inequality
$$a^{3}+b+\frac{c^{3}}{a^{2}}=\frac{a^{4}}{a}+\frac{b^{4}}{b^{3}}+\frac{c^{4}}{c a^{2}} \geqslant \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a+b^{3}+c a^{2}}=\frac{9}{a+b^{3}+c a^{2}}$$
Therefore, we have
$$\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,804 |
Example $6.18$ Given $a, b, c \geqslant 0$, and $a+b+c=2$, prove that
$$\frac{b}{a^{2}+1}+\frac{c}{b^{2}+1}+\frac{a}{c^{2}+1} \geqslant \frac{18}{13}$$ | Prove that by the weighted AM-GM inequality we have
$$\begin{aligned}
36\left(a^{2}+1\right)= & 9 a^{2}+9 a^{2}+9 a^{2}+9 a^{2}+4+4+4+4+4+ \\
& 4+4+4+4 \geqslant 13 \sqrt[13]{9^{4} 4^{9} a^{8}}
\end{aligned}$$
We have
$$\frac{b}{a^{2}+1}=b-\frac{a^{2} b}{a^{2}+1} \geqslant b-\frac{36 a^{2} b}{13 \sqrt[13]{9^{4} 4^{9} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,805 |
Example 6. $20 x_{i}>0, i=1,2, \cdots, n$, where $x_{n+1}=x_{1}, x_{n+2}=x_{2}$. Prove that
$$\sum_{i=1}^{n} \frac{x_{i}}{x_{i+1}+x_{i+2}}>\frac{5 n}{12}$$ | Proof: Given $b + ab = a$, then $b = \frac{a}{1 + a}$ (where $b \neq 1$), then
$$\frac{x_{i}}{x_{i+1} + x_{i+2}} + a = \frac{x_{i} + b x_{i+1}}{x_{i+1} + x_{i+2}} + \frac{a(b x_{i+1} + x_{i+2})}{x_{i+1} + x_{i+2}}$$
Using the AM-GM inequality, we have
$$\begin{aligned}
\sum_{i=1}^{n} \frac{x_{i}}{x_{i+1} + x_{i+2}} = ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,807 |
Example 6.21 (1999 Ukrainian Mathematical Olympiad) $x, y, z \geqslant 0, x+y+z=1$, prove that $a x+b y+c z+2 \sqrt{(x y+y z+z x)(a b+b c+c a)} \leqslant a+b+c$ | $$\begin{array}{l}
a x+b y+c z+\sqrt{2(a b+b c+c a) \cdot 2(x y+y z+z x)} \leqslant \\
\sqrt{x^{2}+y^{2}+z^{2}} \sqrt{a^{2}+b^{2}+c^{2}}+\sqrt{2(a b+b c+c a) \cdot 2(x y+y z+z x)} \leqslant \\
\sqrt{x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x} \sqrt{a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a}= \\
a+b+c
\end{array}$$
Therefore, the o... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,808 |
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