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For example, $1 b_{1}=1$, for integers $n$ greater than 1, let $b_{n}=b_{n-1} \cdot n^{f(n)}, f(n)=2^{1-n}$. Prove: For all positive integers $n, b_{n}<3$.
| Prove that when $n=1$, $b_{1}=1<3$.
When $n=2$, $b_{2}=b_{1} \cdot 2^{2^{1-2}}=2^{\frac{1}{2}}<3$,
When $n=3$, $b_{3}=b_{2} \cdot 3^{2^{1-3}}=2^{\frac{1}{2}} \cdot 3^{2^{1-3}}=2^{\frac{1}{2}} \cdot 3^{\frac{1}{4}}<3$, when $n=4$, $b_{4}=b_{3} \cdot 4^{2^{1-4}}=2^{\frac{1}{2}} \cdot 3^{\frac{1}{4}} \cdot 4^{\frac{1}{8}}... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,040 |
Example 2 Positive sequences $\left\{x_{n}\right\},\left\{y_{n}\right\}$ satisfy the conditions: for all positive integers $n$, we have
$$x_{n+2}=x_{n}+x_{n+1}^{2}, y_{n+2}=y_{n}^{2}+y_{n+1},$$
and $x_{1}, x_{2}, y_{1}, y_{2}$ are all greater than 1. Prove: there exists a positive integer $n$, such that $x_{n}>y_{n}$. | Prove obviously, starting from the second term, the sequence is increasing:
$$x_{n+2}>x_{n+1}^{2}>x_{n+1}, y_{n+2}>y_{n+1}$$
and each sequence from the third term onwards, all terms are greater than 2. Similarly, when $n>3$, we have $x_{n}>3$, $y_{n}>3$.
Note that when $n>1$, $x_{n+2}>x_{n+1}^{2}>x_{n}^{4}$.
On the o... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,041 |
Let $3 a_{1}, a_{2}, \cdots, a_{2002}$ be non-negative integers, satisfying
$$a_{i}+a_{j} \leqslant a_{i+j} \leqslant a_{i}+a_{j}+1,1 \leqslant i, j \leqslant 2002, i+j \leqslant 2002 .$$
Prove: There exists a real number $x, a_{n}=[n x], n=1,2, \cdots, 2002$. | Let $I_{n}=\left(\frac{a_{n}}{n}, \frac{a_{n}+1}{n}\right), n=1,2, \cdots, 2002$. If there exists a real number $x \in \bigcap_{n=1}^{2002} I_{n}$, then the proposition is proved.
For this, let $L=\max _{1 \leqslant n \leqslant 2002} \frac{a_{n}}{n}, U=\min _{1 \leqslant n \leqslant 2002} \frac{a_{n}+1}{n}$. We want t... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,042 |
Example 5 Let two sequences of positive numbers $\left\{a_{n}\right\},\left\{b_{n}\right\}$ satisfy:
(1) $a_{0}=1 \geqslant a_{1}, a_{n}\left(b_{n-1}+b_{n+1}\right)=a_{n-1} b_{n-1}+a_{n+1} b_{n+1}, n \geqslant 1$;
(2) $\sum_{i=0}^{n} b_{i} \leqslant n^{\frac{3}{2}}, n \geqslant 1$.
Find the general term of $\left\{a_{... | From condition (1), we have $a_{n}-a_{n+1}=\frac{b_{n-1}}{b_{n+1}}\left(a_{n-1}-a_{n}\right)$, hence
$$a_{n}-a_{n+1}=\frac{b_{0} b_{1}}{b_{n} b_{n+1}}\left(a_{0}-a_{1}\right)$$
If $a_{1}=a_{0}=1$, then $a_{n}=1$. Below, we discuss $a_{1}a_{0}-a_{n}=b_{0} b_{1}\left(a_{0}-a_{1}\right) \sum_{k=0}^{n-1} \frac{1}{b_{k} b_... | a_{n}=1(n \geqslant 0) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,044 |
Example 1 Let $x, y, z \in \mathbf{R}^{+}$. Prove:
$$(x y+y z+z x)\left(\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}\right) \geqslant \frac{9}{4} .$$ | Prove that since
$$\begin{aligned}
& 4\left(\sum_{\text {cyc }} y z\right)\left(\sum_{\text {cyc }}(x+y)^{2}(z+x)^{2}\right)-9 \prod_{\text {cyc }}(y+z)^{2} \\
= & \sum_{\text {cyc }} y z(y-z)^{2}\left(4 y^{2}+7 y z+4 z^{2}\right) \\
& +\frac{x y z}{x+y+z} \sum_{\text {cyc }}(y-z)^{2}\left(2 y z+(y+z-x)^{2}\right) \geq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,045 |
Example 6 Let $b_{1}, b_{2}, \cdots, b_{n}$ be $n$ positive real numbers, and the system of equations
$$x_{k-1}-2 x_{k}+x_{k+1}+b_{k} x_{k}=0, k=1,2, \cdots, n$$
has a non-zero real solution $x_{1}, x_{2}, \cdots, x_{n}$. Here $x_{0}=x_{n+1}=0$. Prove: $b_{1}+b_{2}+\cdots+b_{n} \geqslant \frac{4}{n+1}$. | Given the conditions, let $a_{i}=-b_{i} x_{i}$, then the system of equations can be rewritten as
$$\begin{array}{c}
-2 x_{1}+x_{2}=a_{1}, \\
x_{1}-2 x_{2}+x_{3}=a_{2}, \\
x_{2}-2 x_{3}+x_{4}=a_{3}, \\
\cdots \\
x_{n-1}-2 x_{n}=a_{n},
\end{array}$$
Thus,
$$\begin{array}{l}
a_{1}+2 a_{2}+\cdots+k a_{k}=-(k+1) x_{k}+k x_... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,046 |
Example 7 Let the bounded sequence $\left\{a_{n}\right\}_{n \geqslant 1}$ satisfy
$$a_{n}<\sum_{k=n}^{2 n+2006} \frac{a_{k}}{k+1}+\frac{1}{2 n+2007}, n=1,2,3, \cdots$$
Prove:
$$a_{n}<\frac{1}{n}, n=1,2,3, \cdots$$ | Proof: Let $b_{n}=a_{n}-\frac{1}{n}$, then
$$b_{n}<\sum_{k=n}^{2 n+2006} \frac{b_{k}}{k+1}, n \geqslant 1$$
We will prove that $b_{n}<0$. Since $a_{n}$ is bounded, there exists a constant $M$ such that $b_{n}<M$. When $n$ is sufficiently large (for example, greater than $10^{6}$), we have
$$\begin{aligned}
b_{n} & <\s... | a_{n}<\frac{1}{n}, n=1,2,3, \cdots | Inequalities | proof | Yes | Yes | inequalities | false | 734,047 |
Example 8 Given a real number $a$ and a positive integer $n$. Prove:
(1) There exists a unique real number sequence $x_{0}, x_{1}, \cdots, x_{n}, x_{n+1}$, satisfying
$$\left\{\begin{array}{l}
x_{0}=x_{n+1}=0, \\
\frac{1}{2}\left(x_{i+1}+x_{i-1}\right)=x_{i}+x_{i}^{3}-a^{3}, i=1,2, \cdots, n ;
\end{array}\right.$$
(2) ... | Proof of (1) Existence: From $x_{i+1}=2 x_{i}+2 x_{i}^{3}-2 a^{3}-x_{i-1}, i=1,2, \cdots, n$, and $x_{0}=0$, we know that each $-x_{i}$ is a real-coefficient polynomial of $x_{1}$ of degree $3^{i-1}$, thus $x_{n+1}$ is a real-coefficient polynomial of $x_{1}$ of degree $3^{n}$. Since $3^{n}$ is an odd number, there exi... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,048 |
Example 9 Let $n$ be a positive integer, and $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}, c_{2}, c_{3}, \cdots, c_{2 n}$ be $4 n-1$ positive real numbers, such that $c_{i+j}^{2} \geqslant a_{i} b_{j}, 1 \leqslant i, j \leqslant n$. Let $m=\max _{2 \leqslant i \leqslant 2 n} c_{i}$, prove:
$$\begin{aligned... | Let $X=\max _{1 \leqslant i \leqslant n} a_{i}, Y=\max _{1 \leqslant i \leqslant n} b_{i}$, and replace $a_{i}, b_{i}, c_{i}$ with $a_{i}^{\prime}=\frac{a_{i}}{X}, b_{i}^{\prime}=\frac{b_{i}}{Y}, c_{i}^{\prime}=\frac{c_{i}}{\sqrt{X \bar{Y}}}$, respectively. Therefore, we can assume $X=Y=1$.
We will prove
$$\begin{arra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,049 |
Example 10 Given the sequence $\left\{a_{n}\right\}$ satisfies the conditions $a_{1}=\frac{21}{16}$,
$$2 a_{n}-3 a_{n-1}=\frac{3}{2^{n-1}}, n \geqslant 2 .$$
Let $m$ be a positive integer, $m \geqslant 2$. Prove: when $n \leqslant m$, we have
$$\left(a_{n}+\frac{3}{2^{n+3}}\right)^{\frac{1}{m}}\left(m-\left(\frac{2}{3... | Prove that from equation (6) we get $2^{n} a_{n}=3 \cdot 2^{n-1} a_{n-1}+\frac{3}{4}$, and let $b_{n}=2^{n} a_{n}, n=1,2, \cdots$
$$b_{n}=3 b_{n-1}+\frac{3}{4}, b_{n}+\frac{3}{8}=3\left(b_{n-1}+\frac{3}{8}\right)$$
Since $b_{1}=2 a_{1}=\frac{21}{8}$, we have $b_{n}+\frac{3}{8}=3^{n-1}\left(b_{1}+\frac{3}{8}\right)=3^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,050 |
Example 11 Let $a_{0}, a_{1}, a_{2}, \cdots$ be any infinite sequence of positive real numbers. Prove that the inequality $1 + a_{n} > \sqrt[n]{2} a_{n-1}$ holds for infinitely many positive integers $n$.
---
The translation maintains the original text's line breaks and format. | Prove that the assumption $1+a_{n}>\sqrt[n]{2} a_{n-1}$ holds only for a finite number of positive integers. Let the largest of these positive integers be $M$, then for any positive integer $n>M$, the above inequality does not hold, i.e., $1+a_{n} \leq \sqrt[n]{2} a_{n-1} (n>M)$, which means $a_{n} \leq \sqrt[n]{2} a_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,051 |
Example 12 Find the largest constant $M>0$, such that for any positive integer $n$, there exist positive real number sequences $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$, satisfying:
(1) $\sum_{k=1}^{n} b_{k}=1, 2 b_{k} \geqslant b_{k-1}+b_{k+1}, k=2,3, \cdots, n-1$;
(2) $a_{k}^{2} \leqslant 1+\sum... | Prove a lemma first: $\max _{1 \leqslant k \leqslant n} a_{k} < 2$ and $\sum_{k=1}^{n} b_{k}=1$, then for any $1 \leqslant m \leqslant n$, we have
$$
\sum_{k=1}^{m} b_{k} \geqslant \frac{m}{2}, \quad \sum_{k=m}^{n} b_{k} \geqslant \frac{n-m+1}{2}
$$
Since $b_{k}>0$, we get from the above
$$
b_{k}>\left\{\begin{array}{... | \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,052 |
Example 13 Let $0<x_{1} \leqslant \frac{x_{2}}{2} \leqslant \cdots \leqslant \frac{x_{n}}{n}, 0<y_{n} \leqslant y_{n-1} \leqslant \cdots \leqslant y_{1}$. Prove:
$$\left(\sum_{k=1}^{n} x_{k} y_{k}\right)^{2} \leqslant\left(\sum_{k=1}^{n} y_{k}\right)\left(\sum_{k=1}^{n}\left(x_{k}^{2}-\frac{1}{4} x_{k} x_{k-1}\right) y... | Prove for $n$ using mathematical induction.
When $n=1$, equation (12) is an identity.
Assume equation (12) holds for $n-1$, i.e.,
$$\left(\sum_{k=1}^{n-1} x_{k} y_{k}\right)^{2} \leqslant\left(\sum_{k=1}^{n-1} y_{k}\right)\left(\sum_{k=1}^{n-1}\left(x_{k}^{2}-\frac{1}{4} x_{k} x_{k-1}\right) y_{k}\right),$$
Therefore,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,053 |
Example 1 (Continuous Convex Function)
Proof: If $\varphi$ is continuous, then for any non-negative real numbers $q_{1}, q_{2}, \cdots, q_{n}, q_{1}+q_{2}+\cdots$ $+q_{n}=1$,
$$\varphi\left(\sum q_{i} x_{i}\right) \leqslant \sum q_{i} \varphi\left(x_{i}\right)$$
This is equivalent to equation (1). | To prove that if $\varphi(x)$ satisfies equation (1), then we have
$$\begin{aligned}
4 \varphi\left(\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}\right) & \leqslant 2 \varphi\left(\frac{x_{1}+x_{2}}{2}\right)+2 \varphi\left(\frac{x_{3}+x_{4}}{2}\right) \\
& \leqslant \varphi\left(x_{1}\right)+\varphi\left(x_{2}\right)+\varphi\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,054 |
Example 2 Proof: For any real number $t$, we have
$$t^{4}-t+\frac{1}{2}>0$$ | Proof of the identity
$$t^{4}-t+\frac{1}{2}=\left(t^{2}-\frac{1}{2}\right)^{2}+\left(t-\frac{1}{2}\right)^{2}$$
shows that equation (3) holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,056 |
Example 1 Let $x_{ij} (i=1,2, \cdots, m, j=1,2, \cdots n)$ be positive real numbers, and positive real numbers $\omega_{1}$, $\omega_{2}, \cdots, \omega_{n}$ satisfy $\omega_{1}+\omega_{2}+\cdots+\omega_{n}=1$, prove
$$\prod_{i=1}^{n}\left(\sum_{i=1}^{m} x_{ij}\right)^{\alpha_{j}} \geqslant \sum_{i=1}^{m}\left(\prod_{i... | Proof: By the homogeneity of equation (1), we can set $x_{1 j}+x_{2 j}+\cdots+x_{m j}=1$, where $j \in\{1,2, \cdots, n\}$. Then we only need to prove
that is, $\square$
$$\prod_{j=1}^{n} 1^{w_{j}} \geqslant \sum_{i=1}^{m} \prod_{j=1}^{n} x_{i j}^{\omega_{j}}$$
$$1 \geqslant \sum_{i=1}^{m} \prod_{j=1}^{n} x_{i j}^{\ome... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,058 |
Example 2 Proof: For positive numbers $x, y, z$, we have
$$\sum_{\text{cyc}}(y+z) \sqrt{\frac{y z}{(z+x)(x+y)}} \geqslant \sum_{\text{cyc}} x .$$ | $$\begin{aligned}
& \text { Prove } \quad \text { by Hölder's inequality } \\
& \left(\sum_{\text {cyc }}(y+z) \sqrt{\frac{y z}{(z+x)(x+y)}}\right)^{2}\left(\sum_{\mathrm{cyc}} y^{2} z^{2}(y+z)(z+x)(z+y)\right) \\
\geqslant & \left(\sum_{\mathrm{cyc}} y z(y+z)\right)^{3},
\end{aligned}$$
Therefore, to prove inequality... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,059 |
Example 1 (Beckenbach Inequality) Let $x_{i}, y_{i}>0$, prove: when $1 \leqslant t \leqslant 2$, we have
$$\frac{\left(M_{t}(x+y)\right)^{t}}{\left(M_{t-1}(x+y)\right)^{t-1}} \leqslant \frac{\left(M_{t}(x)\right)^{t}}{\left(M_{t-1}(x)\right)^{t-1}}+\frac{\left(M_{t}(y)\right)^{t}}{\left(M_{t-1}(y)\right)^{t-1}},$$
whe... | We will prove a more general result: If $t \geqslant 1, p \geqslant 1 \geqslant r$, then
$$\frac{\left(M_{p}(x+y)\right)^{t}}{\left(M_{r}(x+y)\right)^{t-1}} \leqslant \frac{\left(M_{p}(x)\right)^{t}}{\left(M_{r}(x)\right)^{t-1}}+\frac{\left(M_{p}(y)\right)^{t}}{\left(M_{r}(y)\right)^{t-1}},$$
When $0 \leqslant t \leqs... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,062 |
Example 1 If $x_{i} \geqslant 0, \alpha<0<\beta$. Prove:
$$\frac{M_{a}(x)}{M_{a}(x+1)} \leqslant \frac{G(x)}{G(x+1)} \leqslant \frac{M_{\beta}(x)}{M_{\beta}(x+1)},$$
Equality holds if and only if $x_{1}=x_{2}=\cdots=x_{n}$. | Prove that by the monotonicity of power means and Chebyshev's inequality,
$$\begin{aligned}
& \left(\sum_{i=1}^{n}\left(x_{i}+1\right)^{\beta}\right)^{\frac{1}{\beta}}\left(\prod_{i=1}^{n}\left(\frac{x_{i}}{x_{i}+1}\right)\right)^{\frac{1}{n}} \\
\leqslant & \left(\sum_{i=1}^{n}\left(x_{i}+1\right)^{\beta}\right)^{\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,065 |
Example 1 Positive real numbers $a, b, c$ satisfy $a+b+c=3$. Prove:
$$\frac{a \sqrt{a}}{a+b}+\frac{b \sqrt{b}}{b+c}+\frac{c \sqrt{c}}{c+a} \geqslant \frac{a b+b c+c a}{2} .$$ | Proof: By Hölder's inequality, we have
$$\left(\sum_{c y c} \frac{a \sqrt{a}}{a+b}\right)^{2} \sum_{\mathrm{cyc}}(a+b)^{2} \geqslant(a+b+c)^{3},$$
so we only need to prove
$$(a+b+c)^{3} \geqslant\left(\frac{a b+b c+c a}{2}\right)^{2} \sum_{\mathrm{cyc}}(a+b)^{2},$$
In fact, we have
$$\begin{aligned}
(a+b+c)^{3} & \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,066 |
Example 3 Let $x, y, z$ be non-negative real numbers, and the sum of any two of them is not zero. Prove:
$$\sum_{x \mathrm{cc}} \frac{2 x^{2}+y z}{y+z} \geqslant \frac{9\left(x^{2}+y^{2}+z^{2}\right)}{2(x+y+z)}$$ | $$\text { Prove } \begin{aligned}
& \sum_{\mathrm{cyc}} \frac{2 x^{2}+y z}{y+z}-\frac{9\left(x^{2}+y^{2}+z^{2}\right)}{2(x+y+z)} \\
= & \sum_{\mathrm{cyc}} \frac{(y-z)^{2}\left(2(x-y-z)^{2}+y z\right)}{2(x+y)(x+z)(x+y+z)} \\
\geqslant & 0,
\end{aligned}$$
so inequality (5) holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,067 |
Example 2 Non-negative real numbers $a, b, c$ satisfy $a+b+c=3$. Prove:
$$\sqrt{a \sqrt{a}}+\sqrt{b \sqrt{b}}+\sqrt{c \sqrt{c}} \geqslant \sqrt{3(a b+b c+a c)} .$$ | Proof: Let $a=x^{4}, b=y^{4}, c=z^{4}$, where $x, y, z$ are non-negative real numbers, then
(4)
$$\begin{array}{l}
\Leftrightarrow x^{3}+y^{3}+z^{3} \geqslant \sqrt{3\left(x^{4} y^{4}+x^{4} z^{4}+y^{4} z^{4}\right)} \\
\Leftrightarrow\left(x^{4}+y^{4}+z^{4}\right)\left(x^{3}+y^{3}+z^{3}\right)^{4} \geqslant 27\left(x^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,068 |
Example 4 Positive real numbers $a, b, c$ satisfy $a+b+c=3$. Prove:
$$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant a^{2}+b^{2}+c^{2}$$ | Prove that by the Arithmetic Mean-Geometric Mean Inequality,
$$\begin{aligned}
& (a+b+c)^{4}\left(a^{2} b^{2}+a^{2} c^{2}+b^{2} c^{2}\right) \geqslant 81\left(a^{2}+b^{2}+c^{2}\right) a^{2} b^{2} c^{2} \\
\Leftrightarrow & \sum_{\mathrm{sym}}\left(a^{6} b^{2}+4 a^{5} b^{3}+4 a^{5} b^{2} c+3 a^{4} b^{4}+12 a^{4} b^{3} c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,070 |
Example 5 Proof: For positive real numbers $a, b, c$, we have
$$\frac{a^{2}+b c}{b+3 c}+\frac{b^{2}+c a}{c+3 a}+\frac{c^{2}+a b}{a+3 b} \geqslant \frac{a+b+c}{2} .$$ | $$\begin{aligned}
& \frac{a^{2}+b c}{b+3 c}+\frac{b^{2}+c a}{c+3 a}+\frac{c^{2}+a b}{a+3 b} \geqslant \frac{a+b+c}{2} \\
\Leftrightarrow & 2 \sum_{\text {cyc }}\left(a^{2}+b c\right)\left(3 a^{2}+9 a b+a c+3 b c\right) \\
\geqslant & \sum_{\text {cyc }} a \sum_{\text {cyc }}\left(3 a^{2} b+9 a^{2} c+\frac{28}{3} a b c\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,071 |
Example 6 Proof: For non-negative real numbers $a, b, c$, we have
$$\frac{a^{3}+b^{3}}{(a+b)^{3}}+\frac{b^{3}+c^{3}}{(b+c)^{3}}+\frac{c^{3}+a^{3}}{(c+a)^{3}}+\frac{27}{4} \cdot \frac{a b c}{a^{3}+b^{3}+c^{3}} \leqslant 3$$ | Proof
$$\begin{aligned}
& \sum_{\text {cyc }} \frac{a^{3}+b^{3}}{(a+b)^{3}}+\frac{27 a b c}{4\left(a^{3}+b^{3}+c^{3}\right)} \leqslant 3 \\
\Leftrightarrow & \sum_{\text {cyc }}\left(1-\frac{a^{2}-a b+b^{2}}{(a+b)^{2}}\right) \geqslant \frac{27 a b c}{4\left(a^{3}+b^{3}+c^{3}\right)} \\
\Leftrightarrow & \sum_{\text {c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,072 |
Example 7 Positive real numbers $a, b, c, d$ satisfy $a+b+c+d=1$. Prove:
$$\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c} \leqslant \frac{1}{64 a b c d}$$ | Prove $\left(\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\right) a b c d=(a c+b d)(b c+a d)$
$$\begin{array}{l}
\leqslant\left(\frac{a c+b d+b c+a d}{2}\right)^{2}=\left(\frac{(a+b)(c+d)}{2}\right)^{2} \\
\leqslant\left(\frac{\left(\frac{a+b+c+d}{2}\right)^{2}}{2}\right)^{2}=\frac{1}{64}
\end{array}$$
Therefore, th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,073 |
Example 8 Non-negative real numbers $a, b, c$ satisfy $b c+c a+a b=3$. Prove:
$$\frac{1}{1+a^{2}(b+c)}+\frac{1}{1+b^{2}(a+c)}+\frac{1}{1+c^{2}(a+b)} \leqslant \frac{3}{1+2 a b c} .$$ | $$\begin{aligned}
& \sum_{\text {cyc }} \frac{1}{1+a^{2}(b+c)} \leqslant \frac{3}{1+2 a b c} \Leftrightarrow \sum_{\text {cyc }}\left(\frac{1}{1+2 a b c}-\frac{1}{1+a^{2}(b+c)}\right) \geqslant 0 \\
\Leftrightarrow & \sum_{\text {cyc }} \frac{a^{2}(b+c)-2 a b c}{1+a^{2}(b+c)} \geqslant 0 \Leftrightarrow \sum_{\text {cy... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,074 |
Example 9 Positive real numbers $a, b, c$ satisfy $b c+c a+a b=1$. Prove:
$$\sum_{\mathrm{cyc}} \sqrt{a^{3}+a} \geqslant 2 \sqrt{a+b+c}$$ | Proof
$$\begin{aligned}
& \sum_{\text {cyc }} \sqrt{a^{3}+a} \geqslant 2 \sqrt{a+b+c} \\
\Leftrightarrow & \sum_{\text {cyc }} \sqrt{a^{3}+a^{2} b+a^{2} c+a b c} \geqslant 2 \sqrt{(a+b+c)(a b+a c+b c)} \\
\Leftrightarrow & \sum_{\text {cyc }}\left(a^{3}+a^{2} b+a^{2} c+a b c\right) \\
& +2 \sum_{\text {cyc }} \sqrt{\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,075 |
Example 10 Proof: For positive real numbers $a, b$, we have
$$\frac{a}{\sqrt{a^{2}+3 b^{2}}}+\frac{b}{\sqrt{b^{2}+3 a^{2}}} \geqslant 1 .$$ | Prove using Hölder's inequality that
$$\begin{aligned}
& \left(\frac{a}{\sqrt{a^{2}+3 b^{2}}}+\frac{b}{\sqrt{b^{2}+3 a^{2}}}\right)^{2}\left(a\left(a^{2}+3 b^{2}\right)+b\left(b^{2}+3 a^{2}\right)\right) \geqslant(a+b)^{3} \\
= & a\left(a^{2}+3 b^{2}\right)+b\left(b^{2}+3 a^{2}\right),
\end{aligned}$$
so inequality (1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,076 |
Example 11 Proof: For non-negative real numbers $a, b, c$, we have
$$\sum_{\mathrm{cyc}} \sqrt{2\left(a^{2}+b^{2}\right)} \geqslant \sqrt[3]{9 \sum_{\mathrm{cyc}}(a+b)^{3}} .$$ | $$\begin{array}{l}
\sum_{\mathrm{cyc}} \sqrt{2\left(a^{2}+b^{2}\right)}-\sqrt[3]{9 \sum_{\mathrm{cyc}}(a+b)^{3}} \\
=\sum_{\mathrm{cyc}}\left(\sqrt{2\left(a^{2}+b^{2}\right)}-a-b\right)-\left(\sqrt[3]{9 \sum_{\mathrm{cyc}}(a+b)^{3}}-2(a+b+c)\right) \\
=\sum_{\mathrm{cyc}} \frac{(a-b)^{2}}{a+b+\sqrt{2\left(a^{2}+b^{2}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,077 |
Example 4 Let $x, y, z>0$, and $y z+z x+x y=1$. Try to prove:
$$\frac{1+y^{2} z^{2}}{(y+z)^{2}}+\frac{1+z^{2} x^{2}}{(z+x)^{2}}+\frac{1+x^{2} y^{2}}{(x+y)^{2}} \geqslant \frac{5}{2} .$$ | Prove that
$$\begin{aligned}
& \sum_{\text {cyc }} \frac{1+y^{2} z^{2}}{(y+z)^{2}}-\frac{5}{2} \\
= & \frac{(y-z)^{2}(z-x)^{2}(x-y)^{2}}{2(y+z)^{2}(z+x)^{2}(x+y)^{2}} \\
& +\sum_{\text {cyc }} \frac{\left(x(y+z)\left(y^{2}+z^{2}-2 x^{2}\right)+(y-z)^{2}\left(x^{2}+y z\right)\right)^{2}}{6(y+z)^{2}(z+x)^{2}(x+y)^{2}} \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,078 |
Example 12 Proof: For non-negative real numbers $x, y$, we have
$$\left(x^{2}+y^{2}\right)^{2}+x^{3}+y^{3}+4 x y^{2}+x^{2}+y^{2} \geqslant 3 x^{3} y+5 x y^{2}+x y .$$ | $$\begin{aligned}
& \left(x^{2}+y^{2}\right)^{2}+x^{3}+y^{3}+4 x y^{2}+x^{2}+y^{2}-\left(3 x^{3} y+5 x y^{2}+x y\right) \\
= & x^{4}-3 x^{3} y+2 x^{2} y^{2}+y^{4}+x^{3}-x y^{2}+y^{3}+x^{2}-x y+y^{2} \\
\geqslant & x^{4}-3 x^{3} y+2 x^{2} y^{2}+y^{4}=\frac{\left(2 x^{2}-3 x y-y^{2}\right)^{2}+3 y^{2}(x-y)^{2}}{4} \geqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,079 |
Example 13 Proof: For all positive integers $n$, we have
$$0<\sqrt{4 n+2}-\sqrt{n+1}-\sqrt{n}<\frac{1}{16 n \sqrt{n}}$$ | $$\begin{aligned}
& \sqrt{4 n+2}-\sqrt{n+1}-\sqrt{n} \\
= & \sqrt{n+\frac{1}{2}}-\sqrt{n}-\left(\sqrt{n+1}-\sqrt{n+\frac{1}{2}}\right) \\
= & \frac{n+\frac{1}{2}-n}{\sqrt{n+\frac{1}{2}}+\sqrt{n}}-\frac{n+1-\left(n+\frac{1}{2}\right)}{\sqrt{n+1}+\sqrt{n+\frac{1}{2}}} \\
= & \frac{1}{2}\left(\frac{1}{\sqrt{n+\frac{1}{2}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,080 |
Example 14 Proof: For positive real numbers $a, b, c, d$, we have
$$\begin{aligned}
& (a+b)(b+c)(c+d)(d+a)(1+\sqrt[4]{a b c d})^{4} \\
\geqslant & 16 a b c d(1+a)(1+b)(1+c)(1+d) .
\end{aligned}$$ | $$\begin{aligned}
& (a+b)(b+c)(c+d)(d+a)(1+\sqrt[4]{a b c d})^{4} \\
\geqslant & 16 a b c d(1+a)(1+b)(1+c)(1+d) \\
\Leftrightarrow & (a+b)(b+c)(c+d)(d+a)-16 a b c d \\
& +4 \sqrt[4]{a b c d}((a+b)(b+c)(c+d)(d+a) \\
& \left.-4 \sqrt[4]{a^{3} b^{3} c^{3} d^{3}}(a+b+c+d)\right) \\
& +2 \sqrt{a b c d}(3(a+b)(b+c)(c+d)(d+a)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,081 |
Example 15 Proof: For non-negative real numbers $a, b, c, d$, we have
$$\begin{aligned}
& (a+b)(b+c)(c+a)(1+a)(1+b)(1+c) \\
\geqslant \geqslant & a b c(2+a+b)(2+b+c)(2+c+a) .
\end{aligned}$$ | Prove that the original inequality is equivalent to
$$\frac{(a+b)(b+c)(c+a)}{a b c} \geqslant \frac{(a+b+2)(b+c+2)(c+a+2)}{(a+1)(b+1)(c+1)},$$
i.e.,
$$\begin{array}{c}
\frac{(a+b)(b+c)(c+a)}{a b c}-8 \geqslant \frac{(a+b+2)(b+c+2)(c+a+2)}{(a+1)(b+1)(c+1)}-8, \\
\sum_{\mathrm{cyc}}\left(\frac{(a-b)^{2}}{a b}-\frac{(a-b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,082 |
Example 16 When positive real numbers $a, b, c$ satisfy $a b c=1$. Prove:
$$\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{4}{(a+b+c+1)^{2}} \geqslant 1 .$$ | $$\text { Prove } \begin{aligned}
& \frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{4}{(a+b+c+1)^{2}}-1 \\
= & \frac{\left(a^{2}+b^{2}+c^{2}-3\right)^{2}}{(1+a)^{2}(1+b)^{2}(1+c)^{2}(1+a+b+c)^{2}} \geqslant 0,
\end{aligned}$$
so the original inequality holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,083 |
Example 17 Proof: For non-negative real numbers $a, b, c$, we have
$$\sum_{\mathrm{ccc}} \sqrt{\frac{2 a(b+c)}{(2 b+c)(b+2 c)}} \geqslant 2$$ | Prove that by Hölder's inequality,
$$\left(\sum_{\text {cyc }} \sqrt{\frac{a(b+c)}{(2 b+c)(b+2 c)}}\right)^{2} \sum_{\text {cyc }} \frac{a^{2}(2 b+c)(b+2 c)}{b+c} \geqslant(a+b+c)^{3},$$
Therefore, it suffices to prove that
$$\begin{array}{l}
(a+b+c)^{3} \geqslant 2 \sum_{\mathrm{cyc}} \frac{a^{2}(2 b+c)(b+2 c)}{b+c} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,084 |
Example 18 Let $a, b, c$ be the sides of a triangle. Prove:
$$a^{3}+b^{3}+c^{3}-3 a b c \geqslant 2 \max \left(|a-b|^{3},|b-c|^{3},|c-a|^{3}\right) .$$ | Proof: Let $a \geqslant b \geqslant c$ and $a=y+z, b=z+x, c=x+y$, then
$$\begin{aligned}
& a^{3}+b^{3}+c^{3}-3 a b c \geqslant 2 \max \left\{|a-b|^{3},|b-c|^{3},|c-a|^{3}\right\} \\
\Leftrightarrow & (a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-a c-b c\right) \geqslant 2(a-c)^{2} \\
\Leftrightarrow & (x+y+z)\left(x^{2}+y^{2}+z^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,085 |
Example 19 Positive real numbers $a, b, c$ satisfy $a b c=1$. Prove:
$$\sqrt{\frac{1}{1+a}}+\sqrt{\frac{1}{1+b}}+\sqrt{\frac{1}{1+c}} \leqslant 2 \sqrt{1+\frac{1}{(1+a)(1+b)(1+c)}} .$$ | Proof: Let $a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{x}{z}$, where $x, y, z$ are positive real numbers, and satisfy $x + y + z = 1$. Then,
$$\begin{aligned}
& \sum_{c y c} \sqrt{\frac{1}{1+a}} \leqslant 2 \sqrt{1+\frac{1}{(1+a)(1+b)(1+c)}} \\
\Leftrightarrow & \sum_{c y c} \sqrt{\frac{x}{x+y}} \leqslant 2 \sqrt{1+\frac{x ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,086 |
Example 20 Prove: For non-negative real numbers $a, b, c$, we have
$$\left(a^{2}+b^{2}+c^{2}\right)(a+b+c) \geqslant 3 \sqrt{3 a b c\left(a^{3}+b^{3}+c^{3}\right)} .$$ | Prove that the original inequality is equivalent to:
$$\begin{array}{l}
\sum_{\text {sym }}\left(a^{6}+4 a^{5} b+6 a^{4} b^{2}+4 a^{3} b^{3}+8 a^{3} b^{2} c+2 a^{2} b^{2} c^{2}-25 a^{4} b c\right) \geqslant 0 . \\
\sum_{\text {sym }}\left(a^{6}+4 a^{5} b+6 a^{4} b^{2}+4 a^{3} b^{3}+8 a^{3} b^{2} c+2 a^{2} b^{2} c^{2}-2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,087 |
Example 21 Positive real numbers $a, b, c$ satisfy $a+b+c=1$. Prove:
$$\frac{a^{2}+b c}{a^{2}+1}+\frac{b^{2}+a c}{b^{2}+1}+\frac{c^{2}+b a}{c^{2}+1} \leqslant \frac{13}{20}$$ | Prove that
$$
\frac{a^{2}+b c}{a^{2}+1}+\frac{b^{2}+a c}{b^{2}+1}+\frac{c^{2}+b a}{c^{2}+1} \leqslant \frac{13}{20}
$$
$$\begin{aligned}
\Leftrightarrow & \sum_{\text {sym }}\left(3 a^{6}+10 a^{5} b-a^{4} b^{2}-15 a^{3} b^{3}+23 a^{4} b c+114 a^{3} b^{2} c\right. \\
& \left.+32 \frac{2}{3} a^{2} b^{2} c^{2}\right) \geq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,088 |
Example 5 Let $a, b, c$ be three distinct real numbers, prove:
$$\left(\frac{a-b}{b-c}\right)^{2}+\left(\frac{b-c}{c-a}\right)^{2}+\left(\frac{c-a}{a-b}\right)^{2} \geqslant 5$$ | Prove $\quad \begin{aligned} & \left(\frac{a-b}{b-c}\right)^{2}+\left(\frac{b-c}{c-a}\right)^{2}+\left(\frac{c-a}{a-b}\right)^{2} \\ = & 5+\left(1+\frac{a-b}{b-c}+\frac{b-c}{c-a}+\frac{c-a}{a-b}\right)^{2} \geqslant 5 .\end{aligned}$ | 5 | Inequalities | proof | Yes | Yes | inequalities | false | 734,089 |
Example 22 Positive real numbers $a, b, c$ satisfy $a b c=1$. Prove:
$$\frac{a+b}{2\left(a^{7}+b^{7}+c\right)}+\frac{b+c}{2\left(b^{7}+c^{7}+a\right)}+\frac{c+a}{2\left(c^{7}+a^{7}+b\right)} \leqslant 1 .$$ | Prove that by Milhead's theorem, we have
$$a^{7}+b^{7} \geqslant a^{2} b^{2}\left(a^{3}+b^{3}\right),$$
Therefore,
$$\begin{aligned}
\sum_{\text {cyc }} \frac{a+b}{a^{7}+b^{7}+c} & \leqslant \sum_{\text {cyc }} \frac{a+b}{a^{2} b^{2}\left(a^{3}+b^{3}\right)+c} \\
& =\sum_{\text {cyc }} \frac{a+b}{\frac{a^{3}+b^{3}}{c^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,090 |
Example 23 Prove: For positive real numbers $a, b, c$, we have
$$\frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}+\frac{(c+a-b)^{2}}{(c+a)^{2}+a^{2}}+\frac{(a+b-c)^{2}}{(a+b)^{2}+c^{2}} \geqslant \frac{3}{5} .$$ | Prove that by Cauchy-Schwarz inequality,
$$\sum_{\text {cyc }} \frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}=\sum_{\text {cyc }} \frac{\left(b^{2}+b c-a b\right)^{2}}{a^{2} b^{2}+\left(b^{2}+b c\right)^{2}} \geqslant \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{\sum_{\text {cyc }}\left(a^{2} b^{2}+\left(b^{2}+b c\right)^{2}\right)}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,091 |
Example 25 Proof: For positive real numbers $a, b, c$, we have
$$\sum_{\text {cyc }} \frac{a^{5}}{a^{3}+b^{3}}+\sum_{\text {cyc }} \frac{a^{3}}{a+b} \geqslant \sum_{\text {cyc }} \frac{a^{4}}{a^{2}+b^{2}}+\frac{a^{2}+b^{2}+c^{2}}{2}$$ | $$\begin{array}{l}
\sum_{\text {cyc }} \frac{a^{5}}{a^{3}+b^{3}}+\sum_{\text {cyc }} \frac{a^{3}}{a+b}-\sum_{\text {cyc }} \frac{a^{4}}{a^{2}+b^{2}}-\frac{a^{2}+b^{2}+c^{2}}{2} \\
= \sum_{\text {cyc }}\left(\frac{a^{5}}{a^{3}+b^{3}}-\frac{a^{2}+b^{2}}{4}+\frac{a^{3}}{a+b}-\frac{a^{2}+b^{2}}{4}-\frac{a^{4}}{a^{2}+b^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,093 |
Example 26 Prove: For positive real numbers $a, b, c$, we have
$$\frac{a^{5}}{a^{3}+b^{3}}+\frac{b^{5}}{b^{3}+c^{3}}+\frac{c^{5}}{c^{3}+a^{3}} \geqslant \frac{a^{2}+b^{2}+c^{2}}{2}$$ | $$\begin{aligned}
& \sum_{\mathrm{cyc}} \frac{a^{5}}{a^{3}+b^{3}} \geqslant \frac{a^{2}+b^{2}+c^{2}}{2} \Leftrightarrow \sum_{\mathrm{cyc}}\left(\frac{a^{5}}{a^{3}+b^{3}}-\frac{a^{2}+b^{2}}{4}\right) \geqslant 0 \\
\Leftrightarrow & \sum_{\mathrm{cyc}}\left(\frac{(a-b)\left(3 a^{4}+3 a^{3} b+2 a^{2} b^{2}+a b^{3}+b^{4}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,094 |
Example 27 Prove: For non-negative real numbers $a, b, c$, we have
$$\sum_{\mathrm{cyc}} a^{3}+3 a b c \geqslant \sum_{\mathrm{cyc}} a b \sqrt{2\left(a^{2}+b^{2}\right)}$$ | Proof:
$$\begin{aligned}
& \sum_{\text {cyc }}\left(a^{3}+a b c\right) \geqslant \sum_{\text {cyc }} a b \sqrt{2\left(a^{2}+b^{2}\right)} \\
\Leftrightarrow & \sum_{\text {cyc }}\left(a^{6}+2 a^{8} b^{3}+6 a^{4} b c+3 a^{2} b^{2} c^{2}-2 a^{4} b^{2}-2 a^{4} c^{2}\right) \\
\geqslant & 4 \sum_{\text {cyc }} a^{2} b c \s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,095 |
Example 28 Prove: For any two non-zero non-negative real numbers $a, b, c$, we have
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a^{2} b+b^{2} c+c^{2} a}{a b^{2}+b c^{2}+c a^{2}} \geqslant \frac{5}{2} \text {. }$$ | $$\begin{aligned}
& \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a^{2} b+b^{2} c+c^{2} a}{a b^{2}+b c^{2}+c a^{2}} \geqslant \frac{5}{2} \\
\Leftrightarrow & \sum_{\text {cyc }}\left(\frac{a}{b+c}-\frac{1}{2}\right) \geqslant \frac{\sum_{\text {cyc }}\left(a^{2} c-a^{2} b\right)}{\sum_{\text {cyc }} a^{2} c} \\
\Lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,096 |
Example 29 Prove: For non-negative real numbers $a, b, c$, we have
$$a^{3}+b^{3}+c^{3}+12 a b c \leqslant \sum a^{2} \sqrt{a^{2}+24 b c}$$ | Prove that $a^{3}+b^{3}+c^{3}+12 a b c \leqslant \sum_{\mathrm{cyc}} a^{2} \sqrt{a^{2}+24 b c}$
$$\Leftrightarrow \sum_{\mathrm{cyc}}\left(a^{3} b^{3}+24 a^{2} b^{2} c^{2}\right) \leqslant \sum_{\mathrm{cyc}} a^{2} b^{2} \sqrt{\left(a^{2}+24 b c\right)\left(b^{2}+24 a c\right)} .$$
Since
$$\sqrt{\left(a^{2}+24 b c\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,097 |
Example 30 Let non-negative real numbers $a, b, c$ satisfy $a+b+c=1$. Prove:
$$\sqrt{a+b^{2}}+\sqrt{b+c^{2}}+\sqrt{c+a^{2}} \geqslant 2$$ | Prove that the original inequality is equivalent to
$$\sum_{\text {cyc }}\left(\sqrt{a+b^{2}}-b\right) \geqslant 1$$
i.e., $\square$
$$\sum_{\text {cyc }} \frac{a}{b+\sqrt{a+b^{2}}} \geqslant 1$$
By the Arithmetic Mean-Geometric Mean Inequality, we have
$$\begin{aligned}
\frac{a}{b+\sqrt{a+b^{2}}} & =\frac{a(a+b)}{b(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,098 |
Example 6 Proof: For real numbers $x, y, z$,
$$16 \sum_{\text {cyc }} x^{4}-20 \sum_{\text {cyc }} x^{3}(y+z)+9 \sum_{\text {cyc }} y^{2} z^{2}+15 \sum_{\text {cyc }} x^{2} y z \geqslant 0 .$$ | Prove that for any real numbers $u, v, x, y, z$, consider the following two expressions:
$$\begin{array}{c}
u^{2}\left(\sum_{\text {cyc }} x^{4}-\sum_{\text {cyc }} y^{2} z^{2}\right)+v^{2}\left(\sum_{\text {cyc }} y^{2} z^{2}-\sum_{\text {cyc }} x^{2} y z\right), \\
u v\left(\sum_{\text {cyc }} x^{3}(y+z)-2 \sum_{\tex... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,100 |
Example 32 Prove: For non-negative real numbers $a, b, c$, we have
$$\sum_{\text {cyc }} \frac{a^{3}+a b c}{b+c} \geqslant \sum_{\text{cyc}} \frac{a\left(b^{3}+c^{3}\right)}{a^{2}+b c}$$ | $$\begin{aligned}
& \sum_{\text {cyc }}\left(\left(a^{3}+a b c\right)(a+b)(a+c)\left(a^{2}+b c\right)\left(b^{2}+a c\right)\left(c^{2}+a b\right)\right. \\
& \left.-\left(a b^{3}+a c^{3}\right)\left(b^{2}+a c\right)\left(c^{2}+a b\right)(a+b)(a+c)(b+c)\right) \\
= & \sum_{\text {cyc }}\left(a^{5}+a^{4} b+a^{4} c+2 a^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,101 |
Example 33 For non-negative real numbers $a, b, c$, where the sum of any two of them is not zero. Prove:
$$\frac{(b+c)^{2}}{a^{2}+b c}+\frac{(c+a)^{2}}{b^{2}+c a}+\frac{(a+b)^{2}}{c^{2}+a b} \geqslant 6 \text {. }$$ | Prove that the original inequality is equivalent to
$$(a-b)^{2}(a-c)^{2}(b-c)^{2}+\sum_{c y c} a b\left(a^{2}-b^{2}\right)^{2} \geqslant 0 .$$
The above inequality is obviously true, hence the original inequality holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,102 |
\begin{aligned} & \text{Example } 34 \text{ Prove: For positive real numbers } x, y, z, \text{ we have } \\ & \frac{(y+z)^{2}}{x^{2}+y z}+\frac{(z+x)^{2}}{y^{2}+z x}+\frac{(x+y)^{2}}{z^{2}+x y} \\ \geqslant & \frac{2}{3}\left(x^{2}+y z+y^{2}+z x+z^{2}+x y\right)\left(\frac{1}{x^{2}+y z}+\frac{1}{y^{2}+z x}+\frac{1}{z^{... | $$\begin{aligned}
& 3 \sum_{\text {cyc }}(y+z)^{2}\left(y^{2}+z x\right)\left(z^{2}+x y\right)-2 \sum_{\text {cyc }}\left(x^{2}+y z\right) \sum_{\text {cyc }}\left(y^{2}\right. \\
& +z x)\left(z^{2}+x y\right) \\
= & \left(x^{5}(y+z)+y^{5}(z+x)+z^{5}(x+y)\right)-\left(x^{4}\left(y^{2}+z^{2}\right)+y^{4}\left(z^{2}+x^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,103 |
Example 35 Prove: For positive real numbers $a, b, c$, we have
$$\begin{aligned}
& (8 a+5 b+5 c)^{3}+(5 a+8 b+5 c)^{3}+(5 a+5 b+8 c) \\
\geqslant & 1944(a+b+c)(a b+b c+c a) .
\end{aligned}$$ | $$\begin{aligned}
& \sum_{\text {cyc }}(8 a+5 b+5 c)^{3} \geqslant \frac{1}{3} \sum_{\text {cyc }}(8 a+5 b+5 c) \sum_{\text {cyc }}(8 a+5 b+5 c)^{2} \\
= & \sum_{\text {cyc }} 6 a \sum_{\text {cyc }}\left(114 a^{2}+210 a b\right) \geqslant \sum_{\text {cyc }} 6 a \sum_{\text {cyc }} 324 a b \\
= & 1944(a+b+c)(a b+a c+b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,104 |
Example 36 Prove: For non-negative real numbers $a, b, c$, we have
$$\frac{b+c}{2 a^{2}+b c}+\frac{c+a}{2 b^{2}+c a}+\frac{a+b}{2 c^{2}+a b} \geqslant \frac{6}{a+b+c} .$$ | $$\begin{aligned}
& \sum_{\text {cyc }} \frac{b+c}{2 a^{2}+b c} \geqslant \frac{6}{a+b+c} \\
\Leftrightarrow & \sum_{\text {sym }}\left(a^{5} b+3 a^{4} b^{2}-4 a^{3} b^{3}-a^{4} b c+5 a^{3} b^{2} c-4 a^{2} b^{2} c^{2}\right) \geqslant 0 \\
\Leftrightarrow & 4(a-b)^{2}(a-c)^{2}(b-c)^{2}+\sum_{\text {sym }}\left(a^{5} b-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,105 |
Example 37 Prove: For positive real numbers $a, b, c$, we have
$$\frac{a^{2}-b c}{\sqrt{8 a^{2}+(b+c)^{2}}}+\frac{b^{2}-c a}{\sqrt{8 b^{2}+(c+a)^{2}}}+\frac{c^{2}-a b}{\sqrt{8 c^{2}+(a+b)^{2}}} \geqslant 0 .$$ | $$\begin{array}{l}
\sum_{\text {cyc }} \frac{a^{2}-b c}{\sqrt{8 a^{2}+(b+c)^{2}}} \geqslant 0 \\
\Leftrightarrow \sum_{\text {cyc }} \frac{(a-b)(a+c)-(c-a)(a+b)}{\sqrt{8 a^{2}+(b+c)^{2}}} \geqslant 0 \\
\sum_{\text {cyc }} \frac{(a-b)(a+c)-(c-a)(a+b)}{\sqrt{8 a^{2}+(b+c)^{2}}} \\
=\sum_{\mathrm{cyc}}(a-b)\left(\frac{a+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,106 |
Example 38 Prove: For non-negative real numbers $a, b, c$, we have
$$\sum_{\text {cyc }}\left(a^{2}-b c\right) \sqrt{a^{2}+4 b c} \geqslant 0$$ | Prove
$$\begin{aligned}
& \sum_{\text {cyc }}\left(a^{2}-b c\right) \sqrt{a^{2}+4 b c} \geqslant 0 \\
\Leftrightarrow & \sum_{\text {cyc }} a^{2} \sqrt{a^{2}+4 b c} \geqslant \sum_{\text {cyc }} b c \sqrt{a^{2}+4 b c} \\
\Leftrightarrow & \sum_{\text {cyc }}\left(a^{6}+4 a^{4} b c+2 a^{2} b^{2} \sqrt{\left(a^{2}+4 b c\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,107 |
Example 39 When positive real numbers $a, b, c$ satisfy $a+b+c=1$. Prove:
$$a \sqrt{1-b c}+b \sqrt{1-c a}+c \sqrt{1-a b} \geqslant \frac{2 \sqrt{2}}{3} .$$ | Prove that by Hölder's inequality,
$$\left(\sum_{\mathrm{cyc}} a \sqrt{1-\overrightarrow{b c}}\right)^{2} \sum_{\mathrm{cyc}} \frac{a}{1-\overrightarrow{b c}} \geqslant(a+b+c)^{3}=1$$
Therefore, it suffices to prove that
$$\sum_{\text {cyc }} \frac{a}{1-b c} \leqslant \frac{9}{8}$$
In fact,
$$\begin{array}{l}
\sum_{\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,108 |
Example 40 Real numbers $x, y$ satisfy $x+y=1$. Find the maximum value of $\left(x^{3}+1\right)\left(y^{3}+1\right)$. | $\begin{array}{c}\text { Sol }\left(x^{3}+1\right)\left(y^{3}+1\right)=x^{3} y^{3}+x^{3}+y^{3}+1=(x y)^{3}-3 x y+2= \\ \sqrt{\frac{2(x y)^{2}\left(3-(x y)^{2}\right)\left(3-(x y)^{2}\right)}{2}}+2 \leqslant \sqrt{\frac{\left(\frac{6}{3}\right)^{3}}{2}}+2=4 . \\ \text { When } x=\frac{1+\sqrt{5}}{2}, y=\frac{1-\sqrt{5}}... | 4 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,109 |
Example 41 When $1 \leqslant a, b, c, d \leqslant 2$, prove:
$$\frac{4}{3} \leqslant \frac{a}{b+c d}+\frac{b}{c+d a}+\frac{c}{d+a b}+\frac{d}{a+b c} \leqslant 2 .$$ | Proof
$$\begin{aligned}
\sum_{c y c} \frac{a}{b+c d} & =\sum_{c y c} \frac{a^{2}}{a b+a c d} \\
& \geqslant \sum_{c y c} \frac{a^{2}}{a b+2 c d} \\
& \geqslant \frac{(a+b+c+d)^{2}}{3(a b+b c+c d+d a)} \\
& =\frac{(a+b+c+d)^{2}}{3(a+c)(b+d)} \geqslant \frac{4}{3}
\end{aligned}$$
On the other hand, let $f(a, b, c, d)=\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,110 |
Example 8 Real numbers $a, b$ satisfy $a+b=1$. Prove:
$$a b\left(a^{4}+b^{4}\right) \leqslant \frac{5 \sqrt{10}-14}{27}$$ | Prove that since
$$\begin{aligned}
& \frac{5 \sqrt{10}-14}{27}(a+b)^{6}-a b\left(a^{4}+b^{4}\right) \\
= & \frac{\left(2 a^{2}+(6-\sqrt{10}) a b+2 b^{2}\right)\left(a^{2}-(2+\sqrt{10}) a b+b^{2}\right)^{2}}{14+5 \sqrt{10}} \\
\geqslant & 0
\end{aligned}$$
Therefore, equation (15) holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,111 |
Example 42 Prove: For positive real numbers $a, b, c$, we have
$$\frac{8(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}+\frac{3(a+b)(b+c)(c+a)}{a b c} \geqslant 48 .$$ | Prove
$$\begin{aligned}
& \frac{8(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}+\frac{3(a+b)(b+c)(c+a)}{a b c} \geqslant 48 \\
\Leftrightarrow & \sum_{\text {sym }}\left(3 a^{4} b+3 a^{3} b^{2}+11 a^{2} b^{2} c-17 a^{3} b c\right) \geqslant 0 \\
\Leftrightarrow & \sum_{\text {cyc }}(a-b)^{2}\left(6 c^{3}-11 a b c+3 a b(a+b)\right) \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,112 |
Example 43 Let positive real numbers $a, b, c$ satisfy $a b c \geqslant 1$. Prove:
$$\frac{a+1}{a^{2}+a+1}+\frac{b+1}{b^{2}+b+1}+\frac{c+1}{c^{2}+c+1} \leqslant 2$$ | Prove that after expanding and rearranging, we get
$$(a b c-1) \sum_{\text {cyc }}\left(\frac{1}{3}+a+a b+\frac{2}{3} a b c\right)+\frac{1}{2} \sum_{\mathrm{cyc}} c^{2}(a-b)^{2} \geqslant 0$$
Therefore, the original inequality holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,113 |
Example 44 Prove: For non-negative real numbers $a, b, c$, we have
$$\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a} \geqslant \frac{a^{2}}{a^{2}+2 b c}+\frac{b^{2}}{b^{2}+2 c a}+\frac{c^{2}}{c^{2}+2 a b} .$$ | Proof
$$\begin{aligned}
& \frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a} \geqslant \sum_{\text {cyc }} \frac{a^{2}}{a^{2}+2 b c} \\
\Leftrightarrow & \sum_{\text {cyc }}\left(\frac{a^{2}}{a b+a c+b c}-\frac{a^{2}}{a^{2}+2 b c}\right) \geqslant 0 \\
\Leftrightarrow & \sum_{\text {cyc }} \frac{a^{2}(a-b)(a-c)}{a^{2}+2 b c} \geqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,114 |
Example 45 Positive real numbers $a, b, c, d$ satisfy $a+b+c+d=4$. Prove:
$$\frac{a}{1+b^{2} c}+\frac{b}{1+c^{2} d}+\frac{c}{1+d^{2} a}+\frac{d}{1+a^{2} b} \geqslant 2 .$$ | Prove using the Cauchy-Schwarz inequality that
$$\sum_{\text {cyc }} \frac{a}{1+b^{2} c} \geqslant \frac{(a+b+c+d)^{2}}{a+b+c+d+\sum_{\text {cyc }} a b^{2} c}$$
Therefore, it suffices to prove that
$$a b^{2} c+b c^{2} d+c d^{2} a+d a^{2} b \leqslant 4 .$$
In fact,
$$\begin{aligned}
& a b^{2} c+b c^{2} d+c d^{2} a+d a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,115 |
Example 46 Positive real numbers $a, b, c, d$ satisfy $a+b+c+d=4$. Prove: $a^{2} b c+b^{2} c d+c^{2} d a+d^{2} a b \leqslant 4$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Prove that the original inequality is equivalent to
$$a c(a b+c d)+b d(a d+b c) \leqslant 4$$
When $a b+c d \leqslant a d+b c$, we have
$$\begin{aligned}
a c(a b+c d)+b d(a d+b c) & \leqslant(a c+b d)(a d+b c) \\
& \leqslant\left(\frac{a c+b d+a d+b c}{2}\right)^{2} \\
& =\left(\frac{(a+b)(c+d)}{2}\right)^{2} \leqslan... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,116 |
Example 47 Non-negative real numbers $a, b, c$ satisfy $a+b+c=3$. Prove:
$$\sqrt{a(2 a+b)(2 a+c)}+\sqrt{b(2 b+c)(2 b+a)}+\sqrt{c(2 c+a)(2 c+b)} \geqslant 9 .$$ | Prove that using Hölder's inequality we get
$$\left(\sum_{\text {cyc }} \sqrt{a(2 a+b)(2 a+c)}\right)^{2} \sum_{\text {cyc }} \frac{a^{2}}{(2 a+b)(2 a+c)} \geqslant(a+b+c)^{3},$$
Therefore, to prove the original inequality, it suffices to prove
$$\sum_{\text {cyc }} \frac{a^{2}}{(2 a+b)(2 a+c)} \leqslant \frac{1}{3}$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,117 |
Example 48 Prove: For positive real numbers $a, b, c$, we have
$$\frac{a b}{c^{2}+c a}+\frac{b c}{a^{2}+a b}+\frac{c a}{b^{2}+b c} \geqslant \frac{a}{a+c}+\frac{b}{b+a}+\frac{c}{c+b} .$$ | Prove that after eliminating the denominator and rearranging, we get
$$\sum_{\mathrm{cyc}}\left(a^{4} c^{2}-a^{3} c^{2} b\right)+\sum_{\mathrm{cyc}}\left(a^{3} b^{3}-a^{2} b^{2} c^{2}\right) \geqslant 0,$$
By the Arithmetic Mean-Geometric Mean Inequality, it is easy to see that the original inequality holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,118 |
Example 49 Non-negative real numbers $a, b, c$ satisfy $(a+b)(b+c)(c+a)=2$. Prove:
$$\left(a^{2}+b c\right)\left(b^{2}+c a\right)\left(c^{2}+a b\right) \leqslant 1$$ | Prove $\left(a^{2}+b c\right)\left(b^{2}+c a\right)\left(c^{2}+a b\right) \leqslant 1$
$$\begin{array}{l}
\Leftrightarrow(a+b)^{2}(a+c)^{2}(b+c)^{2} \geqslant 4\left(a^{2}+b c\right)\left(b^{2}+c a\right)\left(c^{2}+a b\right) \\
\Leftrightarrow(a-b)^{2}(a-c)^{2}(b-c)^{2}+\sum_{\text {sym }}\left(4 a^{3} b^{2} c+\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,119 |
Example 50 Prove: For positive real numbers $x, y, z$, we have
$$\sum_{\mathrm{cyc}} \sqrt{3 x(x+y)(x+z)} \leqslant \sqrt{4(x+y+z)^{3}} .$$ | Prove
$$\begin{aligned}
& \sum_{\mathrm{cyc}} \sqrt{3 x(x+y)(x+z)} \leqslant \sqrt{4(x+y+z)^{3}} \\
\Leftrightarrow & 4(x+y+z)^{3} \geqslant \sum_{\text {cyc }} 3 x(x+y)(x+z) \\
& +6 \sum_{\mathrm{cyc}}(x+y) \sqrt{x y(x+z)(y+z)}
\end{aligned}$$
Since
$$2 \sqrt{x y(x+z)(y+z)} \leqslant x(y+z)+y(x+z),$$
it suffices to ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,120 |
Example 51 Let $a, b, c$ be the sides of a triangle. Prove:
$$(a+b) \cos \frac{C}{2}+(b+c) \cos \frac{A}{2}+(c+a) \cos \frac{B}{2} \leqslant \sqrt{3}(a+b+c) .$$ | Proof
$$\begin{aligned}
& \sum_{\text {cyc }}(a+b) \cos \frac{C}{2} \leqslant \sqrt{3}(a+b+c) \\
\Leftrightarrow & \sum_{\text {cyc }}(a+b) \sqrt{\frac{(a+b+c)(a+b-c)}{4 a b}} \leqslant \sqrt{3}(a+b+c) \\
\Leftrightarrow & \sum_{\text {cyc }}(a+b) \sqrt{c(a+b-c)} \leqslant 2 \sqrt{3(a+b+c) a b c} \\
\Leftrightarrow & \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,121 |
Example 9 Proof: For real numbers $a, b, c$ and non-negative real numbers $p, q, r$, we have
$$\begin{aligned}
& ((q+r) a+(r+p) b+(p+q) c)^{2} \\
\geqslant & 4(p+q+r)(p b c+q c a+r a b) .
\end{aligned}$$ | To prove without loss of generality, let $a=\max \{a, b, c\}$, because
$$\begin{aligned}
& ((q+r) a+(r+p) b+(p+q) c)^{2}-4(p+q+r)(p b c+q c a+r a b) \\
= & ((q-r) a+(r+p) b-(p+q) c)^{2}+4 q r(a-b)(a-c) \geqslant 0,
\end{aligned}$$
thus, equation (17) holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,122 |
Example 52 Let $a, b, c \in [0,1]$. Prove:
$$\frac{a}{b^{3}+c^{3}+7}+\frac{b}{c^{3}+a^{3}+7}+\frac{c}{a^{3}+b^{3}+7} \leqslant \frac{1}{3}$$ | Prove $\begin{aligned} \sum_{\mathrm{cyc}} \frac{a}{b^{3}+c^{3}+7} & =\sum_{\mathrm{cyc}} \frac{a}{b^{3}+2+c^{3}+2+3} \\ & \leqslant \sum_{\mathrm{cyc}} \frac{a}{3 b+3 c+3 a}=\frac{1}{3} .\end{aligned}$ | \frac{1}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 734,123 |
23. First guess the value of $m$ to be $\frac{3}{2}$, which is achieved if and only if $x=y=z=2$. Let $x=\frac{1}{a}-1, y=\frac{1}{b}-1, z=\frac{1}{c}-1$, then we have $\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1$, which means $a+b+c=1$. Thus, the... | To prove formula (11), that is, to prove
$$\frac{a b}{c(b+c)}+\frac{b c}{a(c+a)}+\frac{c a}{b(a+b)} \geqslant \frac{3}{2} .$$
Using the Cauchy-Schwarz inequality and the Arithmetic Mean-Geometric Mean inequality, it is easy to prove formula (12). In fact, $\square$
$$\begin{aligned}
& \frac{a b}{c(b+c)}+\frac{b c}{a(c... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 734,124 |
$\begin{array}{c}\frac{\sum_{\text {cyc }} x y}{x y z}-3 x y z \geqslant\left(\sum_{\text{cyc}} x\right)^{2}-2 \sum_{\text{cyc}} x y-3 \\ \sum_{\text{cyc}} x=3, \sum_{\text{cyc}} x y=\frac{9-q^{2}}{3}, x y z=r \\ \frac{9-q^{2}}{3 r}-3 r \geqslant q-3-\frac{2}{3}\left(9-q^{2}\right) \\ 9-q^{2} \geqslant q r^{2}+2 q^{2} ... | From Exercise 22, it suffices to prove that
$$9-q^{2} \geqslant 9\left(\frac{(3-q)^{2}(3+2 q)}{27}\right)^{2}+2 q^{2} \frac{(3-q)^{2}(3+2 q)}{27}$$
That is,
$$\begin{array}{l}
3+q-\frac{(3-q)^{2}(3+2 q)^{2}}{81}-\frac{2 q^{2}(3-q)(3+2 q)}{27} \geqslant 0 \\
\frac{q^{2}\left(189+27 q-36 q^{2}+4 q^{3}\right)}{81} \geqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,127 |
Let $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n} \in [1001,2002]$, and
$$a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}$$ | $$\begin{aligned}
& \text { Prove: } \frac{a_{1}^{3}}{b_{1}}+\frac{a_{2}^{3}}{b_{2}}+\cdots+\frac{a_{n}^{3}}{b_{n}} \leqslant \frac{17}{10}\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right) . \\
& \frac{21 a_{m}^{2}-4 b_{m}^{2}}{10}-\frac{a_{m}^{3}}{b_{m}}=\left(2 a_{m}-b_{m}\right)\left(2 b_{m}-a_{m}\right)\left(\frac{... | \frac{a_{1}^{3}}{b_{1}}+\frac{a_{2}^{3}}{b_{2}}+\cdots+\frac{a_{n}^{3}}{b_{n}} \leqslant \frac{17}{10}\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,128 |
Example 10 Let non-negative real numbers $a, b, c, d$ satisfy $a+b+c+d=4$, prove:
$$a^{2}(b+c)+b^{2}(c+d)+c^{2}(d+a)+d^{2}(a+b)+8 a b c d \leqslant 16 .$$ | Prove that by the Arithmetic Mean - Geometric Mean Inequality \((a+b+c+d)(bcd+cda+dab+abc) \geqslant 16abcd\), so \(\sum_{cyc} bcd \geqslant 4abcd\), hence the inequality
$$\sum_{\mathrm{cyc}} a^{2}(b+c)+2 \sum_{\mathrm{cyc}} abc \leqslant 16$$
is stronger than inequality (18).
Because
$$\begin{aligned}
& \frac{(a+b+c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,130 |
Example 3 Let $S_{n}=1+2+3+\cdots+n, n \in \mathrm{N}$, find the maximum value of $f(n)=\frac{S_{n}}{(n+32) S_{n+1}}$. | \begin{aligned} f(n) & =\frac{S_{n}}{(n+32) S_{n+1}}=\frac{n}{(n+32)(n+2)}=\frac{n}{n^{2}+34 n+64} \\ & =\frac{1}{n+34+\frac{64}{n}}=\frac{1}{\left(\sqrt{n}-\frac{8}{\sqrt{n}}\right)^{2}+50} \leqslant \frac{1}{50} .\end{aligned} | \frac{1}{50} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,131 |
If $x_{k}>0(k=1, \cdots, n), x_{1}+x_{2}+\cdots+x_{n}=1$, and $a>0$, then
$$\sum_{k=1}^{n}\left(x_{k}+\frac{1}{x_{k}}\right)^{a} \geqslant \frac{\left(n^{2}+1\right)^{a}}{n^{a-1}}$$
Following the method of Maclaurin in Example 1, generalize the above inequality as follows.
If $x_{k}>0, k=1,2, \cdots, n, n \geqslant 2,... | Prove by the Arithmetic Mean-Geometric Mean Inequality,
$$\left(\frac{1}{n} \sum_{k=1}^{n}\left(x_{k}+\frac{1}{x_{k}}\right)^{a}\right)^{\frac{1}{a}} \geqslant\left(\prod_{k=1}^{n}\left(x_{k}+\frac{1}{x_{k}}\right)\right)^{\frac{1}{n}}$$
Therefore, to prove inequality (28), it suffices to prove:
$$\prod_{k=1}^{n}\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,133 |
Example 1 For any real numbers $a, b, c$, prove:
$$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geqslant 9(a b+b c+c a)$$ | Proof: Let $a=\sqrt{2} \tan A, b=\sqrt{2} \tan B, c=\sqrt{2} \tan C$, where
$A, B, C \in\left(0, \frac{\pi}{2}\right)$. Using $1+\tan ^{2} \theta=\frac{1}{\cos ^{2} \theta}$, equation (1) can be written as
$$\begin{aligned}
\frac{4}{9} \geqslant & \cos A \cos B \cos C(\cos A \sin B \sin C+\sin A \cos B \sin C \\
& +\si... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,135 |
For 3 non-negative real numbers $a, b, c$ satisfying $a^{2}+b^{2}+c^{2}+a b c=4$. Prove:
$$0 \leqslant a b+b c+c a-a b c \leqslant 2$$ | Prove that if $a=\min \{a, b, c\} \leqslant 1$, otherwise $a^{2}+b^{2}+c^{2}+a b c>4$. Then
$$a b+b c+c a-a b c \geqslant(1-a) b c \geqslant 0$$
On the other hand, let $a=2 p, b=2 q, c=2 r$, we get $p^{2}+q^{2}+r^{2}+2 p q r=1$. Thus, we can set $a=2 \cos A, b=2 \cos B, c=2 \cos C$, where $A, B, C \in\left[0, \frac{\p... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,137 |
Example 1 For positive real numbers $a, b, c$, prove:
$$\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1 .$$ | Proof: Let $x=\frac{a}{\sqrt{a^{2}+8 b c}}, y=\frac{b}{\sqrt{b^{2}+8 c a}}, z=\frac{c}{\sqrt{c^{2}+8 a b}}$. Clearly, $x, y, z \in(0,1)$, so equation (1) can be rewritten as:
$$x+y+z \geqslant 1$$
From
$$\frac{a^{2}}{8 b c}=\frac{x^{2}}{1-x^{2}}, \frac{b^{2}}{8 a c}=\frac{y^{2}}{1-y^{2}}, \frac{c^{2}}{8 a b}=\frac{z^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,138 |
Example 2 Let positive real numbers $a, b, c$ satisfy $abc=1$. Prove:
$$\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} .$$ | Proof Let $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$, thus $x y z=1$. Then equation (5) can be rewritten as
$$\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geqslant \frac{3}{2}$$
Using the Cauchy-Schwarz inequality, we get
$$((y+z)+(z+x)+(x+y))\left(\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,139 |
Example 3 Let positive numbers $a, b, c$ satisfy $a+b+c=abc$. Prove:
$$\frac{1}{\sqrt{1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}}+\frac{1}{\sqrt{1+c^{2}}} \leqslant \frac{3}{2} .$$ | Proof Let $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$. From $a+b+c=abc$ we know
$$1=xy+yz+zx$$
Thus, equation (8) can be rewritten as
$$\frac{x}{\sqrt{x^{2}+1}}+\frac{y}{\sqrt{y^{2}+1}}+\frac{z}{\sqrt{z^{2}+1}} \leqslant \frac{3}{2}$$
From equation (9), equation (10) can be rewritten as
$$\frac{x}{\sqrt{x^{2}+xy+yz... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,140 |
Example 4 For any positive real numbers $a, b, c$, prove:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2} .$$ | Proof: Let $x=b+c, y=c+a, z=a+b$, then equation (15) can be rewritten as
i.e. $\square$
$$\begin{array}{c}
\sum_{\mathrm{cyc}} \frac{y+z-x}{2 x} \geqslant \frac{3}{2} \\
\sum_{\mathrm{cyc}} \frac{y+z}{x} \geqslant 6
\end{array}$$
Using the Arithmetic Mean-Geometric Mean Inequality, we get
$$\begin{aligned}
& \sum_{\m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,141 |
Example 4 Let $x, y, z \in \mathbf{R}^{+}$, and $x y z(x+y+z)=1$, find the minimum value of $(x+y)(x+z)$. | From $x y z(x+y+z)=1$, we can get $x^{2}+x y+x z=\frac{1}{y z}$, thus
$$\begin{aligned}
(x+y)(x+z) & =x^{2}+x y+x z+y z=y z+\frac{1}{y z} \\
& =\left(\sqrt{y z}-\frac{1}{\sqrt{y z}}\right)^{2}+2,
\end{aligned}$$
It is easy to derive from equation (8) that the minimum value is 2. | 2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,142 |
Example 5 Let positive real numbers $a, b, c$ satisfy $abc=1$. Prove:
$$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1 .$$ | Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, where $x, y, z>0$. Then the inequality can be rewritten as
$$x y z \geqslant(y+z-x)(z+x-y)(x+y-z)$$
Without loss of generality, assume $z \geqslant y \geqslant x$, and let $y-x=p, z-x=q$, where $p, q \geqslant 0$. Then
$$\begin{aligned}
& x y z-(y+z-x)(z+x-y)(x+y-z) \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,143 |
Example 6 Let positive real numbers $a, b, c$ satisfy $a+b+c=1$. Prove:
$$\frac{a}{a+b c}+\frac{b}{b+c a}+\frac{\sqrt{a b c}}{c+a b} \leqslant 1+\frac{3 \sqrt{3}}{4}$$ | Prove that equation (19) can be rewritten as
$$\frac{1}{1+\frac{b c}{a}}+\frac{1}{1+\frac{c a}{b}}+\frac{\sqrt{\frac{a b}{c}}}{1+\frac{a b}{c}} \leqslant 1+\frac{3 \sqrt{3}}{4}$$
Let \( x = \sqrt{\frac{b c}{a}}, y = \sqrt{\frac{c a}{b}}, z = \sqrt{\frac{a b}{c}} \), then equation (20) can be rewritten as
$$\frac{1}{1+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,144 |
Example 7 Let $a, b, c, d>0$. When $\frac{1}{1+a^{4}}+\frac{1}{1+b^{4}}+\frac{1}{1+c^{4}}+\frac{1}{1+d^{4}}=1$, prove: $a b c d \geqslant 3$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Proof: Let \( A=\frac{1}{1+a^{4}}, B=\frac{1}{1+b^{4}}, C=\frac{1}{1+c^{4}}, D=\frac{1}{1+d^{4}} \), then \( a^{4}=\frac{1-A}{A}, b^{4}=\frac{1-B}{B}, c^{4}=\frac{1-C}{C}, d^{4}=\frac{1-D}{D} \).
Using the Arithmetic Mean-Geometric Mean Inequality, we get
\[
\begin{aligned}
& (B+C+D)(C+D+A)(D+A+B)(A+B+C) \\
\geqslant ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,145 |
Example 8 Real numbers $x, y, z>1$, and satisfy $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. Prove: $\sqrt{x+y+z} \geqslant \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$. | Proof Let $a=\sqrt{x-1}, b=\sqrt{y-1}, c=\sqrt{z-1}$, then
i.e. $\square$
$$\begin{array}{c}
\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}+\frac{1}{1+c^{2}}=2 \\
a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2 a^{2} b^{2} c^{2}=1
\end{array}$$
Thus, equation (25) can be rewritten as
$$\sqrt{a^{2}+b^{2}+c^{2}+3} \geqslant a+b+c$$
i.e. $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,146 |
Example 9 Proof: For any $a, b, c>0$, we have
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant \frac{a+b}{b+c}+\frac{b+c}{a+b}+1$$ | Proof: Let $x=\frac{a}{b}, y=\frac{c}{b}$, thus $\frac{c}{a}=\frac{y}{x}, \frac{a+b}{b+c}=\frac{x+1}{1+y}, \frac{b+c}{a+b}=\frac{1+y}{x+1}$.
Then equation (29) can be rewritten as
$$x^{3} y^{2}+x^{2}+x+y^{3}+y^{2} \geqslant x^{2} y+2 x y+2 x y^{2} \text {. }$$
Using the Arithmetic Mean-Geometric Mean Inequality, we ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,147 |
Example 11 For the three sides $a, b, c$ of a triangle, prove:
$$\left(\sum_{\text {cyc }} a\right)\left(\sum_{\mathrm{cyc}} \frac{1}{a}\right) \geqslant 6\left(\sum_{\mathrm{cyc}} \frac{a}{b+c}\right)$$ | Proof: Let $a=y+z, b=z+x, c=x+y$, then we obtain the equivalent inequality for non-negative numbers $x, y, z$:
$$\left(\sum_{c y c} x\right)\left(\sum_{c y c} \frac{1}{y+z}\right) \geqslant 3\left(\sum_{\mathrm{cyc}} \frac{y+z}{2 x+y+z}\right).$$
Clearing the denominators, subtracting the right side from the left side... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,149 |
Example 12 Real numbers $x, y, z$ satisfy $x y z=8$. Prove:
$$\frac{2}{2+x^{2}}+\frac{2}{2+y^{2}}+\frac{2}{2+z^{2}} \geqslant 1$$ | Proof: Let $x^{2}=u^{3}, y^{2}=v^{3}, z^{2}=w^{3}$, then equation (36) can be rewritten as
$$\frac{v w}{v w+2 u^{2}}+\frac{w u}{w u+2 v^{2}}+\frac{u v}{u v+2 w^{2}} \geqslant 1,$$
where $u, v, w$ are positive real numbers, and satisfy $u v w=4$.
Since
$$\begin{aligned}
& \frac{v w}{v w+2 u^{2}}+\frac{w u}{w u+2 v^{2}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,150 |
Example 13 Positive real numbers $x, y, z$ satisfy $x+y+z+2=x y z$. Prove: $x+y+z+6 \geqslant 2(\sqrt{y z}+\sqrt{z x}+\sqrt{x y})$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Proof: Let $x=\frac{v+w}{u}, y=\frac{w+u}{v}, z=\frac{u+v}{w}$, then equation (38) can be rewritten as
$$\sum_{\mathrm{cyc}} \frac{v+w}{u}+6 \geqslant 2 \sum_{\mathrm{cyc}} \sqrt{\frac{(w+u)(u+v)}{v w}}$$
where $u, v, w$ are positive real numbers.
Since $(2 v w+w u+u v)^{2}-4 v w(w+u)(u+v)=u^{2}(v-w)^{2} \geqslant 0$,... | null | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,151 |
Example 14 Positive real numbers $a, b, c, d$ satisfy $a b c d=1$. Prove:
$$\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(d+1)}+\frac{1}{d(a+1)} \geqslant 2$$ | Proof Let $a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{w}{z}, d=\frac{x}{w}$, where $x, y, z, w>0$, then equation (40) can be rewritten as
$$\sum_{\text {cyc }} \frac{x}{y+z} \geqslant 2$$
which is equivalent to
$$\sum_{\text {cyc }} \frac{x^{2}}{x y+x z} \geqslant 2$$
By the Cauchy-Schwarz inequality, we have
$$\sum_{\mat... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,152 |
Example 5 Let $x, y, z$ be non-negative real numbers, and satisfy $\sum_{\mathrm{cyc}} x=32$. Try to find the maximum value of $\sum_{\mathrm{cyc}} x^{3} y$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note and not part of the translation request, so it is le... | Let $x=\max \{x, y, z\}$, then
$$\begin{aligned}
& 27\left(\sum_{\text {cyc }} x\right)^{4}-256\left(\sum_{\text {cyc }} x^{3} y\right) \\
= & z\left(148\left(x z(x-z)+y^{2}(x-y)\right)+108\left(y z^{2}+x^{3}\right)\right. \\
& \left.+324 x y(x+z)+27 z^{3}+14 x^{2} z+162 y^{2} z+176 x y^{2}\right) \\
& +(x-3 y)^{2}\lef... | 110592 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,153 |
Example 15 Let positive real numbers $a, b, c$ satisfy $abc=1$. Prove:
$$\sqrt{3a^{2}+4}+\sqrt{3b^{2}+4}+\sqrt{3c^{2}+4} \leqslant \sqrt{7}(a+b+c) .$$ | Proof: Let $a=x^{3}, b=y^{3}, c=z^{3}$, then
$$\sum_{\mathrm{cyc}} x \sqrt{\frac{3 x^{4}+4 y^{2} z^{2}}{7}} \leqslant \sum_{\mathrm{cyc}} \frac{x}{7}\left(7 x^{2}+4 y^{2}+4 z^{2}-4 x y-4 x z\right)=\sum_{\mathrm{cyc}} x^{3}.$$
In fact,
$$\begin{aligned}
& \left(7 x^{2}+4 y^{2}+4 z^{2}-4 x y-4 x z\right)^{2}-7\left(3 x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,154 |
Example 1 Given $x, y, z>0$, and $y z+z x+x y=1$. Prove:
$$\sum_{\text {cyc }} \frac{1+y^{2} z^{2}}{(y+z)^{2}} \geqslant \frac{5}{2} .$$ | Proof: By the symmetry of equation (1), without loss of generality, let $x \leqslant y \leqslant z$, and $y=x+s, z=x+s+t$, where $s, t \geqslant 0$. Then,
$$\begin{aligned}
& 2 \sum_{\text {cyc }}(z+x)^{2}(x+y)^{2}\left((y z+z x+x y)^{2}+y^{2} z^{2}\right) \\
& -5(y+z)^{2}(z+x)^{2}(x+y)^{2}(y z+z x+x y) \\
= & 32\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,155 |
Example 2 Real numbers $a, b, c \in\left[\frac{1}{\sqrt{2}}, \sqrt{2}\right]$. Prove:
$$\frac{3}{b+2 c}+\frac{3}{c+2 a}+\frac{3}{a+2 b} \geqslant \frac{2}{b+c}+\frac{2}{c+a}+\frac{2}{a+b}$$ | Prove that
$$\begin{aligned}
& \frac{3}{b+2 c}+\frac{3}{c+2 a}+\frac{3}{a+2 b}-\frac{2}{b+c}-\frac{2}{c+a}-\frac{2}{a+b} \\
\equiv & \frac{F(a, b, c)}{(b+c)(c+a)(a+b)(b+2 c)(c+2 a)(a+2 b)}
\end{aligned}$$
Since \(a, b, c \in\left[\frac{1}{\sqrt{2}}, \sqrt{2}\right]\), it follows that \(b+c > a\), \(c+a > b\), and \(a+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,156 |
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