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values | problem_is_valid stringclasses 1
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35. Let $a, b, c$ be non-negative numbers such that $a^{2}+b^{2}+c^{2}=3$. Then
$$\frac{1}{5-2 a b}+\frac{1}{5-2 b c}+\frac{1}{5-2 c a} \leqslant 1$$ | 35. (2007.04.06) Simplify: The original expression only needs to prove that after removing the denominator, the left side minus the right side is not less than zero. Let
$$s_{1}=\sum a, s_{2}=\sum b c, s_{3}=a b c$$
Then
$$\Pi(5-2 b c)-\sum(5-2 c a)(5-2 a b)=50-30 \sum b c+16 a b c \sum a-8(a b c)^{2}$$
It can be see... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,570 |
36. Let $a, b, c$ be non-negative numbers such that $a^{2}+b^{2}+c^{2}=3$. Then
$$(2-a b)(2-b c)(2-c a) \geqslant 1$$ | $$\begin{array}{l}
\Pi(2-b c)-1=7-4 \sum b c+2 a b c \sum a-(a b c)^{2}= \\
\frac{1}{8}\left[50-30 \sum b c+16 a b c \sum a-8(a b c)^{2}+\frac{1}{4}\left(3-\sum b c\right)\right] \geqslant 0
\end{array}$$
(Applying the conclusion from the previous problem). | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,571 |
37. Let $a, b, c$ be non-negative numbers such that $a+b+c=2$. Prove that
$$\frac{b c}{a^{2}+1}+\frac{c a}{b^{2}+1}+\frac{a b}{c^{2}+1} \leqslant 1$$ | 37. (2007.04.19) Proof: By symmetry, without loss of generality, assume \(a \leqslant b \leqslant c\), then
$$\frac{1}{1+a^{2}} \leqslant \frac{1+c^{2}-a^{2}}{1+c^{2}}, \frac{1}{1+b^{2}} \leqslant \frac{1+c^{2}-b^{2}}{1+c^{2}}$$
Therefore, it suffices to prove
Because
$$\begin{array}{c}
\frac{\left(1+c^{2}-a^{2}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,572 |
38. Let $a, b, c$ be non-negative numbers, no two of which are zero. Then
$$\frac{a^{3}+3 a b c}{(b+c)^{2}}+\frac{b^{3}+3 a b c}{(c+a)^{2}}+\frac{c^{3}+3 a b c}{(a+b)^{2}} \geqslant a+b+c$$ | 38. (2007.04.13) Simplify
$$\begin{aligned}
\text { LHS }- \text { RHS }= & \sum\left[\frac{a^{3}+3 a b c}{(b+c)^{2}}-a\right]=\sum \frac{a\left(a^{2}-b^{2}-c^{2}+b c\right)}{(b+c)^{2}}= \\
& \sum \frac{a(a-b)(a-c)+a b(a-b)-c a(c-a)}{(b+c)^{2}}= \\
& \sum \frac{a(a-b)(a-c)}{(b+c)^{2}}+\sum \frac{a b(a-b)-c a(c-a)}{(b+c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,573 |
Example 2 Given $x_{1}, x_{2}>0, x_{1} y_{1}-z_{1}^{2}>0, x_{2} y_{2}-z_{2}^{2}>0, y_{1}, y_{2} \in \mathbf{R}$, prove that
$$\frac{8}{\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(z_{1}+z_{2}\right)^{2}} \leqslant \frac{1}{x_{1} y_{1}-z_{1}^{2}}+\frac{1}{x_{2} y_{2}-z_{2}^{2}}$$
Equality holds in (2) if and ... | $$\begin{aligned}
\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(z_{1}+z_{2}\right)^{2}= & \left(\sqrt{x_{1} y_{1}-z_{1}^{2}}+\sqrt{x_{2} y_{2}-z_{2}^{2}}\right)^{2}+ \\
& \left[\sqrt{\frac{x_{2}}{x_{1}}\left(x_{1} y_{1}-z_{1}^{2}\right)}-\sqrt{\frac{x_{1}}{x_{2}}\left(x_{2} y_{2}-z_{2}^{2}\right)}\right]^{2}+ ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,575 |
40. If $a, b, c$ are positive numbers, then
$$\frac{a^{3}-b^{3}}{a+b}+\frac{b^{3}-c^{3}}{b+c}+\frac{c^{3}-a^{3}}{c+a} \leqslant \frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{8}$$ | 40. (2007.02.27) Simplify
$$\begin{array}{l}
\sum \frac{a^{3}-b^{3}}{a+b}=\sum \frac{(a-b)\left[(a+b)^{2}-a b\right]}{a+b}=-\sum \frac{a b(a-b)}{a+b}= \\
\frac{a c(a-b+b-c)}{a+b}-\frac{a b(a-b)}{a+b}-\frac{b c(b-c)}{b+c}= \\
{\left[\frac{a b(a-b)}{a+c}-\frac{a b(a-b)}{a+b}\right]+\left[\frac{a c(b-c)}{a+c}-\frac{b c(b-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,576 |
41. Let $a, b, c$ be non-negative numbers, no two of which are zero. Prove that
$$\frac{a^{2}}{(2 a+b)(2 a+c)}+\frac{b^{2}}{(2 b+c)(2 b+a)}+\frac{c^{2}}{(2 c+a)(2 c+b)} \leqslant \frac{1}{3}$$ | 41. (2007.04.03) Simplify: After removing the denominators, we need to prove
$$3 \sum a^{2}(2 b+c)(2 b+a)(2 c+a)(2 c+b) \leqslant \prod(2 a+b)(2 a+c)$$
The left side of equation (1) $=3 \sum a^{2}\left[3 b \sum a+(b-a)(b-c)\right]\left[3 c \sum a+(c-a)(c-b)\right]=$
$$\begin{array}{l}
3 a b c \sum a\left[3 \sum a+\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,577 |
44. Let $a, b, c$ be non-negative numbers such that $a^{2}+b^{2}+c^{2}=1$. Prove that
$$\frac{1}{3+a^{2}-2 b c}+\frac{1}{3+b^{2}-2 c a}+\frac{1}{3+c^{2}-2 a b} \leqslant \frac{9}{8}$$ | 44. (2007.05.02) Simplify: After removing the denominator and rearranging, the original expression is equivalent to
$$\begin{array}{l}
9\left[36\left(\sum a^{2}\right)^{3}-24\left(\sum a^{2}\right)^{2} \sum b c+3 \sum a^{2} \sum b^{2} c^{2}+18 a b c \sum a \sum a^{2}-\right. \\
\left.2 \sum b^{3} c^{3}+4 a b c \sum a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,580 |
45. If $a, b, c$ are positive numbers, then
$$\frac{4 a^{2}-b^{2}-c^{2}}{a(b+c)}+\frac{4 b^{2}-c^{2}-a^{2}}{b(c+a)}+\frac{4 c^{2}-a^{2}-b^{2}}{c(a+b)} \leqslant 3$$ | 45. (2007.03.31) Proof: Since
$$\begin{array}{l}
3-\sum \frac{4 a^{2}-b^{2}-c^{2}}{a(b+c)}=3-\sum \frac{4 a^{2}-\frac{1}{2}(b+c)^{2}-\frac{1}{2}(b-c)^{2}}{a(b+c)}= \\
\left(\sum \frac{b+c}{2 a}-3\right)+\left(6-\sum \frac{4 a}{b+c}\right)+\sum \frac{(b-c)^{2}}{2 a(b+c)}= \\
\frac{1}{2 a b c} \sum a(b-c)^{2}-\frac{2}{\p... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,581 |
46. If $a, b, c$ are positive numbers such that $a b c=1$, then
$$a^{2}+b^{2}+c^{2}+6 \geqslant \frac{3}{2}\left(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$ | 46. (2007.01. 25) Simplify: Let $s_{1}=a+b+c, s_{2}=bc+ca+ab, s_{3}=abc=1$, then the original expression is equivalent to
Since
$$\begin{array}{c}
2 s_{1}^{2}-3 s_{1}-7 s_{2}+12 \geqslant 0 \\
s_{1}^{3}-4 s_{1} s_{2}+9 s_{3} \geqslant 0
\end{array}$$
(see Example 14 in the first chapter "Proving Inequalities by Equiva... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,582 |
47. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive numbers such that $a_{1}+a_{2}+\cdots+a_{n}=n$.
Prove that
$$a_{1} a_{2} \cdots a_{n}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}} \cdots+\frac{1}{a_{n}}-n+3\right) \leqslant 3$$ | 47. (2007.08.29, Excerpt from the original book) Proof: We use induction. When $n=2$, the original inequality holds, because $a_{1} a_{2} \leqslant 1$ and $a_{1}+a_{2}=2$.
Assume $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$, and let $E_{n}\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ denote the left side of th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,583 |
48. Let \(a, b, c\) be the side lengths of a triangle. If \(a^{2}+b^{2}+c^{2}=3\), then
$$a b+b c+c a \geqslant 1+2 a b c$$ | 48. (2007.04.04) Simplify and prove: From $\sum a^{2}=3$, we know that the original inequality is equivalent to
$$\begin{array}{l}
3 \sum a^{2} \cdot \sum b c-\left(\sum a^{2}\right)^{2} \geqslant 6 a b c \cdot \sqrt{3 \sum a^{2}} \Leftrightarrow \\
\sqrt{3 \sum a^{2}}\left(3 \sum b c-\sum a^{2}\right) \geqslant 18 a b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,584 |
49. Let \(a, b, c\) be the side lengths of a triangle. If \(a^{2}+b^{2}+c^{2}=3\), then
$$a+b+c \geqslant 2+abc$$ | 49. (2007.04.04) Simplify and prove: From $\sum a^{3}=3$, we know that the original inequality is equivalent to
$$\begin{array}{l}
3 \sum a \cdot \sum a^{2} \geqslant 2 \sum a^{2} \cdot \sqrt{3 \sum a^{2}}+9 a b c \Leftrightarrow \\
\left(3 \sum a \cdot \sum a^{2}-9 a b c\right)^{2} \geqslant 12\left(\sum a^{2}\right)^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,585 |
2. On August 23, 2006, when I was teaching at South China Normal University High School, student Yu Qiang from South China Normal University High School made the following generalization.
Let \( x_{1}, x_{2}, y_{1}, y_{2}, z_{1}, z_{2} \in \mathbf{R}^{+}, w_{1}, w_{2} \in \mathbf{R} \), and \( x_{1} y_{1} z_{1}-w_{1}^... | Let \( u_{1}^{3}=x_{1} y_{1} z_{1}-w_{1}^{3}, u_{2}^{3}=x_{2} y_{2} z_{2}-w_{2}^{3} \), then
the right side of equation (4) \(\geqslant 2 \sqrt{\frac{1}{u_{1}^{3} u_{2}^{3}}}\)
Therefore, to prove equation (4), it suffices to prove
$$\prod\left(x_{1}+x_{2}\right) \geqslant\left(w_{1}+w_{2}\right)^{3}+8 \sqrt{u_{1}^{3} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,586 |
50. If $a, b, c$ are the side lengths of a non-isosceles triangle, then
(a) $\left|\frac{a+b}{a-b}+\frac{b+c}{b-c}+\frac{c+a}{c-a}\right|>5$;
(b) $\left|\frac{a^{2}+b^{2}}{a^{2}-b^{2}}+\frac{b^{2}+c^{2}}{b^{2}-c^{2}}+\frac{c^{2}+a^{2}}{c^{2}-a^{2}}\right|>3$. | 50. (2007.04.05) Prove briefly: (a) Let $a>b>c$, and $a=c+\alpha+\beta, b=c+\alpha$. $\alpha$, $\beta \in \mathbf{R}^{+}$, then from $b+c>a$ we have $c>\beta$, thus
$$\begin{array}{l}
\left|\sum \frac{a+b}{a-b}\right|-5=\frac{\sum a(b-c)^{2}}{(a-b)(b-c)(a-c)}-5= \\
\frac{(c+\alpha+\beta) \alpha^{2}+(c+\alpha)(\alpha+\b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,587 |
53. If $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6} \in\left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]$, then
$$\frac{a_{1}-a_{2}}{a_{2}+a_{3}}+\frac{a_{2}-a_{3}}{a_{3}+a_{4}}+\cdots+\frac{a_{6}-a_{1}}{a_{1}+a_{2}} \geqslant 0$$ | 53. (2007.08.28, excerpt from the original book) Simplified proof: The original expression is equivalent to
$$\sum\left(\frac{a_{1}-a_{2}}{a_{2}+a_{3}}+\frac{1}{2}\right) \geqslant 3 \Leftrightarrow \sum \frac{2 a_{1}-a_{2}+a_{3}}{a_{2}+a_{3}} \geqslant 6$$
Since \(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6} \in\left[\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,590 |
55. Let \(x, y, z\) be positive numbers such that \(x+y+z \geqslant 3\). Then
\[
\frac{1}{x^{3}+y+z}+\frac{1}{x+y^{3}+z}+\frac{1}{x+y+z^{3}} \leqslant 1
\] | 55. (2007.03.05) Simplify and prove: First prove: If \(a^{2}+b^{2}+c^{2} \geqslant 3\), then
Because
$$\left(\sum a^{3}\right)^{2} \geqslant 3+2 \sum a^{4}$$
$$a^{2}+b^{2}+c^{2} \geqslant 3$$
So
$$3+2 \sum a^{4} \leqslant 3\left(\frac{\sum a^{2}}{3}\right)^{3}+2 \frac{\sum a^{2}}{3} \sum a^{4}$$
Therefore, it suffic... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,592 |
56. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive numbers such that $x_{1} x_{2} \cdots x_{n} \geqslant 1$. If $a>1$, then
$$\sum \frac{x_{1}^{a}}{x_{1}^{\alpha}+x_{2}+\cdots+x_{n}} \geqslant 1$$ | 56. (2007.08. 29 , Excerpt from the original book) Simplified proof: First, we observe that it suffices to prove the following case:
$$x_{1} x_{2} \cdots x_{n}=1$$
To illustrate this case, let \( r=\sqrt[n]{x_{1} x_{2} \cdots x_{n}} \) and \( y_{i}=\frac{x_{i}}{r} \) for \( i=1,2, \cdots, n \), noting that \( r \geqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,593 |
57. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive numbers such that $x_{1}, x_{2} \cdots x_{n} \geqslant 1$. If $n \geqslant 3$ and $\frac{-2}{n-2} \leqslant a < 1$, then
$$\sum \frac{x_{1}^{\alpha}}{x_{1}^{\alpha}+x_{2}+\cdots+x_{n}} \leqslant 1$$ | 57. (2007.08.30, Excerpt from the original book) Proof: The first part of the proof is similar to the inequality proof of the previous problem (Problem $56$). Finally, we need to prove the inequality
$$x_{2}+\cdots+x_{n} \geqslant\left(x_{2} \cdots x_{n}\right)^{\frac{1-p}{n-1}}\left(x_{2}^{p}+\cdots+x_{n}^{p}\right)$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,594 |
58. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive numbers such that $x_{1} x_{2} \cdots x_{n} \geqslant 1$. If $a>1$, then
$$\sum \frac{x_{1}}{x_{1}^{a}+x_{2}+\cdots+x_{n}} \leqslant 1$$ | 58. (2007.08.30, Excerpt from the original book) Proof: We will consider two cases: $10$, according to Bernoulli's inequality, we have
Therefore
$$\begin{array}{c}
x^{\alpha}=[1+(x-1)]^{\alpha} \geqslant 1+\alpha(x-1) \\
x^{\alpha}-x+n \geqslant n-\alpha+1+(\alpha-1) x>0
\end{array}$$
Thus, it suffices to prove
$$\su... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,595 |
59. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive numbers such that $x_{1} x_{2} \cdots x_{n} \geqslant 1$. If $-1- \frac{2}{n-2} \leqslant a < 1$, then
$$\sum \frac{x_{1}}{x_{1}^{\alpha}+x_{2}+\cdots+x_{n}} \geqslant 1$$ | 59. (2007.08.30, Excerpt from the original book) Simplified proof: It suffices to prove the case when $x_{1} x_{2} \cdots x_{n}=1$ (similar to problem 58). According to the Cauchy-Schwarz (Cauchy) inequality, we have
$$\begin{array}{l}
\sum \frac{x_{1}}{x_{1}^{\alpha}+x_{2}+\cdots+x_{n}} \geqslant \frac{\left(x_{1}+x_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,596 |
Let $x_{i}, y_{i}>0, x_{i} y_{i}-z_{i}^{2}>0, z_{i} \in \mathbf{R}, i=1,2, \cdots, n$, then
$$\frac{n^{3}}{\sum x_{1} \cdot \sum y_{1}-\left(\sum z_{1}\right)^{2}} \leqslant \sum \frac{1}{x_{1} y_{1}-z_{1}^{2}}$$ | $$\begin{array}{l}
{\left[\sum x_{1} \cdot \sum y_{1}-\left(\sum z_{1}\right)^{2}\right] \cdot \sum \frac{1}{x_{1} y_{1}-z_{1}^{2}} \geqslant\left[\left(\sum \sqrt{x_{1} y_{1}}\right)^{2}-\left(\sum z_{1}\right)^{2}\right] \cdot} \\
\sum \frac{1}{x_{1} y_{1}-z_{1}^{2}}=\sum\left(\sqrt{x_{1} y_{1}}+z_{1}\right) \cdot \s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,597 |
60. Let $n \geqslant 3$ be an integer and let $p$ be a real number such that $1<p<n-1$. If $0<x_{1}, x_{2}, \cdots, x_{n} \leqslant \frac{p n-p-1}{p(n-p-1)}$ such that $x_{1} x_{2} \cdots x_{n}=1$, then
$$\frac{1}{1+p x_{1}}+\frac{1}{1+p x_{2}}+\cdots+\frac{1}{1+p x_{n}} \geqslant \frac{n}{1+p}$$ | 60. (2007.08.30, Excerpt from the original book) Proof: We will prove by induction that
$$\frac{1}{1+q x_{1}}+\frac{1}{1+q x_{2}}+\cdots+\frac{1}{1+q x_{n}} \geqslant \frac{n}{1+q}$$
for any $q \geqslant p$.
For $n=2$, the inequality simplifies to
$$\frac{(q-1)\left(x_{1}-1\right)^{2}}{\left(1+q x_{1}\right)\left(1+q ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,598 |
62. Let \(a, b, c\) be positive numbers such that \(a b c=1\). Prove that
$$a^{2}+b^{2}+c^{2}+9(a b+b c+c a) \geqslant 10(a+b+c)$$ | 62. (2007.03.10) Proof: From Problem 55 in Chapter 1, we have $\sum b c \geqslant \frac{9 \sum a}{\sum a+6}$. Therefore, we only need to prove
$$\begin{array}{l}
\sum a^{2}+9 \sum b c-10 \sum a=\left(\sum a\right)^{2}+7 \sum b c-10 \sum a \geqslant \\
\left(\sum a\right)^{2}+7 \cdot \frac{9 \sum a}{\sum a+6}-10 \sum a ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,600 |
63. Let $a, b, c$ be non-negative numbers such that $a b+b c+c a=3$. Prove that
$$\frac{a\left(b^{2}+c^{2}\right)}{a^{2}+b c}+\frac{b\left(c^{2}+a^{2}\right)}{b^{2}+c a}+\frac{c\left(a^{2}+b^{2}\right)}{c^{2}+a b} \geqslant 3$$ | 63. (2007.06.03) Proof: Since
$$\left[\sum \frac{a\left(b^{2}+c^{2}\right)}{a^{2}+b c}\right]^{2} \geqslant 3 \sum \frac{a b\left(b^{2}+c^{2}\right)\left(a^{2}+c^{2}\right)}{\left(a^{2}+b c\right)\left(b^{2}+c a\right)}$$
To prove the original inequality, it suffices to prove
$$\begin{array}{l}
\sum \frac{a b\left(b^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,601 |
64. If $a, b, c$ are positive numbers, then
$$a+b+c+\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \geqslant \frac{6\left(a^{2}+b^{2}+c^{2}\right)}{a+b+c}$$ | 64. (2007.03.10) Proof: Since
$$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}=\sum\left[-b+2 a+\frac{(a-b)^{2}}{b}\right]=\sum a+\sum \frac{(a-b)^{2}}{b}$$
Therefore,
$$\begin{array}{l}
\sum a \cdot\left(2 \sum a+\sum \frac{a^{2}}{b}\right)=2\left(\sum a\right)^{2}+\sum a \cdot \sum \frac{(a-b)^{2}}{b} \geqslant \\
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,602 |
65. If $a, b, c$ are positive numbers, then
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{3\left(a^{3}+b^{3}+c^{3}\right)}{2\left(a^{2}+b^{2}+c^{2}\right)}$$ | 65. (2007.02.24) Simplify
$$\begin{array}{l}
\sum \frac{a^{2}}{b+c}-\frac{3 \sum a^{3}}{2 \sum a^{2}}=\sum\left[a-\frac{1}{4}(b+c)+\frac{(2 a-b-c)^{2}}{4(b+c)}\right]-\frac{3 \sum a^{3}}{2 \sum a^{2}}= \\
\frac{1}{2} \sum a+\frac{1}{4} \sum \frac{[(a-b)+(a-c)]^{2}}{(b+c)}-\frac{3 \sum a^{3}}{2 \sum a^{2}}= \\
\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,603 |
66. If $a, b, c$ are given non-negative numbers, find the minimum value $E(a, b, c)$ of the expression
$$E=\frac{a x}{y+z}+\frac{b y}{z+x}+\frac{c z}{x+y}$$
for any positive numbers $x, y, z$. | 66. (2007.08.30, Excerpt from the original book) Proof: Assume $a=\max \{a, b, c\}$, then
$$\begin{aligned}
E= & \sum \frac{a x}{y+z}=\sum \frac{a(x+y+z)-a(y+z)}{y+z}= \\
& (x+y+z) \sum \frac{a}{y+z}-\sum a= \\
& \frac{1}{2}\left[\sum(y+z)\right]\left(\sum \frac{a}{y+z}\right)-\sum a
\end{aligned}$$
According to the C... | null | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,604 |
68. Let $a, b, c$ be non-negative numbers such that $a+b+c=3$. Prove that
$$\left(a^{2}-a b+b^{2}\right)\left(b^{2}-b c+c^{2}\right)\left(c^{2}-c a+a^{2}\right) \leqslant 12$$ | 68. (2007.04. 17) Proof: Without loss of generality, assume $a \geqslant b \geqslant c \geqslant 0$, then $c^{2}-b c \leqslant 0, c^{2}-c a \leqslant 0, a+b \leqslant 3$, then
$$\begin{array}{l}
\left(a^{2}-a b+b^{2}\right)\left(b^{2}-b c+c^{2}\right)\left(c^{2}-c a+a^{2}\right) \leqslant \\
{\left[(a+b)^{2}-3 a b\righ... | 12 | Inequalities | proof | Yes | Yes | inequalities | false | 734,606 |
Example 3 In $\triangle A B C$, prove
$$\left(\sum \sin \frac{A}{2}\right)^{2} \geqslant \frac{\sqrt{3}}{2} \sum \sin A$$ | To prove that by making angle transformations such as $\frac{A}{2} \rightarrow \frac{\pi}{2}-A$, etc., the equation (6) transforms into proving
$$\left(\sum \cos A\right)^{2} \geqslant \frac{\sqrt{3}}{2} \sum \sin 2 A$$
where $A, B, C$ are the three interior angles of an acute $\triangle ABC$. At this point,
$$\begin{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,608 |
71. If $a, b, c$ are non-negative numbers, then
$$\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right) \geqslant \frac{15}{16}(1+a+b+c)^{2}$$ | 71. (2007.06. 20) Briefly prove: In Example 33 of Chapter 7 "Other Inequality Proof Examples", take $n=5$, and let
$$\begin{array}{l}
\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right)=\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right) \\
\left(b_{1}, b_{2}, b_{3}, b_{4}, b_{5}\right)=\left(a^{2},... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,610 |
72. Let \(a, b, c, d\) be positive real numbers such that \(a b c d=1\). Prove that
\[
\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)\left(1+d^{2}\right) \geqslant(a+b+c+d)^{2}
\] | $$\begin{array}{l}
\prod\left(1+a^{2}\right)-\left(\sum a\right)^{2}=1+\sum a^{2}+\left(a^{2} b^{2}+a^{2} c^{2}+a^{2} d^{2}+b^{2} c^{2}+b^{2} d^{2}+c^{2} d^{2}\right)+ \\
\sum b^{2} c^{2} d^{2}+1-\sum a^{2}-2(a b+a c+a d+b c+b d+c d)= \\
(a b-1)^{2}+(a c-1)^{2}+(a d-1)^{2}+(b c-1)^{2}+ \\
(b d-1)^{2}+(c d-1)^{2}+\sum b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,611 |
73. If $x_{1}, x_{2}, \cdots, x_{n}$ are non-negative numbers, then
$$x_{1}+x_{2}+\cdots+x_{n} \geqslant(n-1) \sqrt[n]{x_{1} x_{2} \cdots x_{n}}+\sqrt{\frac{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}{n}}$$ | 73. (2007.08.28, Excerpt from the original book) Simplified proof: Let $x=\frac{1}{n} \sum x_{1}, y=\sqrt{\frac{2 \sum_{i<i<j<} x_{i} x_{j}}{n(n-1)}}$, $z=\sqrt[n]{x_{1} x_{2} \cdots x_{n}}$, then the original expression is to prove
$$\begin{array}{l}
n x-(n-1) z \geqslant \sqrt{\frac{(n x)^{2}-n(n-1) y^{2}}{n}} \Leftr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,612 |
74. If $k$ is a real number and $x_{1}, x_{2}, \cdots, x_{n}$ are positive numbers, then
$$\begin{array}{l}
(n-1)\left(x_{1}^{n+k}+x_{2}^{n+k}+\cdots+x_{n}^{n+k}\right)+x_{1} x_{2} \cdots x_{n}\left(x_{1}^{k}+x_{2}^{k}+\cdots+x_{n}^{k}\right) \geqslant \\
\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(x_{1}^{n+k-1}+x_{2}^{... | 74. (2007.08.30, Excerpt from the original book) Proof: Using induction, when $n=2$, the original inequality obviously holds.
When $n=3$, we obtain the Schur inequality
$$\sum x_{1}^{k+1}\left(x_{1}-x_{2}\right)\left(x_{1}-x_{2}\right) \geqslant 0$$
Assume the original inequality holds for $n$ numbers, we will prove ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,613 |
75. Let $a, b, c$ be non-negative numbers, no two of which are zero. Prove that
$$\frac{a^{4}}{a^{3}+b^{3}}+\frac{b^{4}}{b^{3}+c^{3}}+\frac{c^{4}}{c^{3}+a^{3}} \geqslant \frac{a+b+c}{2}$$ | 75. (2007.04.13) Prove: (1) If $a \geqslant b \geqslant c$, then
$$\begin{array}{l}
\left(a^{6} b^{4}+b^{6} c^{4}+c^{6} a^{4}\right)-\left(a^{4} b^{6}+b^{4} c^{6}+c^{4} a^{6}\right)= \\
\left(a^{2}-b^{2}\right)\left(b^{2}-c^{2}\right)\left(a^{2}-c^{2}\right) \sum b^{2} c^{2} \geqslant 0
\end{array}$$
Thus, we have
$$\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,614 |
Theorem 1 Let $x, y, z, w \in \mathbf{R}^{+}$, and angles $\alpha, \beta, \gamma, \theta$ satisfy
$$\alpha+\beta+\gamma+\theta=(2 k+1) \pi \quad(n \in \mathbf{Z})$$
Then
$$x \sin \alpha+y \sin \beta+z \sin \gamma+w \sin \theta \leqslant \sqrt{\frac{(x y+z w)(y z+x w)(z x+y w)}{x y z w}}$$
Equality in (2) holds if and... | Proof: Let \( u = x \sin \alpha + y \sin \beta, v = z \sin \gamma + w \sin \theta \), then
\[
\begin{aligned}
u^{2} = & (x \sin \alpha + y \sin \beta)^{2} \leqslant (x \sin \alpha + y \sin \beta)^{2} + (x \cos \alpha - y \cos \beta)^{2} = \\
& x^{2} + y^{2} - 2 x y \cos (\alpha + \beta)
\end{aligned}
\]
Since \( x, y ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,615 |
Example 4 (Self-created, $1992,03,25$) Let $E, F$ be any two points on the rays $A C$, $A B$ starting from $A$ in $\triangle A B C$, then
$$|A B-A C|+|A E-A F| \geqslant|B E-C F|$$
Equality in (7) holds if and only if $A B=A C$ and $A E=A F$. | Proof: Without loss of generality, assume $A C \geqslant A B$ (Figure 1). Take a point $D$ on $A C$ such that $A D=A B$. Take a point $G$ on $A B$ (or its extension) such that $A G=A E$, then
$$\begin{aligned}
|A B-A C|+ & |A E-A F|=|C D|+|F G| \geqslant \\
& |C F-F D|+|F D-D G| \geqslant \\
& |C F-F D+F D-D G|= \\
& |... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,619 |
Example 1 Find the maximum value of the function $y=\sin \alpha+2 \sin \beta+3 \sin \gamma+4 \sin \theta$, where $\alpha+\beta+$ $\gamma+\theta=\pi$. | Solving: In equation (2), taking $x=1, y=2, z=3, w=4$, we get
$$\sin \alpha+2 \sin \beta+3 \sin \gamma+4 \sin \theta \leqslant \sqrt{\frac{385}{6}}$$
That is, $y$ has a maximum value of $\sqrt{\frac{385}{6}}$. | \sqrt{\frac{385}{6}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,627 |
Example 2 Find the maximum value of the function $y=\sqrt{2} \sin \alpha+\sqrt{5} \sin \beta+\sqrt{10} \sin \gamma$, where $\alpha+\beta+$ $\gamma=\pi$, and find the angles $\alpha, \beta, \gamma$ at which the maximum value is achieved. | In Theorem 2, take $x=\sqrt{2}, y=\sqrt{5}, z=\sqrt{10}$, substitute into equation ( $※$ ), and simplify to get
$$w^{3}+4 w^{2}-5=0$$
This equation has a positive root $w=1$. Substituting $x=\sqrt{2}, y=\sqrt{5}, z=\sqrt{10}, w=1$ into inequality $\left(10^{\prime}\right)$, we get
$$\sqrt{2} \sin \alpha+\sqrt{5} \sin ... | 6 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,628 |
Example 3 Find the maximum value of the function $\gamma=\sqrt{2} \sin \alpha+\sqrt{3} \sin \beta+\sqrt{6} \sin \gamma$, where $\alpha+\beta+\gamma=$ $180^{\circ}$, and find the angles $\alpha, \beta, \gamma$ at which the maximum value is achieved. | In Theorem 2, let $x=\sqrt{2}, y=\sqrt{3}, z=\sqrt{6}$, and substitute into equation (※), and simplify to get $w^{3}+3 w^{2}-3=0$. According to the formula for the roots of a cubic equation, the positive root can be found as $w=\frac{\sqrt{3}}{2} \sec 10^{\circ}$. Substituting $x=\sqrt{2}, y=\sqrt{3}, z=\sqrt{6}, w=\fr... | \frac{\sqrt{2}}{4}\left(\csc 10^{\circ}\right)^{\frac{3}{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,629 |
Example 5 Let $x, y, z \in \mathbf{R}^{+}$, then
$$x^{2} y+y^{2} z+z^{2} x+x y z \leqslant \frac{4}{27}(x+y+z)^{3}$$ | To prove that $x^{2} y+y^{2} z+z^{2} x-\left(x y^{2}+y z^{2}+z x^{2}\right)=(x-y)(y-z)(x-z)$, we know that when $x \geqslant y \geqslant z$, $x^{2} y+y^{2} z+z^{2} x \geqslant x y^{2}+y z^{2}+z x^{2}$. Therefore, we only need to prove that equation (8) holds when $x \geqslant y \geqslant z$. In this case,
$$\begin{alig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,630 |
Theorem In three-dimensional Euclidean space, for any vectors $a_{i}, b_{i}(i=1,2, \cdots, n)$, the inequality always holds:
$$\begin{array}{l}
\sum_{i=1}^{n}\left(\left|a_{i}\right|^{2} \cdot\left|b_{i}\right|^{2}\right)+2 \sum_{i<i<j<n}\left(a_{i} \cdot a_{j}\right)\left(b_{i} \cdot b_{j}\right) \geqslant \\
\left.\l... | Prove that in a three-dimensional Cartesian coordinate system, let $a_{i}=\left(x_{i 1}, x_{i 2}, x_{\mathcal{B}}\right)(i=1,2, \cdots$,
$$\begin{array}{l}
n), c_{i}=x_{1 j} b_{1}+x_{2 j} b_{2}+\cdots+x_{v} b_{n}(i=1,2,3) \text {, then } \\
\left|c_{1}\right|^{2}+\left|c_{2}\right|^{2}+\left|c_{3}\right|^{2}= \\
\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,634 |
Inference 3 The spatial vectors $a_{i}$ and $b_{i}(i=1,2,3,4)$ satisfy $a_{1}+a_{2}+a_{3}+a_{4}=0, b_{1}+$ $b_{2}+b_{3}+b_{4}=0$, then
$$\begin{array}{l}
-\left(b_{1} \cdot b_{2}\right)\left(a_{3} a_{4}\right)^{2}-\left(b_{1} \cdot b_{3}\right)\left(a_{2} a_{4}\right)^{2}- \\
\left(b_{1} \cdot b_{4}\right)\left(a_{2} a... | Substitute $a_{4}=-a_{1}-a_{2}-a_{3}$ into the left side of equation (4), expand and rearrange, and note that $b_{1}+b_{2}+b_{3}+b_{4}=0$, then we get
the left side of equation (4) $=b_{1}^{2}\left(a_{2} a_{3}\right)^{2}+b_{2}^{2}\left(a_{3} a_{1}\right)^{2}+b_{3}^{2}\left(a_{1} a_{2}\right)^{2}+$
$$\begin{array}{l}
2\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,637 |
Inference 4 The spatial vectors $b_{i}$ satisfy $b_{1}+b_{2}+b_{3}+b_{4}=0, \lambda_{i}(i=1,2,3,4)$ are any real numbers such that the sum of any two is positive, then
$$\begin{array}{l}
\lambda_{1} b_{1}^{2}+\lambda_{2} b_{2}^{2}+\lambda_{3} b_{3}^{2}+\lambda_{4} b_{4}^{2} \geqslant \\
\left.3\left(\lambda_{2} \lambda... | To prove the case where $n=3$ in the theorem, and let $a_{1} \cdot a_{2}=a_{1} \cdot a_{3}=a_{2} \cdot a_{3}=\lambda_{4}$, and $a_{i}^{2}=\lambda_{i}+\lambda_{4}(i=1,2,3)$, at this time
the left side of equation (1) $=\left|a_{1}\right|^{2}\left|b_{1}\right|^{2}+\left|a_{2}\right|^{2}\left|b_{2}\right|^{2}+\left|a_{3}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,638 |
Proposition 1 (Generalization of Pedoe's Inequality in Three-Dimensional Space) For tetrahedra \(A_{1} A_{2} A_{3} A_{4}\) and \(B_{1} B_{2} B_{3} B_{4}\) with volumes \(A\) and \(B\) respectively, denote the edge lengths \(A_{i} A_{j} = a_{ij} (i, j = 1, 2, 3, 4, i \neq j\), with the convention \(a_{ij} = a_{ji}\)), t... | To prove: In three-dimensional space, let the points $A_{i}(i=1,2,3,4)$ correspond to vectors $\overrightarrow{O A_{i}}=a_{i}(i=1,2,3,4)$, then the vector $\overrightarrow{A_{i} A_{j}}=a_{j}-a_{i}(i, j=1,2,3,4)$, and
$$6 A=\left|\left(\overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}\right) \cdot \overri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,639 |
Proposition 4 The volume of the spatial tetrahedron $B_{1} B_{2} B_{3} B_{4}$ is $B, P$ is any point inside this tetrahedron, the orthogonal projections of point $P$ on the faces opposite to vertices $B_{i}(i=1,2,3,4)$ are $C_{i}(i=1,2,3,4)$ respectively, the volume of the tetrahedron formed by $C_{1}, C_{2}$, $C_{3}, ... | Prove that if the area of the face opposite vertex $B_{i}(i=1,2,3,4)$ is denoted as $s_{i}(i=1,2,3,4)$, and in equation (8) we take $x_{i}=\frac{\left|P C_{i}\right|}{s_{i}}(i=1,2,3,4)$, then the left side of equation (8) is
$$\begin{array}{l}
x_{1} s_{1}^{2}+x_{2} s_{2}^{2}+x_{3} s_{3}^{2}+x_{4} s_{4}^{2}= \\
\left|P ... | C \leqslant \frac{1}{27} B | Geometry | proof | Yes | Yes | inequalities | false | 734,640 |
Example 6 Let $a, b, c \in \mathbf{R}^{-}$, prove that
$$\begin{array}{l}
1 \leqslant \sum \sqrt{\frac{a^{3}}{a^{3}+(b+c)^{3}}} \\
\sum \sqrt{\frac{a^{3}}{a^{3}+(b+c)^{3}}} \leqslant \sqrt{2}
\end{array}$$
Equality in (9) holds if and only if $a=b=c$; equality in (10) holds if and only if one of $a, b, c$ is zero and ... | To prove:
$$\sqrt{\frac{a^{3}}{a^{3}+(b+c)^{3}}} \geqslant \frac{a^{2}}{\sum a^{2}}, \sqrt{\frac{a^{3}}{a^{3}+(b+c)^{3}}} \leqslant \frac{\sqrt{2} a^{\frac{3}{2}}}{\sum a^{\frac{3}{2}}}$$
It is easy to prove.
Note: The exponents 2 and $\frac{3}{2}$ on the right side of the inequalities can be determined using the meth... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,641 |
Lemma 1 Let $\alpha, \beta, \gamma, \theta \in[0, \pi]$, and $\alpha+\beta+\gamma+\theta=\pi$, then
$$\sin \alpha+\sin \beta+\sin \gamma+\sin \theta \leqslant 2 \sqrt{2}$$
Equality holds if and only if $\alpha=\beta=\gamma=\theta=\frac{\pi}{4}$. | $$\begin{array}{l}
\sin \alpha+\sin \beta+\sin \gamma+\sin \theta= \\
2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}+2 \sin \frac{\gamma+\theta}{2} \cos \frac{\gamma-\theta}{2} \leqslant \\
2 \sin \frac{\alpha+\beta}{2}+2 \sin \frac{\gamma+\theta}{2}= \\
2\left(\sin \frac{\alpha+\beta}{2}+\cos \frac{\alpha+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,646 |
Lemma 2 For a cyclic convex quadrilateral $ABCD$ with side lengths $a, b, c, d$, its area is $\Delta$, and the radius of the circle is $R$, then
$$\Delta=\frac{\sqrt{(a b+c d)(a c+b d)(a d+b c)}}{4 R}$$ | Proof: Let $A B=a, B C=b, C D=c, D A=d$, then
$$\cos B=\frac{a^{2}+b^{2}-A C^{2}}{2 a b}, \cos D=\frac{c^{2}+d^{2}-A C^{2}}{2 c d}$$
But, $B+D=\pi$, so we have
$$\cos B+\cos D=0$$
Therefore,
$$\begin{array}{c}
\frac{a^{2}+b^{2}-A C^{2}}{2 a b}+\frac{c^{2}+d^{2}-A C^{2}}{2 c d}=0 \\
(a b+c d) A C^{2}=\left(a^{2}+b^{2}... | proof | Geometry | proof | Yes | Yes | inequalities | false | 734,647 |
Theorem 1 If the semi-major and semi-minor axes of an ellipse are $a, b$, then the maximum area of an inscribed triangle in the ellipse is $\frac{3 \sqrt{3}}{4} a b$. | Theorem 1 Proof: In the Cartesian coordinate system, let $A_{i}\left(x_{i}, y_{i}\right)(i=1,2,3)$ be three points on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Then $\left(x_{i}, y_{i}\right)(i=1,2,3)$ satisfy the equation
$$\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1$$
According to Lemm... | \frac{3 \sqrt{3}}{4} a b | Geometry | proof | Yes | Yes | inequalities | false | 734,649 |
Lemma 1 Suppose $\left(x_{i}, y_{i}\right)(i=1,2,3)$ satisfy the equation $x^{2}+y^{2}=1$, then
$$\left\|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right\| \leqslant \frac{3 \sqrt{3}}{2}$$ | Let $A_{i}\left(x_{i}, y_{i}\right)(i=1,2,3)$, since $\left(x_{i}, y_{i}\right)(i=1,2,3)$ satisfy the equation $x^{2}+$ $y^{2}=1$, therefore the points $A_{i}(i=1,2,3)$ are all on the unit circle $x^{2}+y^{2}=1$. Additionally, according to plane geometry, among inscribed triangles in a circle, the area of the inscribed... | \frac{3 \sqrt{3}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 734,651 |
Example 7 Let $x_{1}, x_{2}, x_{3}, x_{4} \in \mathbf{R}^{+}$, prove that
$$\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\frac{x_{3}^{2}}{x_{4}}+\frac{x_{4}^{2}}{x_{1}} \geqslant 2 \sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}$$
Equality holds in (11) if and only if $x_{1}=x_{2}=x_{3}=x_{4}$. | To prove that
$$\left(\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\frac{x_{3}^{2}}{x_{4}}+\frac{x_{4}^{2}}{x_{1}}\right)\left(x_{1}^{2} x_{2}+x_{2}^{2} x_{3}+x_{3}^{2} x_{4}+x_{4}^{2} x_{1}\right) \geqslant\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}\right)^{2}$$
it suffices to prove that
$$\begin{array}{c}
\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,652 |
Lemma 2 The volume of tetrahedron $ABCD$ is $V$, and the radius of its circumscribed sphere is $R$, then
$$V \leqslant \frac{8 \sqrt{3}}{27} R^{3}$$
Equality holds if and only if tetrahedron $ABCD$ is a regular tetrahedron.
In the article "On the Inequality Relations of Tetrahedra" in Issue 12, 1984 of *Mathematical C... | Proof: Let the edge lengths $AB = a, CD = a'$, and the angle between $AB$ and $CD$ be $\alpha$. Let $O$ be the center of the circumsphere of the tetrahedron. Draw perpendiculars from $O$ to $AB$ and $CD$, with feet at $Q$ and $Q'$, respectively. Then $Q$ and $Q'$ are the midpoints of segments $AB$ and $CD$, respectivel... | V \leq \frac{8 \sqrt{3}}{27} R^3 | Inequalities | proof | Yes | Yes | inequalities | false | 734,653 |
Lemma 3 Let $\left(x_{i}, y_{i}, z_{i}\right)(i=1,2,3,4)$ satisfy the equation $x^{2}+y^{2}+z^{2}=1$, then
$$\left\|\begin{array}{llll}
x_{1} & y_{1} & z_{1} & 1 \\
x_{2} & y_{2} & z_{2} & 1 \\
x_{3} & y_{3} & z_{3} & 1 \\
x_{4} & y_{4} & z_{4} & 1
\end{array}\right\| \leqslant \frac{16 \sqrt{3}}{9}$$ | Proof: Let $A_{i}\left(x_{i}, y_{i}, z_{i}\right)(i=1,2,3,4)$. Since $\left(x_{i}, y_{i}, z_{i}\right)(i=1,2,3,4)$ satisfy the equation $x^{2}+y^{2}+z^{2}=1$, it follows that the points $A_{i}(i=1,2,3,4)$ are all on the sphere $x^{2}+y^{2}+z^{2}=1$, meaning that the tetrahedron $A_{1} A_{2} A_{3} A_{4}$ is an inscribed... | \left\|\begin{array}{llll}
x_{1} & y_{1} & z_{1} & 1 \\
x_{2} & y_{2} & z_{2} & 1 \\
x_{3} & y_{3} & z_{3} & 1 \\
x_{4} & y_{4} & z_{4} & 1
\end{array}\right\| \leqslant \frac | Inequalities | proof | Yes | Yes | inequalities | false | 734,654 |
Theorem The maximum area of an inscribed convex $n$-sided polygon in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $\frac{1}{2} n a b \sin \frac{2 \pi}{n}(n \in$ $\mathrm{N}, n \geqslant 3)$. If the vertices of this $n$-sided polygon, in counterclockwise order, are $A_{k}\left(x_{k}, y_{k}\right)(k=1,2, \c... | Below we will prove the theorem proposed in this paper.
Let $A_{k}\left(a \cos \theta_{k}, b \sin \theta_{k}\right)(k=1,2, \cdots, n, n \geqslant 3)$, be the vertices of an inscribed $n$-sided polygon in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ in counterclockwise order (i.e., the eccentric angles $\thet... | proof | Geometry | proof | Yes | Yes | inequalities | false | 734,655 |
Lemma Let $0 \leqslant \theta_{1}<\theta_{2}<\cdots<\theta_{n} \leqslant 2 \pi$, then
$$\sin \left(\theta_{2}-\theta_{1}\right)+\sin \left(\theta_{3}-\theta_{2}\right)+\cdots+\sin \left(\theta_{n}-\theta_{n-1}\right)+\sin \left(\theta_{1}-\theta_{n}\right) \leqslant n \sin \frac{2 \pi}{n}$$
Equality holds in (1) if an... | Let $A_{k}\left(\cos \theta_{k}, \sin \theta_{k}\right)\left(k=1,2, \cdots, n, n \geqslant 3\right.$, and $0 \leqslant \theta_{1}<\theta_{2}<\cdots<\theta_{n} \leqslant 2 \pi$) be $n$ points on the unit circle $x^{2}+y^{2}=1$, forming a counterclockwise inscribed $n$-sided polygon $B_{1} B_{2} \cdots B_{n}$, whose area... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,656 |
Theorem For a tetrahedron $ABCD$ with the circumradius $R$, the areas of the triangles opposite to vertices $A, B, C, D$ are $S_{A}, S_{B}, S_{C}, S_{D}$, respectively, then
$$R^{4} \geqslant \frac{3}{16}\left(S_{A}^{2}+S_{B}^{2}+S_{C}^{2}+S_{D}^{2}\right)$$ | We know that if the three sides of a triangle are $a, b, c$, and the area is $\Delta$, then we have
$$\Delta^{2}=\frac{1}{16}\left[2\left(b^{2} c^{2}+c^{2} a^{2}+a^{2} b^{3}\right)-\left(a^{4}+b^{4}+c^{4}\right)\right]$$
From this, we can obtain
$$S_{A}^{2}=\frac{1}{16}\left[2\left(B C^{2} \cdot B D^{2}+B D^{2} \cdot ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,659 |
Lemma Assume that real numbers $x^{\prime}, y^{\prime}, z^{\prime}$ and $x, y, z$ simultaneously satisfy $x^{\prime}+y^{\prime}+z^{\prime}>0, x+y+z>$ $0, y^{\prime} z^{\prime}+z^{\prime} x^{\prime}+x^{\prime} y^{\prime}>0, y z+z x+x y>0$, then
$$\begin{array}{l}
\left(y^{\prime}+z^{\prime}\right) x+\left(z^{\prime}+x^{... | Prove that
$$
\begin{array}{c}
\left(x^{\prime}+y^{\prime}+z^{\prime}\right)^{2}=x^{\prime 2}+y^{\prime 2}+z^{\prime 2}+\left[\sqrt{\left(y^{\prime} z^{\prime}+z^{\prime} x^{\prime}+x^{\prime} y^{\prime}\right)}\right]^{2} \\
(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+[\sqrt{(y z+z x+x y)}]^{2}
\end{array}
$$
(Note that here \( y^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,660 |
Proposition $1 \triangle A B C$ has side lengths $a, b, c$, and $\Delta$ is its area. Real numbers $x, y, z$ simultaneously satisfy $x+y+z>0, y z+z x+x y>0$. Then,
$$x a^{2}+y b^{2}+z c^{2} \geqslant 4 \sqrt{u v+v \lambda+\lambda u} \Delta$$
Equality in (2) holds if and only if $\frac{u+v}{a^{2}}=\frac{v+\lambda}{b^{2... | To prove that in equation (1), by setting $\frac{y^{\prime}+z^{\prime}}{2}=a^{2}, \frac{z^{\prime}+x^{\prime}}{2}=b^{2}, \frac{x^{\prime}+y^{\prime}}{2}=c^{2}$, we have
$$\begin{array}{c}
\left\{\begin{array}{l}
x^{\prime}=-a^{2}+b^{2}+c^{2} \\
y^{\prime}=a^{2}-b^{2}+c^{2} \\
z^{\prime}=a^{2}+b^{2}-c^{2}
\end{array}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,661 |
Proposition 2 Pedoe's Inequality.
Let the side lengths of $\triangle ABC$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ be $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$, and their areas be $\Delta$ and $\Delta^{\prime}$, respectively. Then,
$$\left(-a^{\prime 2}+b^{\prime 2}+c^{\prime 2}\right) a^{2}+\left(a^{\... | To prove that in equation (1), let
and
$$\begin{array}{c}
\frac{y^{\prime}+z^{\prime}}{2}=a^{2}, \frac{z^{\prime}+x^{\prime}}{2}=b^{2}, \frac{x^{\prime}+y^{\prime}}{2}=c^{2} \\
x=-a^{2}+b^{2}+c^{2}, y=a^{2}-b^{2}+c^{2}, z=a^{2}+b^{2}-c^{2} \\
\sqrt{\left(y^{\prime} z^{\prime}+z^{\prime} x^{\prime}+x^{\prime} y^{\prime... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,662 |
Example 8 Let $x, y, z \in \mathbf{R}^{-}$, prove that
$$\left(\sum x\right)^{5} \geqslant 81 x y z \cdot \sum x^{2}$$
Equality in (13) holds if and only if $x=y=z$. | Let $u=\sum x, v=\sum y z, w=x y z$, then
$$v^{2} \geqslant 3 u w$$
Thus,
$$\begin{aligned}
u^{5}+162 v w= & u^{5}+81 v w+81 v w \geqslant 3\left(u^{5} \cdot 81 v w \cdot 81 v w\right)^{\frac{1}{3}} \geqslant \\
& 3\left(u^{5} \cdot 81^{2} \cdot w^{2} \cdot 3 u w\right)^{\frac{1}{3}}= \\
& 81 u^{2} w
\end{aligned}$$
... | u^{5} \geqslant 81 w\left(u^{2}-2 v\right) | Inequalities | proof | Yes | Yes | inequalities | false | 734,664 |
Proposition 3 Let $a, b, c$ be the side lengths of triangle $ABC$, $\Delta$ its area, and $\lambda, u, v$ any real numbers, then
$$(u v+v \lambda+\lambda u) a b c \geqslant 4 \sqrt{\lambda u v\left(\lambda a^{2}+u b^{2}+v c^{2}\right)} \Delta$$
Equality in (4) holds if and only if $\lambda a^{2}\left(-a^{2}+b^{2}+c^{2... | To prove in the theorem, let $x^{\prime}=\frac{u v}{a^{2}}, y^{\prime}=\frac{v \lambda}{b^{2}}, z^{\prime}=\frac{\lambda u}{c^{2}}, x=-a^{2}+b^{2}+c^{2}, y=$ $a^{2}-b^{2}+c^{2}, z=a^{2}+b^{2}-c^{2}$, then
$$\sqrt{\left(y^{\prime} z^{\prime}+z^{\prime} x^{\prime}+x^{\prime} y^{\prime}\right)}=4 \Delta$$
Substituting in... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,665 |
Proposition 4 Let the side lengths of $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ be $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$, and their areas be $\Delta$ and $\Delta^{\prime}$, respectively, then
$$a^{\prime}(-a+b+c)+b^{\prime}(a-b+c)+c^{\prime}(a+b-c) \geqslant 4 \sqrt{3} \sqrt{\Delta... | To prove the theorem, let $\frac{y^{\prime}+z^{\prime}}{2}=a^{\prime}, \frac{z^{\prime}+x^{\prime}}{2}=b^{\prime}, \frac{x^{\prime}+y^{\prime}}{2}=c^{\prime}$, and $x=-a+b+c, y=a-b+c, z=a+b-c$, then
$$\begin{aligned}
\sqrt{y^{\prime} z^{\prime}+z^{\prime} x^{\prime}+x^{\prime} y^{\prime}}= & {\left[\left(a^{\prime}-b^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,666 |
Proposition 5 Let the side lengths and semi-perimeters of $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ be $a, b, c, p$ and $a^{\prime}, b^{\prime}, c^{\prime}$, $p^{\prime}$, and the areas be $\Delta$ and $\Delta^{\prime}$, respectively, then
$$\begin{array}{l}
a^{\prime}\left(p^{\prime}-a^{\prime... | To prove, in the theorem, let
$$\begin{array}{l}
x^{\prime}= 4\left(p^{\prime}-b^{\prime}\right)\left(p^{\prime}-c^{\prime}\right) \\
y^{\prime}= 4\left(p^{\prime}-c^{\prime}\right)\left(p^{\prime}-a^{\prime}\right) \\
z^{\prime}=4\left(p^{\prime}-a^{\prime}\right) \cdot\left(p^{\prime}-b^{\prime}\right) \\
x= 4(p-b)(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,667 |
Proposition 6 Let the side lengths of $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ be $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$, and their areas be $\Delta$ and $\Delta^{\prime}$, respectively, then
$$\begin{array}{l}
(a-b+c)(a+b-c) a^{\prime 2}+(a+b-c)(-a+b+c) b^{\prime 2}+ \\
(-a+b+c)(a... | To prove in the theorem, let
then $\square$
$$\begin{array}{c}
x^{\prime}=-a^{\prime 2}+b^{\prime 2}+c^{\prime 2} \\
y^{\prime}=a^{\prime 2}-b^{\prime 2}+c^{\prime 2} \\
z^{\prime}=a^{\prime 2}+b^{\prime 2}-c^{\prime 2} \\
x=(a-b+c)(a+b-c) \\
y=(a+b-c)(-a+b+c) \\
z=(-a+b+c)(a-b+c) \\
\sqrt{\left(y^{\prime} z^{\prime}+z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,668 |
Proposition 7 Let $a, b, c$ be the side lengths of triangle $ABC$, $\Delta$ its area, and $\lambda, u, v$ any positive numbers, then
$$\begin{array}{l}
\lambda b c(-a+b+c)+u c a(a-b+c)+v a b(a+b-c) \geqslant \\
4 \sqrt{u v a^{2}+v \lambda b^{2}+\lambda u c^{2}} \Delta
\end{array}$$
Equality in (8) holds if and only if... | To prove that the inequality (1) can also be written as
$$\begin{array}{l}
\frac{\left(y^{\prime}+z^{\prime}\right)}{2} x+\frac{\left(z^{\prime}+x^{\prime}\right)}{2} y+\frac{\left(x^{\prime}+y^{\prime}\right)}{2} z \geqslant \\
\sqrt{\left(y^{\prime} z^{\prime}+z^{\prime} x^{\prime}+x^{\prime} y^{\prime}\right)(y z+z ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,669 |
Proposition 8 Let $P$ be any point inside (including the boundary) $\triangle ABC$, draw perpendiculars from $P$ to the three sides $BC$, $CA$, $AB$, with the feet of the perpendiculars being $A'$, $B'$, $C'$, respectively. Connect $A'$, $B'$, $C'$ to form $\triangle A'B'C'$. If the areas of $\triangle ABC$ and $\trian... | Let $BC = a$, $CA = b$, $AB = c$, $PA' = \lambda$, $PB' = u$, $PC' = v$. In the theorem, let
$$\begin{array}{c}
x' = -a^2 + b^2 + c^2 \\
y' = a^2 - b^2 + c^2 \\
z' = a^2 + b^2 - c^2 \\
x = \frac{\lambda}{a}, y = \frac{u}{b}, z = \frac{v}{c}
\end{array}$$
Then the left side of equation $\left(1'\right)$ is
$$\frac{(y' ... | \Delta' \leqslant \frac{1}{4} \Delta | Geometry | proof | Yes | Yes | inequalities | false | 734,670 |
Theorem In a non-obtuse $\triangle A B C$, with circumradius $R$, $B C$ being the shortest side, $m_{d}, w_{a}, h_{\mathrm{a}}$ being the median, angle bisector, and altitude to side $B C$, respectively, then
$$m_{\mathrm{a}} w_{\mathrm{a}}+w_{\mathrm{a}} h_{a}+m_{\mathrm{a}} h_{a} \geqslant \frac{27}{4} R^{2}$$
Equal... | Prove that from formula (20) in Example 12 of Chapter 6 "Trigonometric Geometric Inequalities," if the sides of $\triangle ABC$ are $BC = a$, $CA = b$, $AB = c$, and the area is $\Delta$, then we have
$$m_{a} w_{a} \geqslant \frac{1}{4}(a+b+c)(-a+b+c)$$
Therefore,
$$\begin{array}{l}
m_{a} w_{a} + w_{a} h_{a} + m_{a} h... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,671 |
Example 9 (Original Problem, 2007.07.15) Let $x, y, z \in \mathbf{R}^{+}$, then
$$\begin{aligned}
\frac{x(z+x)}{(y+z)(3 x+y+2 z)}+ & \frac{y(x+y)}{(z+x)(2 x+3 y+z)}+ \\
& \frac{z(y+z)}{(x+y)(x+2 y+3 z)} \geqslant \frac{1}{2}
\end{aligned}$$ | To prove:
$$\begin{aligned}
\text { Eq. (14) Left side }= & \sum \frac{3 x(z+x)(3 x+2 y+z)}{(3 y+3 z)(3 x+y+2 z)(3 x+2 y+z)} \geqslant \\
& \sum \frac{3 x(z+x)(3 x+2 y+z)}{8(x+y+z)^{3}}= \\
& \frac{9 \sum x^{3}+12 \sum x y^{2}+9 \sum x^{2} y+18 x y z}{8\left(\sum x\right)^{3}}
\end{aligned}$$
Therefore, to prove Eq. (... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,672 |
Example 13 Let $x, y, z > -1$, prove
$$\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \geqslant 2$$
Equality in (19) holds if and only if $x=y=z=1$. | Let $1+x^{2}=u, 1+y^{2}=v, 1+z^{2}=w$.
Since $0<1+y+z^{2} \leqslant \frac{v}{2}+w, 0<1+z+x^{2} \leqslant \frac{w}{2}+u, 0<1+x+y^{2} \leqslant$ $\frac{u}{2}+v$, it suffices to prove for positive $u, v, w$ that
$$\frac{u}{v+2 w}+\frac{v}{w+2 u}+\frac{w}{u+2 v} \geqslant 1$$
By the Cauchy-Schwarz inequality, we have
$$\b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,674 |
Example 16 Let $a, b, c, d \in \mathbf{R}^{+}$, and $a b c d=1$, prove that
$$\frac{1}{1+a+a^{2}+a^{3}}+\frac{1}{1+b+b^{2}+b^{3}}+\frac{1}{1+c+c^{2}+c^{3}}+\frac{1}{1+d+d^{2}+d^{3}} \geqslant 1$$
Equality in (23) holds if and only if $a=b=c=d=1$. | First, prove
$$\frac{1}{1+a+a^{2}+a^{3}}+\frac{1}{1+b+b^{2}+b^{3}} \geqslant \frac{1}{1+\sqrt{(a b)^{3}}}$$
Equation (24) is equivalent to, for $x, y \in \mathbf{R}^{+}$, we have
$$\frac{1}{1+x^{2}+x^{4}+x^{6}}+\frac{1}{1+y^{2}+y^{4}+y^{6}} \geqslant \frac{1}{1+x^{3} y^{3}}$$
Equation $(※) \Leftrightarrow\left(1+y^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,677 |
Example 17 (Original Problem, 2005.07.17) Let $a, b, c, \geqslant 0, x \geqslant y \geqslant z \geqslant 0, a \geqslant x, a y+$ $b x \geqslant 2 x y, a y z+b x z+c x y \geqslant 3 x y z, n$ be a positive integer, then
$$a^{n}+b^{n}+c^{n} \geqslant x^{n}+y^{n}+z^{n}$$ | Prove that if $x \geqslant 0, y=z=0$, then (26) holds.
If $x \geqslant y>0, z=0$, let $\frac{a}{x}=\lambda, \frac{b}{y}=\mu$, then $\lambda \geqslant 1, \lambda+\mu \geqslant 2$, and
$$\begin{array}{l}
a^{n}+b^{n}+c^{n}-x^{n}-y^{n}-z^{n} \geqslant a^{n}+b^{n}-x^{n}-y^{n}= \\
(\lambda x)^{n}+(\mu y)^{n}-x^{n}-y^{n}=\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,678 |
Example 3 (2006 National Training Team Test Question) Let $x, y, z \in \overline{\mathbf{R}^{-}}$, and $x+y+z=1$, prove that
$$\frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+z x}}+\frac{z x}{\sqrt{z x+x y}} \leqslant \frac{\sqrt{2}}{2}$$ | To prove that by the Cauchy-Schwarz inequality, to show that inequality (3) holds, it suffices to prove:
$$\begin{array}{l}
(x y+y z+z x) \cdot\left(\frac{x y}{x y+y z}+\frac{y z}{y z+z x}+\frac{z x}{z x+x y}\right) \leqslant \frac{1}{2} \\
\text { Equation (1) } \Leftrightarrow \sum y z \cdot \sum \frac{x}{z+x}=[x z+y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,682 |
Example 5 Given $a_{i} \in \mathbf{R}^{+}, i=1,2, \cdots, n$, and $\sum_{i=1}^{n} a_{i}=1$, prove that
$$\frac{1}{a_{1}\left(1+a_{2}\right)}+\frac{1}{a_{2}\left(1+a_{3}\right)}+\cdots+\frac{1}{a_{n-1}\left(1+a_{n}\right)}+\frac{1}{a_{n}\left(1+a_{1}\right)} \geqslant \frac{n^{3}}{n+1}$$
Equality in (6) holds if and on... | Prove that
$$\frac{1}{a_{1}\left(1+a_{2}\right)}+\frac{n^{3}}{n+1} a_{1}+\frac{n^{3}\left(1+a_{2}\right)}{(n+1)^{2}} \geqslant \frac{3 n^{2}}{n+1}$$
Therefore,
$$\begin{array}{c}
\sum \frac{1}{a_{1}\left(1+a_{2}\right)} \geqslant \frac{3 n^{3}}{n+1}-\frac{n^{3}}{n+1} \sum a_{1}-\frac{n^{3}}{(n+1)^{2}} \sum\left(1+a_{2... | \frac{n^{3}}{n+1} | Inequalities | proof | Yes | Yes | inequalities | false | 734,684 |
Example 7 (Original Problem, 1988.07.07) In $\triangle A B C$, prove that
$$\sum \tan ^{2} \frac{A}{2} \geqslant \frac{16}{9}\left(\sum \sin ^{2} \frac{A}{2}\right)^{2}$$
Equality holds in (8) if and only if $\triangle A B C$ is an equilateral triangle. | $$\begin{aligned}
\sum \tan ^{2} \frac{A}{2} \geqslant & \frac{4}{9}\left(\sum \sin ^{2} A\right) \cdot\left(\sum \tan ^{2} \frac{A}{2}\right)= \\
& \frac{16}{9}\left(\sum \sin ^{2} \frac{A}{2} \cos ^{2} \frac{A}{2}\right) \cdot\left(\sum \frac{\sin ^{2} \frac{A}{2}}{\cos ^{2} \frac{A}{2}}\right) \geqslant \\
& \frac{1... | \sum \tan ^{2} \frac{A}{2} \geqslant \frac{16}{9}\left(\sum \sin ^{2} \frac{A}{2}\right)^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 734,686 |
Example 8 (Original Problem, 1987.07.20) Let $x, y, z, w \in \mathbf{R}^{+}$, then
$$8+\sum x \cdot \sum \frac{1}{x} \geqslant \frac{9\left(\sum x\right)^{2}}{x y+x z+x w+y z+y w+z w}$$
Equality in (9) holds if and only if $x=y=z=w$. | Prove
$$\begin{aligned}
\sum x \cdot \sum \frac{1}{x}= & 4+\left[\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)+\left(\frac{x}{w}+\frac{w}{x}\right)+\right. \\
& \left.\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{y}{w}+\frac{w}{y}\right)+\left(\frac{z}{w}+\frac{w}{z}\right)\right]= \\
& -... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,687 |
Example 9 (Self-created problem, 2007.07.20) Let $x, y, z, w \in \mathbf{R}^{+}$, then
$$\sum x \cdot \sum \frac{1}{x} \leqslant 4+\frac{(x y+x z+x w+y z+y w+z w)^{2}}{3 x y z w}$$
Equality in (11) holds if and only if three of $x, y, z, w$ are equal. | $$\begin{array}{l}
\sum x \cdot \sum \frac{1}{x}=4+\sum\left(\frac{x}{y}+\frac{y}{x}\right)= \\
4+\left[\left(\frac{x}{w}+\frac{w}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)\right]+ \\
{\left[\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{z}{w}+\frac{w}{z}\right)\right]+\left[\left(\frac{x}{z}+\frac{z}{x}\right)+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,688 |
Example 10 (Self-created problem, 1985.07.29) Let $x_{1}, x_{2}, x_{3}, x_{4} \in \mathbf{R}^{+}$, then
$$\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{3}}+\frac{x_{3}}{x_{4}}+\frac{x_{4}}{x_{1}} \geqslant \frac{x_{1}}{x_{2}+x_{3}}+\frac{x_{2}}{x_{3}+x_{4}}+\frac{x_{3}}{x_{4}+x_{1}}+\frac{x_{4}}{x_{1}+x_{2}}+2$$
Equality in (12... | Proof
The left side of equation (12) - the right side $=$
$$\begin{array}{l}
\frac{x_{1} x_{3}}{x_{2}\left(x_{2}+x_{3}\right)}+\frac{x_{2} x_{4}}{x_{3}\left(x_{3}+x_{4}\right)}+\frac{x_{3} x_{1}}{x_{4}\left(x_{4}+x_{1}\right)}+\frac{x_{4} x_{2}}{x_{1}\left(x_{1}+x_{2}\right)}-2= \\
x_{1} x_{2} x_{3} x_{4}\left[\frac{\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,689 |
Example 12 (2005 National Eighteen Schools Olympiad Collaborative Body Test) Let $a, b, c \in \mathbf{R}^{+}$, and $b c + c a + a b = 1$, prove that
$$\sqrt[3]{\frac{1}{a}+6 b}+\sqrt[3]{\frac{1}{b}+6 c}+\sqrt[3]{\frac{1}{c}+6 a} \leqslant \frac{1}{a b c}$$ | To prove the stronger inequality:
$$\sqrt[3]{\frac{1}{a}+6 b}+\sqrt[3]{\frac{1}{b}+6 c}+\sqrt[3]{\frac{1}{c}+6 a} \leqslant \frac{3}{\sqrt[3]{a b c}}$$
From inequality (17), we can derive:
$$\sqrt[3]{b c+6 a b^{2} c}+\sqrt[3]{c a+6 a b c^{2}}+\sqrt[3]{a b+6 a^{2} b c} \leqslant 3$$
Now, we prove inequality (※).
Since... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,691 |
Example 9 (Self-created problem, 2003.07.24) Let $x, y, z \in \overline{\mathbf{R}^{-}}, 0<\lambda \leqslant 3$, then
$$\Pi\left(3 \lambda x^{2}+\sum y z\right) \geqslant(1+\lambda)^{3}\left(\sum y z\right)^{3}$$
Equality in (20) holds if and only if $x=y=z$. | Let $s_{1}=\sum x, s_{2}=\sum y z, s_{3}=x y z$, then
$$\begin{array}{c}
\prod\left(3 \lambda x^{2}+\sum y z\right)-(1+\lambda)^{3}\left(\sum y z\right)^{3}= \\
\left(\sum y z\right)^{3}+3 \lambda\left(\sum y z\right)^{2} \sum x^{2}+9 \lambda^{2} \sum y z \cdot \sum y^{2} z^{2}+ \\
27 \lambda^{3} x^{2} y^{2} z^{2}-(1+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,692 |
Example 13 (2005, 17th Asia Pacific Mathematical Olympiad) Let $x, y, z \in \mathbf{R}^{+}$, and $x y z=8$, then
$$\sum \frac{x^{2}}{\sqrt{\left(1+x^{3}\right)\left(1+y^{3}\right)}} \geqslant \frac{4}{3}$$
Equality in (18) holds if and only if $x=y=z=2$. | $$\begin{aligned}
\sum \frac{x^{2}}{\sqrt{\left(1+x^{3}\right)\left(1+y^{3}\right)}}= & \sum \frac{x^{2}}{\sqrt{(1+x)\left(1-x+x^{2}\right)} \cdot \sqrt{(1+y)\left(1-y+y^{2}\right)}} \geqslant \\
& \sum \frac{x^{2}}{\frac{2+x^{2}}{2} \cdot \frac{2+y^{2}}{2}}= \\
& \sum \frac{4 x^{2}}{\left(2+x^{2}\right)\left(2+y^{2}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,693 |
Example 14 (Self-created problem, 2005.12.16) For $\triangle A B C$ with sides $a, b, c$, and circumradius $R$, then
$$1+\frac{|(a-b)(b-c)(a-c)|}{a b c} \leqslant \frac{3 \sqrt{3} R}{\sum a}$$
Equality in (19) holds if and only if $\triangle A B C$ is an equilateral triangle. | Given $a \geqslant b \geqslant c$, then equation (19) becomes
$$\begin{array}{l}
\frac{a b c \cdot \sum a}{a b c}+\frac{\left(\sum a\right)(a-b)(b-c)(a-c)}{a b c} \leqslant 3 \sqrt{3} R \Leftrightarrow \\
\frac{a b\left(a^{2}-b^{2}+c^{2}\right)+b c\left(a^{2}+b^{2}-c^{2}\right)+c a\left(-a^{2}+b^{2}+c^{2}\right)}{a b c... | \sum \sin A \cos B \leqslant \frac{3 \sqrt{3}}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 734,694 |
Example 15 (2004 China Mathematical Winter Camp Selection Contest, Changsha, Hunan) Let $a, b, c, d$ be positive real numbers, satisfying $a b+c d=1$, and let points $p_{i}\left(x_{i}, y_{i}\right)(i=1,2,3,4)$ be four points on the unit circle centered at the origin. Prove that
$$\begin{array}{l}
\left(a y_{1}+b y_{2}+... | Let $\alpha^{2}=a y_{1}+b y_{2}+c y_{3}+d y_{4}, \beta^{2}=a x_{4}+b x_{3}+c x_{2}+d x_{1}$, then
$$\alpha^{2} \leqslant\left(a d y_{1}^{2}+b c y_{2}^{2}+b c y_{3}^{2}+a d y_{4}^{2}\right)\left(\frac{a}{d}+\frac{b}{c}+\frac{c}{b}+\frac{d}{a}\right)$$
Similarly, we have
$$\beta^{2} \leqslant\left(a d x_{4}^{2}+b c x_{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,695 |
Example 16 Let $a, b, c \in \mathbf{R}^{+}$, prove that
$$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}} \geqslant \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{2}}$$
Equality in (23) holds if and only if $a=b=c$. | $$\begin{array}{l}
2\left(\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\right)^{2}-(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}= \\
2 \sum \frac{a^{2}}{a+b}+4 \sum \frac{a b}{\sqrt{a+b} \cdot \sqrt{b+c}}-\sum a-2 \sum \sqrt{b c}= \\
\sum \frac{a^{2}+b^{2}}{a+b}+\sum \frac{a^{2}-b^{2}}{a+b}+4 \sum \frac{a b}{\sqrt{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,696 |
Example 21 (from the "Inequality Research Website", "Competition Inequality" column, January 6, 2007, proposed by Mr. Chen Shengli, Fujian) Let $a, b, c > 0$, and $abc = 1$. Prove that
$$\sqrt[3]{\Pi\left(a^{6}+1\right)} \geqslant 2+\frac{1}{3} \sum\left(a-\frac{1}{a}\right)^{2}$$ | $$\begin{array}{c}
\text { Noting the condition } a b c=1, \text { we have } \\
\text { (27) } \Leftrightarrow \sqrt[3]{\prod\left(a^{4}+b^{2} c^{2}\right)} \geqslant 2+\frac{1}{3} \sum(a-b c)^{2} \Leftrightarrow \\
\sqrt[3]{\prod\left(a^{4}+b^{2} c^{2}\right)} \geqslant \frac{1}{3}\left(\sum a^{2}+\sum b^{2} c^{2}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,701 |
Example 22 In the 12th issue of "Middle School Mathematics Research" in 2006, the author proposed the following conjecture at the end of the article on "A Cyclic Symmetric Inequality with Four Variables."
Conjecture Let
$$\begin{array}{c}
n \in \mathbf{N}^{+}, n \geqslant 3, a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{R}^... | $$\begin{array}{c}
f\left(a_{1}, a_{2}\right) \cdot f\left(a_{2}, a_{3}\right) \cdots f\left(a_{n-1}, a_{n}\right) \cdot f\left(a_{n}, a_{1}\right)= \\
\left(a_{1}^{n-1}+a_{1}^{n-2} a_{2}+\cdots+a_{1}^{3} a_{2}^{n-4}+a_{1}^{2} a_{2}^{n-3}+a_{1} a_{2}^{n-2}+a_{2}^{n-1}\right) \cdot \\
\left(a_{2}^{n-1}+a_{2}^{n-2} a_{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,702 |
Example 10 (2006 China Mathematical Winter Camp Problem 1. Fujian. Fuzhou) Real numbers $a_{1}, a_{2}, \cdots, a_{k}$ satisfy $a_{1}+a_{2}+\cdots+a_{k}=0$, prove that
$$\max \left(a_{i}^{2}\right) \leqslant \frac{k}{2}\left[\left(a_{1}-a_{2}\right)^{2}+\left(a_{2}-a_{3}\right)^{2}+\cdots+\left(a_{k-1}-a_{k}\right)^{2}\... | Let $\sum a=a_{1}+a_{2}+\cdots+a_{k}$, then
$$\begin{array}{l}
k a_{1}=(k-1)\left(a_{1}-a_{2}\right)+(k-2)\left(a_{2}-a_{3}\right)+(k-3)\left(a_{3}-a_{4}\right)+\cdots+ \\
3\left(a_{k-3}-a_{k-2}\right)+2\left(a_{k-2}-a_{k-1}\right)+\left(a_{k-1}-a_{k}\right)+\sum a \\
k a_{2}= a_{2}-a_{1}+(k-2)\left(a_{2}-a_{3}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,703 |
Example 23 (Recorded from "China Inequality Research Group Website", hljhdf, 2007.09.08 provided) Let $a$, $b, c \in \mathbf{R}^{+}$, then
$$\sum \frac{a^{3}}{b^{2}-b c+c^{2}} \geqslant \sum a$$
Equality in (29) holds if and only if $a=b=c$. | Prove that by the Cauchy-Schwarz inequality,
$$\sum a\left(b^{2}-b c+c^{2}\right) \cdot \sum \frac{a^{3}}{b^{2}-b c+c^{2}} \geqslant\left(\sum a^{2}\right)^{2}$$
Therefore, it suffices to prove that
$$\begin{array}{l}
\left(\sum a^{2}\right)^{2} \geqslant \sum a \cdot \sum a\left(b^{2}-b c+c^{2}\right) \Leftrightarrow... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,704 |
Example 26 (Self-created problem, 2007.07.12) Let $a, b, c \in \overline{\mathbf{R}^{-}}$, and $\sum a^{2}=1$, then
$$\sum \sqrt{1-b c} \geqslant \sqrt{6}$$
Equality in (32) holds if and only if $a=b=c=\frac{\sqrt{3}}{3}$. | By symmetry, without loss of generality, assume $a$ is the largest. Since
$$\begin{aligned}
(\sqrt{1-a b}+\sqrt{1-a c})^{2}= & \frac{1}{2}\left(\sqrt{\sum a^{2}+c^{2}+(a-b)^{2}}+\right. \\
& \left.\sqrt{\sum a^{2}+b^{2}+(a-c)^{2}}\right)^{2} \geqslant \\
& \frac{1}{2}\left[4 \sum a^{2}+(b+c)^{2}+(2 a-b-c)^{2}\right]= \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,707 |
Theorem 1 Let $\lambda, u, v \in \mathbf{R}^{+}$, and denote $s_{1}=\lambda+u+v, s_{2}=u v+v \lambda+\lambda u, s_{3}=\lambda u v$, $x=\frac{s_{1}}{3 \sqrt[3]{s_{3}}}, y=\frac{s_{2}}{3 \sqrt[3]{s_{3}^{2}}}$, then
i ) $3(x y)^{2}+6 x y-1-(x y-1) \sqrt{(9 x y-1)(x y-1)} \stackrel{(1)}{\leqslant} 8 x^{3} \stackrel{(2)}{\l... | Prove i) Since
$$\begin{array}{l}
(u-v)^{2}(v-\lambda)^{2}(\lambda-u)^{2}=-4 s_{1}^{3} s_{3}+s_{1}^{2} s_{2}^{2}+18 s_{1} s_{2} s_{3}-4 s_{2}^{3}-27 s_{3}^{2}= \\
\frac{s_{3}}{16 s_{1}^{3}}\left[\left(s_{1} s_{2}-s_{3}\right)\left(s_{1} s_{2}-9 s_{3}\right)^{3}-\left(8 s_{1}^{3} s_{3}-s_{1}^{2} s_{2}^{2}-18 s_{1} s_{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,708 |
Theorem 2 Let $\lambda, u, v \in \mathbf{R}$, and denote $s_{1}=\lambda+u+v, s_{2}=u v+v \lambda+\lambda u, s_{3}=\lambda u v, w=$ $\sqrt{s_{1}^{2}-3 s_{2}}\left(0 \leqslant w \leqslant s_{1}\right)$, then
$$\begin{array}{l}
\frac{s_{1}^{3}-3 s_{1} w^{2}-2 w^{3}}{27}=\frac{\left(s_{1}-2 w\right)\left(s_{1}+w\right)^{2}... | Prove that
$$\begin{aligned}
(u-v)^{2}(v-\lambda)^{2}(\lambda-u)^{2}= & -4 s_{1}^{3} s_{3}+s_{1}^{2} s_{2}^{2}+18 s_{1} s_{2} s_{3}-4 s_{2}^{3}-27 s_{3}^{2}= \\
& -27 s_{3}^{2}+2\left(9 s_{1} s_{2}-2 s_{1}^{3}\right) s_{3}+\left(s_{1}^{2} s_{2}^{2}-4 s_{2}^{3}\right) \geqslant \\
& 0
\end{aligned}$$
From this, we obta... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,712 |
Example 11 (Self-created, $1999,05,14$) In $\triangle A B C$, the lengths of the three sides are $B C=a, C A=b, A B=$ $c, S=\frac{1}{2}(a+b+c), R$ and $r$ are the circumradius and inradius of $\triangle A B C$, respectively, then
$$\sum \cos \frac{B-C}{2} \geqslant 1+\sqrt{\frac{s^{2}+2 R r+r^{2}}{2 R^{2}}}$$
The equa... | Prove that
$$\begin{aligned}
\frac{s^{2}+2 R r+r^{2}}{2 R^{2}}= & \frac{1}{2}\left(\frac{R+r}{R}\right)^{2}+\frac{1}{2}\left(\frac{S}{R}\right)^{2}-\frac{1}{2}= \\
& \frac{1}{2}\left(\sum \cos A\right)^{2}+\frac{1}{2}\left(\sum \sin A\right)^{2}-\frac{1}{2}= \\
& \frac{3}{2}+\sum \cos (B-C)-\frac{1}{2}= \\
& 1+\sum \co... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,714 |
Example 1 Let $a, b, c \in \mathbf{R}^{+}$, and $abc=1$, prove that
$$\sum a^{3}+3 \geqslant 2 \sum a^{2}$$
Equality in (17) holds if and only if $a=b=c=1$. | Let $s_{1}=\sum a, s_{2}=\sum b c, s_{3}=a b c=1$, then equation (17) is equivalent to
$$s_{1}^{3}-3 s_{1} s_{2}+6-2 s_{1}^{2}+4 s_{2} \geqslant 0 \Leftrightarrow s_{1}^{3}-2 s_{1}^{2}+6 \geqslant\left(3 s_{1}-4\right) s_{2}$$
Let $s_{1}=3 x, s_{2}=3 y$, then the above equation is equivalent to
$$9 x^{3}-6 x^{2}+2 \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,715 |
Example 2 (Self-created problem, 2007.08.08) Let $a, b, c \in \mathbf{R}^{+}$, and $abc=4$, then
$$108 \sum bc \geqslant -\left(\sum a\right)^{4} + 6\left(\sum a\right)^{3} + 27\left(\sum a\right)^{2}$$
Equality in (18) holds if and only if one of $a, b, c$ is equal to 4, and the other two are equal to 1. | Let $s_{1}=\sum a, s_{2}=\sum b c, s_{3}=a b c=4, w=\sqrt{s_{1}^{2}-3 s_{2}}$, then we have
$$a b c \leqslant \frac{\left(s_{1}-w\right)^{2}\left(s_{1}+2 w\right)}{27}$$
Additionally, we have
$$\begin{array}{l}
\left(s_{1}-w\right)^{2}\left(s_{1}+2 w\right) \cdot\left(s_{1}+2 w\right)^{3}= \\
\left(s_{1}-w\right)\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,716 |
Example 3 (Recorded from Vasile Cirtoaje's edited book "Algebraic Inequalities. Old and New Methods" 8 8.1. Applications Problem 70) Let $a, b, c \in \mathbf{R}^{+}$, prove that
$$a^{3}+b^{3}+c^{3}+3 a b c \geqslant \sum b c \sqrt{2\left(b^{2}+c^{2}\right)}$$
Equality in (19) holds if and only if $a=b=c$. | Prove that by the Cauchy-Schwarz inequality, we have
$$\left[\sum b c \sqrt{2\left(b^{2}+c^{2}\right)}\right]^{2} \leqslant 2 \sum b c \cdot \sum b c\left(b^{2}+c^{2}\right)$$
Therefore, it suffices to prove
$$\left(a^{3}+b^{3}+c^{3}+3 a b c\right)^{2} \geqslant 2 \sum b c \cdot \sum b c\left(b^{2}+c^{2}\right)$$
Now... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,717 |
Example 4 (Recorded from Vasile Cirtoaje's edited book *Algebraic Inequalities. Old and New Methods*) Let $a, b, c \in \mathbf{R}^{+}$, prove that
$$\frac{1}{5\left(a^{2}+b^{2}\right)-a b}+\frac{1}{5\left(b^{2}+c^{2}\right)-b c}+\frac{1}{5\left(c^{2}+a^{2}\right)-c a} \geqslant \frac{1}{a^{2}+b^{2}+c^{2}}$$ | After removing the denominator and rearranging, the equivalent form of formula (20) is obtained as follows:
$$\begin{array}{l}
\sum a^{2}\left[25\left(\sum a^{2}\right)^{2}+25 \sum b^{2} c^{2}-5 \sum a^{2} \sum b c-4 a b c \sum a\right] \geqslant \\
125 \sum a^{2} \sum b^{2} c^{2}-25 a b c \sum a^{3}-25 \sum b c \sum b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,718 |
Example 5 (Self-created problem, 2005. 12.04) Let $a, b, c \in \mathbf{R}^{+}$, and $a+b+c=1$, then
$$(2 \sqrt{3}-3)-9(6 \sqrt{3}-5) a b c+108 a b c \sum b c \geqslant 0$$
Equality in (21) holds if and only if $a=b=c=\frac{1}{3}$, or one of $a, b, c$ equals $\frac{3-\sqrt{3}}{3}$, and the other two are both $\frac{\sq... | Let $w=\sqrt{\left(\sum a\right)^{2}-3 \sum b c}=\sqrt{1-3 \sum b c}$, i.e., $\sum b c=\frac{1-w^{2}}{3}(0 \leqslant w \leqslant 1)$, such that
$$(\lambda-1)-27 \lambda a b c+81 a b c \sum b c \geqslant 0$$
holds. By Corollary 3 of Theorem 2, we have
$$\lambda \geqslant \frac{4-2 w-3 w^{2}+2 w^{3}}{3-2 w}$$
Using der... | \frac{6 \sqrt{3}-5}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 734,719 |
Example 6 (Self-created problem, 2007.09.18) Let $a, b, c \in \mathbf{R}^{+}$, and $a+b+c=1$, then
$$\frac{7}{a b c}-4 \sum \frac{1}{a^{2}} \leqslant 81$$
Equality in (22) holds if and only if $a=b=c=\frac{1}{3}$, or one of $a, b, c$ equals $\frac{2}{3}$, and the other two are both $\frac{1}{6}$. | To prove:
$$\text{Equation (22)} \Leftrightarrow 4\left(\sum bc\right)^{2}-15abc+81(abc)^{2} \geqslant 0$$
Let \( w = \sqrt{1 - 3 \sum bc} \), i.e., \( \sum bc = \frac{1}{3}(1 - w^2), 0 \leqslant w < 1 \). Then, by Corollary 3 of Theorem 2, we have
$$abc \leqslant \frac{(1 + 2w)(1 - w)^2}{27}$$
Thus,
$$\begin{aligned... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,720 |
Example 7 (Provided by Mr. Chen Ji, Zhejiang, 2008.01.21) Let real numbers $a, b, c$ satisfy $\sum a^{2}=9$, then
$$6 \sum a \leqslant a b c+26$$
Equality in (23) holds if and only if one of $a, b, c$ is 1 and the other two are 2. | Given the problem, we can rewrite equation (23) as a homogeneous form:
$$6 \sum a \cdot \frac{\sum a^{2}}{9} \leqslant a b c + 26\left(\sqrt{\left(\frac{\sum a^{2}}{9}\right)^{3}}\right.$$
From this, to prove equation (23), it suffices to prove:
$$\left(18 \sum a \cdot \sum a^{2} - 27 a b c\right)^{2} \leqslant 676\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,721 |
Example 8 (Self-created, 2006.02.22) Let $a, b, c \in \mathbf{R}^{-}$, then
$$\left(\sum a^{2}\right)^{3} \geqslant 8 \sum b^{3} c^{3}+3 a^{2} b^{2} c^{2}$$
Equality holds if and only if $a=b=c$, or one of $a, b, c$ is zero and the other two are equal. | Let $s_{1}=\sum a=1, s_{2}=\sum b c, s_{3}=a b c$, then
$$\sum a^{2}=1-2 s_{2}, \sum b^{3} c^{3}=s_{2}^{3}-3 s_{2} s_{3}+3 s_{3}^{2}$$
Substitute into the original equation, and after rearrangement, we get the equivalent form of the original equation:
$$1-6 s_{2}+12 s_{2}^{2}-16 s_{2}^{3}+24 s_{2} s_{3}-27 s_{3}^{2} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,722 |
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