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Example 10 Let $n \geqslant 2, a_{1}, a_{2}, \cdots, a_{n}$ be $n$ positive real numbers, satisfying:
$$\left(a_{1}+\cdots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right) \leqslant\left(n+\frac{1}{2}\right)^{2} .$$
Prove: $\max \left\{a_{1}, \cdots, a_{n}\right\} \leqslant 4 \min \left\... | To prove, without loss of generality, let
$$m=a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}=M$$
We need to prove $M \leqslant 4 m$.
When $n=2$, the condition is
$$(m+M)\left(\frac{1}{m}+\frac{1}{M}\right) \leqslant \frac{25}{4}$$
This is equivalent to
$$4(m+M)^{2} \leqslant 25 m M$$
i.e., $\square$
$$(4 M-m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,061 |
Example 11 Let
$$f(x, y, z)=\frac{x(2 y-z)}{1+x+3 y}+\frac{y(2 z-x)}{1+y+3 z}+\frac{z(2 x-y)}{1+z+3 x}$$
where $x, y, z \geqslant 0$, and $x+y+z=1$. Find the maximum and minimum values of $f(x, y, z)$. | First, we prove that \( f \leqslant \frac{1}{7} \), with equality holding if and only if \( x = y = z = \frac{1}{3} \). Because
\[ f = \sum \frac{x(x + 3y - 1)}{1 + x + 3y} = 1 - 2 \sum \frac{x}{1 + x + 3y} \]
By the Cauchy-Schwarz inequality,
\[ \sum \frac{x}{1 + x + 3y} \geqslant \frac{\left( \sum x \right)^2}{\sum ... | f_{\text{max}} = \frac{1}{7}, \, f_{\text{min}} = 0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,062 |
Example 13 Let $x, y, z, w \in \mathbf{R}^{+}$, prove:
$$\frac{x}{y+2 z+3 w}+\frac{y}{z+2 w+3 x}+\frac{z}{w+2 x+3 y}+\frac{w}{x+2 y+3 z} \geqslant \frac{2}{3} .$$ | $$\begin{array}{c}
\text { Prove the left side }=\sum \frac{x}{y+2 z+3 w}=\sum \frac{x^{2}}{x(y+2 z+3 w)} \\
\geqslant \frac{\left(\sum x\right)^{2}}{\sum x(y+2 z+3 w)} \\
=\frac{\left(\sum x\right)^{2}}{4 \sum x y}, \\
(x-y)^{2}+(x-z)^{2}+(x-w)^{2}+(y-z)^{2}+(y-w)^{2}+(z-w)^{2} \\
=3\left(x^{2}+y^{2}+z^{2}+w^{2}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,064 |
20. Let $a, b, c, d$ be non-negative real numbers, satisfying $ab + bc + cd + da = 1$. Prove:
$$\frac{a^{3}}{b+c+d}+\frac{b^{3}}{a+c+d}+\frac{c^{3}}{a+d+b}+\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3} .$$ | 20. Since $\frac{a^{3}}{b+c+d}+\frac{b+c+d}{18}+\frac{1}{12} \geqslant 3 \sqrt[3]{\frac{a^{3}}{a+c+d} \cdot \frac{b+c+d}{18} \cdot \frac{1}{12}}=$ $\frac{a}{2}$, i.e., $\frac{a^{3}}{b+c+d} \geqslant \frac{a}{2}-\frac{b+c+d}{18}-\frac{1}{12}$, so the left $\geqslant \frac{a+b+c+d}{2}-\frac{1}{18}(3 a+$ $3 b+3 c+3 d)-\fr... | \frac{1}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 738,065 |
Example 14 Let $x_{1}, x_{2}, \cdots, x_{n}$ be any real numbers, prove:
$$\frac{x_{1}}{1+x_{1}^{2}}+\frac{x_{2}}{1+x_{1}^{2}+x_{2}^{2}}+\cdots+\frac{x_{n}}{1+x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}<\sqrt{n} .$$ | Prove that by the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(\frac{x_{1}}{1+x_{1}^{2}}+\frac{x_{2}}{1+x_{1}^{2}+x_{2}^{2}}+\cdots+\frac{x_{n}}{1+x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}\right)^{2} \\
\leqslant & {\left[\left(\frac{x_{1}}{1+x_{1}^{2}}\right)^{2}+\left(\frac{x_{2}}{1+x_{1}^{2}+x_{2}^{2}}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,066 |
Example 1 Arrange 1650 students in a 22 by 75 matrix. It is known that for any two columns, the number of pairs of students in the same row who have the same gender does not exceed 11. Prove: the number of boys does not exceed 928. | Let the number of boys in the $i$-th row be $x_{i}$, then the number of girls is $75-x_{i}$. According to the problem, we have
$$\begin{array}{c}
\sum_{i=1}^{22}\left(\mathrm{C}_{x_{i}}^{2}+\mathrm{C}_{75-x_{i}}^{2}\right) \leqslant 11 \times \mathrm{C}_{75}^{2} \\
\sum_{i=1}^{22}\left(x_{i}^{2}-75 x_{i}\right) \leqsla... | 928 | Combinatorics | proof | Yes | Yes | inequalities | false | 738,068 |
Example 2 In a group of mathematicians, each one has some friends (the relationship is mutual). Prove: there exists a mathematician whose friends' average number of friends is not less than the average number of friends of the group. | Let $M$ be the set of mathematicians, $n=|M|, F(m)$ denote the set of friends of mathematician $m$, and $f(m)$ denote the number of friends of mathematician $m$ $(f(m)=|F(m)|)$. The proposition is equivalent to proving: there must be an $m_{0}$ such that
$$\frac{1}{f\left(m_{0}\right)} \sum_{m \in F\left(m_{0}\right)} ... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 738,069 |
Example 3 Suppose there are $2 n(n \geqslant 2)$ points in space, where no four points are coplanar. Connect $N$ line segments arbitrarily between them, and these line segments must form at least one triangle. Find the minimum value of $N$.
| Solve: Divide $2 n$ known points into two groups $A$ and $B$:
$$A=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}, B=\left\{B_{1}, B_{2}, \cdots, B_{n}\right\}$$
Now connect each pair of points $A_{i}$ and $B_{i}$ with a line segment $A_{i} B_{i}$, while no lines are drawn between any two points within the same group. Thus... | 2n+1 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 738,070 |
Example 4 In an $m \times m$ grid paper, how many small squares must be selected at least, so that among these small squares, there exist four small squares whose centers form the 4 vertices of a rectangle, with the sides of the rectangle parallel to the sides of the original square. | The required minimum value is $\left[\frac{m}{2}(1+\sqrt{4 m-3})-1\right]+1$. Suppose the maximum number of small squares that can be selected such that no four of these squares have their midpoints forming the four vertices of a rectangle (with sides parallel to the original square's sides) is $k$. Assume that the num... | \left[\frac{m}{2}(1+\sqrt{4 m-3})-1\right]+1 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 738,071 |
Example 6 Given $n$ distinct points in the plane. Prove: the number of pairs of points at unit distance apart is less than $2 \sqrt{n^{3}}$ pairs. | Prove that for a set of points $\left\{P_{1}, P_{2}, \cdots, P_{n}\right\}$ in the plane, let $a_{i}$ be the number of points $P_{i}$ that are a unit distance from $P_{i}$. Assume $a_{i} \geqslant 1$, then the number of pairs of points that are a unit distance apart is
$$A=\frac{a_{1}+a_{2}+\cdots+a_{n}}{2}$$
Let $C_{... | A<2\sqrt{n^{3}} | Combinatorics | proof | Yes | Yes | inequalities | false | 738,073 |
Example 7 Given a point $O$ in three-dimensional space and a finite set $A$ of several line segments with a total length of 1988, prove: there exists a plane that does not intersect set $A$ and is at a distance of no more than 574 from point $O$.
| Prove: Establish a rectangular coordinate system with point $O$ as the origin, and project the given line segments onto the 3 coordinate axes. Suppose $A$ contains $n$ line segments, and their projections on the 3 axes have lengths
$$x_{i}, y_{i}, z_{i}, i=1,2, \cdots, n$$
Let $x=\sum x_{i}, y=\sum y_{i}, z=\sum z_{i}... | proof | Geometry | proof | Yes | Yes | inequalities | false | 738,074 |
Example 8 Let $O x y z$ be a spatial rectangular coordinate system, $S$ be a finite set of points in space, and $S_{x}, S_{y}, S_{z}$ be the sets formed by the orthogonal projections of all points in $S$ onto the $O y z$ plane, $O z x$ plane, and $O x y$ plane, respectively. Prove:
$$|S|^{2} \leqslant\left|S_{x}\right|... | Proof: Let there be $n$ planes parallel to the $Oxy$ plane containing points from $S$, denoted as $M_{1}, M_{2}, \cdots, M_{n}$. For the plane $M_{i}, 1 \leqslant i \leqslant n$, let it intersect the $Ozx, Ozy$ planes at lines $l_{y}$ and $l_{x}$, respectively, and let $M_{i}$ contain $m_{i}$ points from $S$. Clearly, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,075 |
3 Given $0<a, b, c<1$, and $a b+b c+c a=1$. Prove: $\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+$ $\frac{c}{1-c^{2}} \geqslant \frac{3 \sqrt{3}}{2}$. | 3. Let $A=a\left(1-a^{2}\right)$, then $A^{2}=\frac{1}{2} \cdot 2 a^{2}\left(1-a^{2}\right)\left(1-a^{2}\right) \leqslant \frac{4}{27}$, so $A \leqslant \frac{2}{3 \sqrt{3}}$, hence $\frac{a}{1-a^{2}}=\frac{a^{2}}{a\left(1-a^{2}\right)} \geqslant \frac{3 \sqrt{3}}{2} a^{2}$, similarly for the other two expressions, the... | \frac{3 \sqrt{3}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 738,076 |
21 Let $n$ be a given natural number, $n \geqslant 3$, and for $n$ given real numbers $a_{1}, a_{2}, \cdots, a_{n}$, denote the minimum value of $\left|a_{i}-a_{j}\right|(1 \leqslant i<j \leqslant n)$ as $m$. Find the maximum value of $m$ when
$$a_{1}^{2}+\cdots+a_{n}^{2}=1$$ | 21. Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$, then $a_{i}-a_{j} \geqslant(i-j) m$. $\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2}=$ $(n-1) \sum_{i=1}^{n} a_{i}^{2}-2 \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}=(n-1) \sum_{i=1}^{n} a_{i}^{2}-\left[\left(\sum_{i=1}^{n} a_{i}\right)... | \sqrt{\frac{12}{n\left(n^{2}-1\right)}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,077 |
Example 1 Given positive real numbers $a, b, c, d$ satisfying
$$a\left(c^{2}-1\right)=b\left(b^{2}+c^{2}\right),$$
and $d \leqslant 1$. Prove:
$$d\left(a \sqrt{1-d^{2}}+b^{2} \sqrt{1+d^{2}}\right) \leqslant \frac{(a+b) c}{2} \text {. }$$ | Proof: Let the parameter $\lambda>1$, by the Cauchy-Schwarz inequality we have
$$\begin{aligned}
& d\left(a \sqrt{1-d^{2}}+b^{2} \sqrt{1+d^{2}}\right) \\
\leqslant & d \sqrt{\left(\frac{a^{2}}{\lambda}+b^{4}\right)\left[\left(1-d^{2}\right) \lambda+\left(1+d^{2}\right)\right]} \\
= & \sqrt{\left(\frac{a^{2}}{\lambda}+b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,078 |
Example 2 Let $p, q \in \mathbf{R}^{+}, x \in\left(0, \frac{\pi}{2}\right)$, try to find
$$\frac{p}{\sqrt{\sin x}}+\frac{q}{\sqrt{\cos x}}$$
the minimum value. | By Cauchy-Schwarz inequality, we have
$$(\sqrt{p m}+\sqrt{q n})^{2} \leqslant\left(\frac{p}{\sqrt{\sin x}}+\frac{q}{\sqrt{\cos x}}\right)(m \sqrt{\sin x}+n \sqrt{\cos x})$$
Equality holds if and only if $\frac{\frac{p}{\sqrt{\sin x}}}{m \sqrt{\sin x}}=\frac{\frac{q}{\sqrt{\cos x}}}{n \sqrt{\cos x}}$. Hence,
$$\tan x=\... | \left(p^{\frac{4}{5}}+q^{\frac{4}{5}}\right)^{\frac{5}{4}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,079 |
Example 3 (1) Let 3 positive real numbers $a, b, c$ satisfy
$$\left(a^{2}+b^{2}+c^{2}\right)^{2}>2\left(a^{4}+b^{4}+c^{4}\right)$$
Prove: $a, b, c$ must be the lengths of the 3 sides of some triangle;
(2) Let $n$ positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\r... | (1) Without loss of generality, let $a \geqslant b \geqslant c>0$. From the given condition, we have
$$\left(a^{2}+b^{2}+c^{2}\right)^{2}-2\left(a^{4}+b^{4}+c^{4}\right)>0$$
Factoring, we get
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)>0,$$
Thus, $b+c-a>0$, which means $b+c>a$, so $a, b, c$ are the lengths of the three sides of s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,080 |
Example 4 Let $a=\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ and $b=\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ be two non-proportional real number sequences, and let $x=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ be any real number sequence such that
$$\sum_{i=1}^{n} a_{i} x_{i}=0, \sum_{i=1}^{n} b_{i} x_{i}=1$$
Prove:
$$\... | Prove that for any real number $\lambda$, by the Cauchy-Schwarz inequality, we have
$$\left(\sum_{i=1}^{n} x_{i}^{2}\right) \sum_{i=1}^{n}\left(a_{i} \lambda-b_{i}\right)^{2} \geqslant\left(\lambda \sum_{i=1}^{n} a_{i} x_{i}-\sum_{i=1}^{n} b_{i} x_{i}\right)^{2}=1 .$$
Thus,
$$\left(\sum_{i=1}^{n} x_{i}^{2}\right)\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,081 |
Example 1 Let $a, b, c$ be real numbers, satisfying $a^{2}+2 b^{2}+3 c^{2}=\frac{3}{2}$, prove:
$$3^{-a}+9^{-b}+27^{-c} \geqslant 1$$ | Prove that by the AM-GM inequality, we have
$$3^{-a}+9^{-b}+27^{-c} \geqslant 3 \sqrt[3]{3^{-a-2 b-3 c}}=3^{\frac{3-a-2 b-3 c}{3}}$$
Then by the Cauchy-Schwarz inequality, we get
$$\begin{aligned}
(a+2 b+3 c)^{2} & =(a+\sqrt{2} \cdot \sqrt{2} b+\sqrt{3} \cdot \sqrt{3} c)^{2} \\
& \leqslant(1+2+3)\left(a^{2}+2 b^{2}+3 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,082 |
Example 2 Find the maximum value of
$$x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}$$ | By Cauchy-Schwarz inequality, we have
$$\left|x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right|^{2} \leqslant\left(x^{2}+y^{2}\right)\left(2-x^{2}-y^{2}\right) .$$
By the arithmetic mean inequality, we have
$$\left|x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right| \leqslant \frac{x^{2}+y^{2}+2-x^{2}-y^{2}}{2}=1 .$$
If $x=\frac{1}{2}, ... | 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,083 |
Example 3 Let $a, b, c$ be positive numbers, and satisfy $a b c=1$, prove:
$$\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} .$$ | Prove that by Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& {\left[\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1}{c^{3}(a+b)}\right] \cdot[a(b+c)+b(a+c)+c(a+b)] } \\
\geqslant & \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}=(a b+b c+a c)^{2}
\end{aligned}$$
Therefore, by the AM-GM inequality, we ha... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 738,084 |
Example 4 Let $x_{i}, i=1,2, \cdots, n$ be positive numbers, and satisfy $\sum_{i=1}^{n} x_{i}=a, a \in \mathbf{R}^{+}, m, n \in$ $\mathbf{N}^{*}, n \geqslant 2$, prove:
$$\sum_{i=1}^{n} \frac{x_{i}^{m}}{a-x_{i}} \geqslant \frac{a^{m-1}}{(n-1) n^{m-2}}$$ | Prove that when $m=1$, it is to prove
$$\sum_{i=1}^{n} \frac{x_{i}}{a-x_{i}} \geqslant \frac{n}{n-1}$$
Since
$$\sum_{i=1}^{n} \frac{x_{i}}{a-x_{i}}=\sum_{i=1}^{n}\left[\left(\frac{a}{a-x_{i}}\right)-1\right]=\sum_{i=1}^{n} \frac{a}{a-x_{i}}-n$$
By the Cauchy-Schwarz inequality, we get
i.e. $\square$
$$\begin{array}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,085 |
Example 5 Let real numbers $x_{i}$ satisfy $\left|x_{i}\right|<1(i=1,2, \cdots, n), n \geqslant 2$, prove:
$$\sum_{i=1}^{n} \frac{1}{1-\left|x_{i}\right|^{n}} \geqslant \frac{n}{1-\prod_{i=1}^{n} x_{i}}$$ | Prove that by the Cauchy-Schwarz inequality, we have
$$\sum_{i=1}^{n} \frac{1}{1-\left|x_{i}\right|^{n}} \cdot \sum_{i=1}^{n}\left(1-\left|x_{i}\right|^{n}\right) \geqslant n^{2} \text {. }$$
Therefore, to prove the original inequality, it suffices to prove
That is, to prove $\square \square$
$$\begin{array}{c}
\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,086 |
Example 6 Given positive numbers $x_{i}$ satisfying $\sum_{i=1}^{n} \frac{1}{1+x_{i}}=1$, prove:
$$\prod_{i=1}^{n} x_{i} \geqslant(n-1)^{n}$$ | Prove that by Cauchy-Schwarz inequality, we have
i.e. $\square$
$$\begin{array}{l}
\sum_{i=1}^{n} \frac{1}{1+x_{i}} \cdot \sum_{i=1}^{n} \frac{1+x_{i}}{x_{i}} \geqslant\left(\sum_{i=1}^{n} \frac{1}{\sqrt{x_{i}}}\right)^{2} \\
\quad \sum_{i=1}^{n} \frac{1}{x_{i}}+n \geqslant \sum_{i=1}^{n} \frac{1}{x_{i}}+2 \sum_{1 \l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,087 |
Example 1 Let $f(x)=\frac{a}{a^{2}-1}\left(a^{x}-a^{-x}\right)(a>0, a \neq 1)$, prove: for positive integer $n \geqslant$ 2, we have
$$f(n)>n$$ | Prove that when $n \geqslant 2$, by the AM-GM inequality, we have
$$\begin{aligned}
f(n) & =\frac{a}{a^{2}-1}\left(a^{n}-a^{-n}\right)=\frac{a}{a^{2}-1}\left(a^{n}-\frac{1}{a^{n}}\right) \\
& =\frac{a}{a^{2}-1}\left(a-\frac{1}{a}\right)\left(a^{n-1}+a^{n-2} \frac{1}{a}+a^{n-3} \frac{1}{a^{2}}+\cdots+a \frac{1}{a^{n-2}}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,088 |
Example 7 Let $x, y, z \geqslant 0$, and $x^{2}+y^{2}+z^{2}=1$, prove that:
$$\frac{x}{1-y z}+\frac{y}{1-x z}+\frac{z}{1-x y} \leqslant \frac{3 \sqrt{3}}{2} .$$ | Let $S=\frac{x}{1-y z}+\frac{y}{1-x z}+\frac{z}{1-x y}$. If $x=0$ (or $y=0$ or $z=0$), then
$$S=y+z<2<\frac{3}{2} \sqrt{3}$$
So, assume $x y z \neq 0$, such that $x, y, z \in(0,1)$. Since
$$\frac{x}{1-y z}=x+\frac{z y x}{1-y z}$$
Therefore,
$$S=x+y+z+x y z\left(\frac{1}{1-y z}+\frac{1}{1-z x}+\frac{1}{1-x y}\right)$$... | \frac{3 \sqrt{3}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 738,089 |
Example 8 Let $a, b, c>0$, prove:
$$\sum \sqrt{\frac{5 a^{2}+8 b^{2}+5 c^{2}}{4 a c}} \geqslant 3 \sqrt[9]{\frac{8(a+b)^{2}(b+c)^{2}(c+a)^{2}}{(a b c)^{2}}}$$ | Prove that by Cauchy-Schwarz inequality and AM-GM inequality,
$$\begin{aligned}
5 a^{2}+8 b^{2}+5 c^{2} & \geqslant 4\left(a^{2}+b^{2}\right)+4\left(b^{2}+c^{2}\right) \\
& \geqslant 2(a+b)^{2}+2(b+c)^{2} \\
& \geqslant 4(a+b)(b+c),
\end{aligned}$$
Therefore,
$$\sum \sqrt{\frac{5 a^{2}+8 b^{2}+5 c^{2}}{4 a c}} \geqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,090 |
Example 9 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}>0, a_{2}>0, a_{n+2}=\frac{2}{a_{n}+a_{n+1}} . M_{n}=$ $\max \left\{a_{n}, \frac{1}{a_{n}}, \frac{1}{a_{n+1}}, a_{n+1}\right\}$. Prove:
$$M_{n+3} \leqslant \frac{3}{4} M_{n}+\frac{1}{4}$$ | To prove that
$$M_{n+3}=\max \left\{a_{n+3}, a_{n+4}, \frac{1}{a_{n+3}}, \frac{1}{a_{n+4}}\right\},$$
we need to show that
$$\begin{array}{l}
a_{n+3} \leqslant \frac{3}{4} M_{n}+\frac{1}{4} \\
a_{n+4} \leqslant \frac{3}{4} M_{n}+\frac{1}{4} \\
\frac{1}{a_{n+3}} \leqslant \frac{3}{4} M_{n}+\frac{1}{4} \\
\frac{1}{a_{n+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,091 |
Example 11 Let $n \geqslant 2$ be a positive integer. Find the maximum value of the constant $C(n)$, such that for all real numbers $x_{1}$, $x_{2}, \cdots, x_{n}$ satisfying $x_{i} \in (0,1)(i=1,2, \cdots, n)$, and $\left(1-x_{i}\right)\left(1-x_{j}\right) \geqslant \frac{1}{4}(1 \leqslant i<j \leqslant n)$, we have
$... | First, take $x_{i}=\frac{1}{2}(i=1,2, \cdots, n)$. Substituting into equation (45) gives
$$\frac{n}{2} \geqslant C(n) \mathrm{C}_{n}^{2}\left(\frac{1}{2}+\frac{1}{2}\right)$$
Thus, $C(n) \leqslant \frac{1}{n-1}$.
Next, we prove that $C(n)=\frac{1}{n-1}$ satisfies the condition.
From $1-x_{i}+1-x_{j} \geqslant 2 \sqrt{... | \frac{1}{n-1} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,093 |
Example 12 Given an integer $n \geqslant 2$ and a positive real number $a$, positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1} x_{2} \cdots x_{n}=1$. Find the smallest real number $M=M(n, a)$, such that
$$\sum_{i=1}^{n} \frac{1}{a+S-x_{i}} \leqslant M$$
always holds, where $S=x_{1}+x_{2}+\cdots+x_{n}$. | First, consider the case where $a \geqslant 1$. Let $x_{i}=y_{i}^{n}, y_{i}>0$, then $y_{1} y_{2} \cdots y_{n}=1$. We have
$$\begin{array}{l}
S-x_{i}=\sum_{j \neq i} y_{j}^{n} \geqslant(n-1)\left(\frac{\sum_{j \neq i} y_{j}}{n-1}\right)^{n} \text{ (Power Mean Inequality)} \\
\geqslant(n-1)\left(\frac{\sum_{i \neq i} y_... | \left\{\begin{array}{ll}
\frac{n}{a-1+n}, & \text { if } a \geqslant 1 \\
\frac{1}{a}, & \text { if } 0<a<1
\end{array}\right.} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,094 |
■ Given non-negative real numbers $a_{1}, a_{2}, \cdots, a_{100}$ satisfy $a_{1}^{2}+a_{2}^{2}+\cdots+a_{100}^{2}=1$. Prove:
$$a_{1}^{2} a_{2}+a_{2}^{2} a_{3}+\cdots+a_{100}^{2} a_{1}<\frac{12}{25} .$$ | 1. Let $S=\sum_{k=1}^{100} a_{k}^{2} a_{k+1}$, where $a_{101}=a_{1}, a_{102}=a_{2}$. By the Cauchy-Schwarz inequality and the AM-GM inequality, we have $(3 S)^{2}=\left[\sum_{k=1}^{100} a_{k+1}\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\right)\right]^{2} \leqslant\left(\sum_{k=1}^{100} a_{k+1}^{2}\right) \sum_{k=1}^{100}\left(a_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,095 |
2 Let $x, y, z \in \mathbf{R}^{+}$, and $x+y+z \geqslant 6$. Find
$$M=\sum x^{2}+\sum \frac{x}{y^{2}+z+1}$$
the minimum value, where " $\sum$ " denotes the cyclic sum. | 2. By the AM-GM inequality, we have $\frac{x^{2}}{14}+\frac{x}{y^{2}+z+1}+\frac{2}{49}\left(y^{2}+z+1\right) \geqslant 3 \sqrt[3]{\frac{x^{3}}{7^{3}}}=\frac{3}{7} x$. Thus, $\frac{1}{14} \sum x^{2}+\sum \frac{x}{y^{2}+z+1}+\frac{2}{49} \sum x^{2}+\frac{2}{49} \sum x+\frac{6}{49} \geqslant \frac{3}{7} \sum x$. Therefore... | \frac{90}{7} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,096 |
3 Let $x, y, z$ be positive real numbers, satisfying
$$x y+y z+z x=x+y+z .$$
Prove: $\frac{1}{x^{2}+y+1}+\frac{1}{y^{2}+z+1}+\frac{1}{z^{2}+x+1} \leqslant 1$, and determine the condition for equality. | 3. By the Cauchy inequality, we have $\frac{1}{x^{2}+y+1} \leqslant \frac{1+y+z^{2}}{(x+y+z)^{2}}, \frac{1}{y^{2}+z+1} \leqslant$ $\frac{1+z+x^{2}}{(x+y+z)^{2}}, \frac{1}{z^{2}+x+1} \leqslant \frac{1+x+y^{2}}{(x+y+z)^{2}}$. Therefore, $\frac{1}{x^{2}+y+1}+\frac{1}{y^{2}+z+1}+$ $\frac{1}{z^{2}+x+1} \leqslant \frac{3+x+y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,097 |
5 Let $a, b, c>0$ and $a+b+c=3$. Prove:
$$\frac{a^{2}}{a+b^{2}}+\frac{b^{2}}{b+c^{2}}+\frac{c^{2}}{c+a^{2}} \geqslant \frac{3}{2} .$$ | 5. By Cauchy-Schwarz inequality, we have $\sum \frac{a^{2}}{a+b^{2}} \sum a^{2}\left(a+b^{2}\right) \geqslant\left(a^{2}+b^{2}+c^{2}\right)^{2}$, thus it suffices to prove $2\left(a^{2}+b^{2}+c^{2}\right)^{2} \geqslant 3\left[a^{2}\left(a+b^{2}\right)+b^{2}\left(b+c^{2}\right)+c^{2}\left(c+a^{2}\right)\right]$, which i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,100 |
6 Given that $\lambda$ is a positive real number. Find the maximum value of $\lambda$ such that for all positive real numbers $u, v, w$ satisfying the condition
$$u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v} \geqslant 1$$
we have
$$u+v+w \geqslant \lambda .$$ | 6. First, it is easy to observe that when $u=v=w=\frac{\sqrt{3}}{3}$, $u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v}=1$ and $u+v+w=\sqrt{3}$. Therefore, the maximum value of $\lambda$ does not exceed $\sqrt{3}$. Below, we prove that for all $u, v, w > 0$, and satisfying $u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v} \geqslant 1$, we... | \sqrt{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,101 |
7 Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, $x_{n+1}=x_{1}+x_{2}+\cdots+x_{n}$, prove:
$$x_{n+1} \sum_{i=1}^{n}\left(x_{n+1}-x_{i}\right) \geqslant\left(\sum_{i=1}^{n} \sqrt{x_{i}\left(x_{n+1}-x_{i}\right)}\right)^{2} .$$ | 7. Since $\sum_{i=1}^{n}\left(x_{n+1}-x_{i}\right)=n x_{n+1}-\sum_{i=1}^{n} x_{i}=(n-1) x_{n+1}$, it suffices to prove that $x_{n+1} \sqrt{n-1} \geqslant \sum_{i=1}^{n} \sqrt{x_{i}\left(x_{n+1}-x_{i}\right)}$, which is equivalent to proving $\sum_{i=1}^{n} \sqrt{\frac{x_{i}}{x_{n+1}}\left(1-\frac{x_{i}}{x_{n+1}}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,102 |
8. Let \(x, y, z, w \in \mathbf{R}^{+}, \alpha, \beta, \gamma, \theta\) satisfy \(\alpha+\beta+\gamma+\theta=(2 k+1) \pi, k \in \mathbf{Z}\). Prove:
$$(x \sin \alpha+y \sin \beta+z \sin \gamma+w \sin \theta)^{2} \leqslant \frac{(x y+z w)(x z+y w)(x w+y z)}{x y z w},$$
with equality if and only if \(x \cos \alpha=y \co... | 8. Let $u=x \sin \alpha+y \sin \beta, v=z \sin \gamma+w \sin \theta$, then $u^{2}=$ $(x \sin \alpha+y \sin \beta)^{2} \leqslant(x \sin \alpha+y \sin \beta)^{2}+(x \cos \alpha-y \cos \beta)^{2}=x^{2}+y^{2}-$ $2 x y \cos (\alpha+\beta)$. Therefore, $\cos (\alpha+\beta) \leqslant \frac{x^{2}+y^{2}-u^{2}}{2 x y}$. Similarl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,103 |
9 Given $0<a_{1}<a_{2}<\cdots<a_{n}$, for any permutation $b_{1}, b_{2}, \cdots$, $b_{n}$ of $a_{1}, a_{2}, \cdots, a_{n}$. Let $M=\prod_{i=1}^{n}\left(a_{i}+\frac{1}{b_{i}}\right)$, find the permutation $b_{1}, b_{2}, \cdots, b_{n}$ that maximizes $M$. | 9. Let $A=a_{1} a_{2} \cdots a_{n}$, then $M=\frac{1}{A} \prod_{i=1}^{n}\left(a_{i} b_{i}+1\right)$. From $\left(a_{i} b_{i}+1\right)^{2} \leqslant\left(a_{i}^{2}+1\right)$ $\left(b_{i}^{2}+1\right)$, we know that equality holds $\Leftrightarrow a_{i}=b_{i}$. This leads to $M \leqslant \frac{1}{A} \prod_{i=1}^{n}\left(... | b_{1}=a_{1}, b_{2}=a_{2}, \cdots, b_{n}=a_{n} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,104 |
10. Let the complex number $z_{k}=x_{k}+\mathrm{i} y_{k}, k=1,2, \cdots, n, x_{i}$ and $y_{i}$ be real numbers, $\mathrm{i}=\sqrt{-1}$. Let $r$ denote the absolute value of the real part of $\sqrt{z_{1}^{2}+z_{2}^{2}+\cdots+z_{n}^{2}}$, prove that:
$$r \leqslant\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\r... | 10. Let $a+\mathrm{i} b=\sqrt{\sum_{i=1}^{n} z_{i}^{2}}, a, b \in \mathbf{R}$, then $a^{2}-b^{2}=\sum_{k=1}^{n} x_{k}^{2}-\sum_{k=1}^{n} y_{k}^{2}, a b=$ $\sum_{k=1}^{n} x_{k} y_{k}$. If $r=|a|>\sum_{k=1}^{n}\left|x_{k}\right|$, since $\sum_{k=1}^{n}\left|x_{k}\right| \geqslant\left(\sum_{k=1}^{n} x_{k}^{2}\right)^{\fr... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,105 |
11 Let $A_{n}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}, a_{i}>0, i=1,2, \cdots, n$. Prove: | \begin{aligned} \text { 11. } n \sum_{i=1}^{n}\left(a_{i}-\frac{1}{a_{i}}\right)^{2} & =n \sum_{i=1}^{n} a_{i}^{2}+n \sum_{i=1}^{n} \frac{1}{a_{i}^{2}}-2 n^{2} \geqslant\left(\sum_{i=1}^{n} a_{i}\right)^{2}+\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right)^{2}- \\ 2 n^{2} \geqslant n^{2}\left(A_{n}^{2}+\frac{1}{A_{n}^{2}}-2\... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,106 |
12 For all real numbers $r, s, t$ satisfying $1 \leqslant r \leqslant s \leqslant t$. Find
$$w=(r-1)^{2}+\left(\frac{s}{r}-1\right)^{2}+\left(\frac{t}{s}-1\right)^{2}+\left(\frac{4}{t}-1\right)^{2}$$
the minimum value. | 12. By Cauchy-Schwarz inequality, we have \( w \geqslant \frac{1}{4}\left[(r-1)+\left(\frac{s}{r}-1\right)+\left(\frac{t}{s}-1\right)+\left(\frac{4}{t}-1\right)\right]^{2} \) \( =\frac{1}{4}\left(r+\frac{s}{r}+\frac{t}{s}+\frac{4}{t}-4\right)^{2} \). Also, \( r+\frac{s}{r}+\frac{t}{s}+\frac{4}{t} \geqslant 4 \sqrt[4]{r... | 4(\sqrt{2}-1)^{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,107 |
13 Let $a_{i}>0, b_{i}>0, a_{i} b_{i}-c_{i}^{2}>0(i=1,2, \cdots, n)$, then
$$\frac{n^{3}}{\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} b_{i}\right)-\left(\sum_{i=1}^{n} c_{i}\right)^{2}} \leqslant \sum_{i=1}^{n} \frac{1}{a_{i} b_{i}-c_{i}^{2}}$$ | 13. Let $a_{i} b_{i}-c_{i}^{2}=d_{i}^{2}>0$, then by the Cauchy-Schwarz inequality, we get $\left(\sum a_{i}\right)\left(\sum b_{i}\right) \geqslant$ $\left(\sum \sqrt{a_{i} b_{i}}\right)^{2}=\sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{a_{i} b_{i}} \sqrt{a_{j} b_{j}}=\sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{c_{i}^{2}+d_{i}^{2}} \sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,108 |
14 Let $\frac{1}{2} \leqslant p \leqslant 1, a_{i} \geqslant 0,0 \leqslant b_{i} \leqslant p$ and $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}=1$, prove:
$$\sum_{i=1}^{n} b_{i} \prod_{\substack{1 \leq j \leq n \\ j \neq i}} a_{j} \leqslant \frac{p}{(n-1)^{n-1}}$$ | 14. Let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}$. Let $A_{i}=\prod_{\substack{j=1 \\ j \neq i}}^{n} a_{j}$, then $A_{1} \geqslant A_{2} \geqslant \cdots \geqslant A_{n} \geqslant 0$. By the rearrangement inequality, we get $\sum_{i=1}^{n} b_{i} A_{... | \frac{p}{(n-1)^{n-1}} | Inequalities | proof | Yes | Yes | inequalities | false | 738,109 |
Example 3 Let $a_{i}>0, i=1,2, \cdots, n$ satisfy $a_{1} a_{2} \cdots a_{n}=1$. Prove:
The translation maintains the original text's line breaks and format. | $$\left(2+a_{1}\right)\left(2+a_{2}\right) \cdots\left(2+a_{n}\right) \geqslant 3^{n}$$
Proof: Since for any $i$,
$$2+a_{i}=1+1+a_{i} \geqslant 3 \sqrt[3]{a_{i}} .$$
Thus $\quad\left(2+a_{1}\right)\left(2+a_{2}\right) \cdots\left(2+a_{n}\right) \geqslant 3^{n} \sqrt[3]{a_{1} a_{2} \cdots a_{n}}=3^{n}$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,110 |
15 Given a natural number $n \geqslant 2$, find the smallest positive number $\lambda$, such that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, and any $n$ numbers $b_{1}, b_{2}, \cdots, b_{n}$ in $\left[0, \frac{1}{2}\right]$, if
$$\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}=1,$$
then
$$\prod_{i=1}^{n} a_{i} ... | 15. By Cauchy's inequality, we have \(1 = \sum_{i=1}^{n} b_{i} \leqslant \left(\sum_{i=1}^{n} \frac{b_{i}}{a_{i}}\right)^{\frac{1}{2}} \left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{\frac{1}{2}}\), thus \(\frac{1}{\sum_{i=1}^{n} a_{i} b_{i}} \leqslant \sum_{i=1}^{n} \frac{b_{i}}{a_{i}}\). Let \(M = \prod_{i=1}^{n} a_{i}, A_{... | \frac{1}{2} \left(\frac{1}{n-1}\right)^{n-1} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,111 |
16 Given two natural numbers $n$ and $m$ greater than 1, find all natural numbers $l$ such that for any positive numbers $a_{1}$, $a_{2}, \cdots, a_{n}$, we have
$$\sum_{k=1}^{n} \frac{1}{S_{k}}\left(l k+\frac{1}{4} l^{2}\right)<m^{2} \sum_{k=1}^{n} \frac{1}{a_{k}},$$
where, $S_{k}=\sum_{i=1}^{k} a_{i}$. | 16. $\sum_{k=1}^{n} \frac{1}{S_{k}}\left(l k+\frac{1}{4} l^{2}\right)=\sum_{k=1}^{n}\left[\frac{1}{S_{k}}\left(\frac{l}{2}+k\right)^{2}-\frac{k^{2}}{S_{k}}\right]=\left(\frac{l}{2}+1\right)^{2} \frac{1}{S_{1}}-\frac{n^{2}}{S_{n}}+$ $\sum_{k=2}^{n}\left[\frac{1}{S_{k}}\left(\frac{l}{2}+k\right)^{2}-\frac{(k-1)^{2}}{S_{k... | 1,2, \cdots, 2(m-1) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,112 |
17 Let $u, v$ be positive real numbers, and for a given positive integer $n$, find the necessary and sufficient conditions for $u, v$ such that there exist real numbers $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}>0$ satisfying
$$\sum_{i=1}^{n} a_{i}=u, \sum_{i=1}^{n} a_{i}^{2}=v,$$
When these numbers exist... | 17. If there exist $a_{1}, a_{2}, \cdots, a_{n}$, by the Cauchy-Schwarz inequality, we have $\left(\sum_{i=1}^{n} a_{i}\right)^{2} \leqslant n \sum_{i=1}^{n} a_{i}^{2}$. Also, $\left(\sum_{i=1}^{n} a_{i}\right)^{2} \geqslant \sum_{i=1}^{n} a_{i}^{2}$, so the necessary condition for $u$ and $v$ is $v \leqslant u^{2} \le... | \frac{k u+\sqrt{k\left[(k+1) v-u^{2}\right]}}{k(k+1)} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,113 |
18 Let the sum of $m$ distinct positive even numbers and $n$ distinct positive odd numbers be 1987. For all such $m$ and $n$, what is the maximum value of $3m + 4n$? | 18. Let \(a_{1}+a_{2}+\cdots+a_{m}+b_{1}+b_{2}+\cdots+b_{n}=1987\), where \(a_{i} (1 \leqslant i \leqslant m)\) are distinct positive even numbers, and \(b_{j} (1 \leqslant j \leqslant n)\) are distinct positive odd numbers. Clearly, \(n\) is odd, and \(a_{1}+a_{2}+\cdots+a_{m} \geqslant 2+4+\cdots+2 m = m(m+1)\), \(b_... | 221 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 738,114 |
19 Let $x_{i} \in \mathbf{R} (i=1,2, \cdots, n)$ and $\sum_{i=1}^{n} x_{i}^{2}=1$, prove: for any integer $k \geqslant 3$ there exist integers $a_{i},\left|a_{i}\right| \leqslant k-1$. such that
$$\left|\sum_{i=1}^{n} a_{i} x_{i}\right| \leqslant \frac{(k-1) \sqrt{n}}{k^{n}-1}$$ | 19. Let $x_{i} \geqslant 0$, consider $A=\left\{\sum_{i=1}^{n} e_{i} x_{i} \mid e_{i} \in\{0,1,2, \cdots, k-1\}\right\}$. If all numbers in $A$ are distinct, then $|A|=k^{n}$. By the Cauchy-Schwarz inequality, we get $0 \leqslant \sum e_{i} x_{i} \leqslant(k-1) \sum x_{i} \leqslant(k-1) \sqrt{\sum x_{i}^{2}} \cdot \sqr... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,115 |
20. Let $s, t, u \in\left(0, \frac{\pi}{2}\right)$, satisfying $s+t+u+v=\pi$, prove: $\frac{\sqrt{2} \sin s-1}{\cos s}+$
$$\frac{\sqrt{2} \sin t-1}{\cos t}+\frac{\sqrt{2} \sin u-1}{\cos u}+\frac{\sqrt{2} \sin v-1}{\cos v} \geqslant 0$$ | 20. Let $a=\tan s, b=\tan t, c=\tan u, d=\tan v$, then $a, b, c, d \in \mathbf{R}^{+}$, from $s+t+u+v=\pi$, we get $\tan (s+t)+\tan (u+v)=0$. That is, $\frac{a+b}{1-a b}+\frac{c+d}{1-c d}=0$. Multiplying both sides by $(1-a b)(1-c d)$, we get $a+b+c+d=a b c+b c d+c d a+d a b$. This leads to $(a+b)(a+c)(a+d)=\left(a^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,116 |
21 Proof:
$$\sqrt{\frac{A B_{1}}{A B}}+\sqrt{\frac{B C_{1}}{B C}}+\sqrt{\frac{C A_{1}}{C A}} \leqslant \frac{3}{\sqrt{2}}$$
where, $A_{1} 、 B_{1} 、 C_{1}$ are the points of tangency of the incircle of $\triangle A B C$ with sides $B C 、 A C 、 A B$, respectively. | 21. Let $x=A B_{1}, y=B C_{1}, z=C A_{1}$. To prove $\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}} \leqslant \frac{3}{\sqrt{2}}$, it suffices to prove $\frac{1}{\sqrt{1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}}+\frac{1}{\sqrt{1+c^{2}}} \leqslant \frac{3}{\sqrt{2}}$, where $a, b, c$ are positive real numbers, an... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,117 |
$22 a_{1}, a_{2}, a_{3}, a_{4}$ are the side lengths of a quadrilateral with perimeter $2 s$, prove: $\sum_{i=1}^{4} \frac{1}{a_{i}+s} \leqslant$
$$\frac{2}{9} \sum_{1 \leq i<j \leqslant 4} \frac{1}{\sqrt{\left(s-a_{i}\right)\left(s-a_{j}\right)}}$$ | 22. Proof: Since $\frac{2}{9} \sum_{1 \leqslant i<j \leqslant 4} \frac{1}{\sqrt{\left(s-a_{i}\right)\left(s-a_{j}\right)}} \geqslant \frac{4}{9} \sum_{1 \leqslant i<j \leqslant 4} \frac{1}{\left(s-a_{i}\right)\left(s-a_{j}\right)} \cdots$ (1). Therefore, it suffices to prove: $\sum_{i=1}^{4} \frac{1}{a_{i}+s} \leqslant... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,118 |
23 Let $a, b, c$ be the lengths of the three sides of a triangle. Prove:
$$\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}} \leqslant 3 .$$ | 23. Let $a \geqslant b \geqslant c$. Then, $\sqrt{a+b-c}-\sqrt{a}=\frac{(a+b-c)-a}{\sqrt{a+b-c}+\sqrt{a}} \leqslant$ $\frac{b-c}{\sqrt{b}+\sqrt{c}}=\sqrt{b}-\sqrt{c}$. Therefore, $\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}} \leqslant 1 \cdots$ (1). Let $p=\sqrt{a}+\sqrt{b}, q=\sqrt{a}-\sqrt{b}$. Then $a-b=p q, p \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,119 |
Example 4 Let $a>b>0$, prove: $\sqrt{2} a^{3}+\frac{3}{a b-b^{2}} \geqslant 10$. | Prove that since $a b-b^{2}=b(a-b) \leqslant \frac{[b+(a-b)]^{2}}{4}=\frac{a^{2}}{4}$, therefore
$$\begin{aligned}
& \sqrt{2} a^{3}+\frac{3}{a b-b^{2}} \geqslant \sqrt{2} a^{3}+\frac{12}{a^{2}} \\
= & \frac{\sqrt{2}}{2} a^{3}+\frac{\sqrt{2}}{2} a^{3}+\frac{4}{a^{2}}+\frac{4}{a^{2}}+\frac{4}{a^{2}} \\
\geqslant & 5 \sqr... | 10 | Inequalities | proof | Yes | Yes | inequalities | false | 738,120 |
Example 5 Let $a, b, c>0$, prove:
$$\frac{c}{a}+\frac{a}{b+c}+\frac{b}{c} \geqslant 2$$ | Prove that by the AM-GM inequality, we have
$$\begin{aligned}
\frac{c}{a}+\frac{a}{b+c}+\frac{b}{c} & =\frac{c}{a}+\frac{a}{b+c}+\frac{b+c}{c}-1 \\
& \geqslant 3 \sqrt[3]{\frac{c}{a} \cdot \frac{a}{b+c} \cdot \frac{b+c}{c}}-1 \\
& =3-1=2
\end{aligned}$$
Thus, the proposition is proved. | 2 | Inequalities | proof | Yes | Yes | inequalities | false | 738,121 |
Example 6 Let $x+y+z=0$, prove:
$$6\left(x^{3}+y^{3}+z^{3}\right)^{2} \leqslant\left(x^{2}+y^{2}+z^{2}\right)^{3} .$$ | Prove that from $x+y+z=0$ and its symmetry, without loss of generality, assume $x, y \geqslant 0, z \leqslant 0$, since $x+y=-z$, we get $z^{2}=(x+y)^{2}$, thus
$$\left(x^{2}+y^{2}+z^{2}\right)^{3}=8\left(x^{2}+x y+y^{2}\right)^{3}$$
By $A_{3} \geqslant G_{3}$, we have
$$\begin{aligned}
x^{2}+x y+y^{2} & =\frac{x(x+y)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,122 |
Example 7 Let $a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{R}^{+}, S=a_{1}+a_{2}+\cdots+a_{n}$. Prove:
$$\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right) \leqslant 1+S+\frac{S^{2}}{2!}+\cdots+\frac{S^{n}}{n!} .$$ | Prove that since $G_{n} \leqslant A_{n}$, we have
$$\begin{aligned}
& \left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right) \\
\leqslant & \left(\frac{n+a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^{n}=\left(1+\frac{S}{n}\right)^{n} \\
= & 1+\mathrm{C}_{n}^{1}\left(\frac{S}{n}\right)+\mathrm{C}_{n}^{2}\left(\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,123 |
Example 8 Let $k, n$ be positive integers, and $1 \leqslant k \leqslant n, a_{i} \in \mathbf{R}^{+}$, satisfying $a_{1}+a_{2}+\cdots+$ $a_{k}=a_{1} a_{2} \cdots a_{k}$. Prove:
$$a_{1}^{n-1}+a_{2}^{n-1}+\cdots+a_{k}^{n-1} \geqslant k n$$
and determine the necessary and sufficient conditions for equality. | Prove that $a=a_{1}+a_{2}+\cdots+a_{k}=a_{1} a_{2} \cdots a_{k}$. By the AM-GM inequality, we have $a \geqslant k a^{\frac{1}{k}}$, i.e., $a \geqslant k^{\frac{k}{k-1}}$.
Also,
$$a_{1}^{n-1}+a_{2}^{n-1}+\cdots+a_{k}^{n-1} \geqslant k\left(a_{1} a_{2} \cdots a_{k}\right)^{\frac{n-1}{k}}=k a^{\frac{n-1}{k}} \geqslant k ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,124 |
Example 9 Let $a_{i}>0(i=1,2, \cdots, n)$, prove:
$$\sum_{k=1}^{n} k a_{k} \leqslant \frac{n(n-1)}{2}+\sum_{k=1}^{n} a_{k}^{k} .$$ | Prove that since $\frac{n(n-1)}{2}=\sum_{k=1}^{n}(k-1)$, by the AM-GM inequality, we have
$$\begin{array}{l}
\frac{n(n-1)}{2}+\sum_{k=1}^{n} a_{k}^{k} \\
=\sum_{k=1}^{n}\left[(k-1)+a_{k}^{k}\right] \\
=\sum_{k=1}^{n}\left(1+1+\cdots+1+a_{k}^{k}\right) \\
\geqslant \sum_{k=1}^{n} k \sqrt[k]{1^{k-1} \cdot a_{k}^{k}}=\sum... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,125 |
Example 10 Let $a_{i}>0, b_{i}>0$ and satisfy $a_{1}+a_{2}+\cdots+a_{n} \leqslant 1, b_{1}+b_{2}+\cdots+$ $b_{n} \leqslant n$. Prove:
$$\left(\frac{1}{a_{1}}+\frac{1}{b_{1}}\right)\left(\frac{1}{a_{2}}+\frac{1}{b_{2}}\right) \cdots\left(\frac{1}{a_{n}}+\frac{1}{b_{n}}\right) \geqslant(n+1)^{n}$$ | Prove that from the given conditions and the AM-GM inequality, we have
$$\begin{aligned}
a_{1} a_{2} \cdots a_{n} & \leqslant\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^{n} \leqslant \frac{1}{n^{n}} \\
b_{1} b_{2} \cdots b_{n} & \leqslant\left(\frac{b_{1}+b_{2}+\cdots+b_{n}}{n}\right)^{n} \leqslant 1 \\
\frac{1}{a_... | (n+1)^{n} | Inequalities | proof | Yes | Yes | inequalities | false | 738,127 |
Example 11 Assume $a, b, c$ are all positive numbers, prove:
$$a b c \geqslant(a+b-c)(b+c-a)(c+a-b) .$$ | Prove that if $a+b-c, b+c-a, c+a-b$ contain a negative number, without loss of generality, assume $a+b-c<0$. Therefore, $b+c-a$ and $c+a-b$ are both positive, and the conclusion is obviously true.
If $a+b-c, b+c-a, c+a-b$ are all non-negative, then by the arithmetic mean inequality, we get
$$\sqrt{(a+b-c)(b+c-a)} \leq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,128 |
Example 12 Assume positive numbers $a, b, c$ satisfy $(1+a)(1+b)(1+c)=8$. Prove: $abc \leqslant 1$. | Proof: From the assumption, we have
$$1+(a+b+c)+(a b+b c+c a)+a b c=8 .$$
By the AM-GM inequality, we get
$$a+b+c \geqslant 3(a b c)^{\frac{1}{3}}, a b+b c+c a \geqslant 3(a b c)^{\frac{2}{3}} .$$
Equality holds if and only if $a=b=c$. Therefore,
$$8 \geqslant 1+3(a b c)^{\frac{1}{3}}+3(a b c)^{\frac{2}{3}}+a b c=\le... | abc \leqslant 1 | Inequalities | proof | Yes | Yes | inequalities | false | 738,129 |
Example 13 Let $n$ be a positive integer, prove:
$$n\left[(n+1)^{\frac{1}{n}}-1\right] \leqslant 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} \leqslant n-(n-1)\left(\frac{1}{n}\right)^{\frac{1}{n-1}}$$ | Prove only the left side of the inequality; the right side can be handled similarly.
Let \( A = \frac{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} + n}{n} \), then the left side of the inequality is equivalent to
\[ A \geqslant (n+1)^{\frac{1}{n}} \]
By the AM-GM inequality, we have
\[ \begin{aligned}
A & = \f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,130 |
Example 15 Let $a, b, c \in \mathbf{R}^{+}$, and $a^{2}+b^{2}+c^{2}=1$. Prove:
$$\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\frac{c}{1-c^{2}} \geqslant \frac{3 \sqrt{3}}{2} .$$ | Prove the original inequality
$$\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\frac{c}{1-c^{2}} \geqslant \frac{3 \sqrt{3}}{2}$$
is equivalent to $\quad \frac{a^{2}}{a\left(1-a^{2}\right)}+\frac{b^{2}}{b\left(1-b^{2}\right)}+\frac{c^{2}}{c\left(1-c^{2}\right)} \geqslant \frac{3 \sqrt{3}}{2}$.
Given that $a^{2}+b^{2}+c^{2}=1$, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,132 |
Example 16 Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of $1,2, \cdots, n$. Prove that:
$$\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1}{n} \leqslant \frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}$$ | Prove that since $a_{1}, a_{2}, \cdots, a_{n}$ is a permutation of $1,2, \cdots, n$, we have
$$\begin{aligned}
& \left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n-1}\right) \\
\geqslant & (1+1)(1+2) \cdots[1+(n-1)] \\
= & a_{1} a_{2} \cdots a_{n} .
\end{aligned}$$
Thus,
$$\begin{aligned}
& \frac{a_{1}}{a_{2}... | \frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}} \geqslant \frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1}{n} | Inequalities | proof | Yes | Yes | inequalities | false | 738,133 |
Example 17 Let $a, b, c$ be positive real numbers, prove that:
$$\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1 .$$ | It is easy to see that if we can prove $\frac{a}{\sqrt{a^{2}+8 b c}} \geqslant \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$, then by adding them up, we obtain the inequality to be proved. Because
$$\begin{array}{c}
\frac{a}{\sqrt{a^{2}+8 b c}} \geqslant \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,134 |
Example 19 Given $n \geqslant 2, n \in \mathbf{Z}^{+}$, find all $m \in \mathbf{Z}^{+}$, such that for $a_{i} \in \mathbf{R}^{+}, i=1,2$, $\cdots, n$, satisfying $a_{1} a_{2} \cdots a_{n}=1$, then
$$a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} .$$ | Let $x=a_{1}=a_{2}=\cdots=a_{n-1}>0, a_{n}=\frac{1}{x^{n-1}}$, then
$$(n-1) x^{m}+\frac{1}{x^{(n-1) m}} \geqslant \frac{n-1}{x}+x^{n-1}$$
From this, we get $m \geqslant n-1$. Now, assume $m \geqslant n-1$, then
$$\begin{aligned}
& (n-1)\left(a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}\right)+n(m-n+1) \\
= & \left(a_{1}^{m}+a... | m \geqslant n-1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,136 |
5 positive real numbers $x, y, z$ satisfy $xyz=1$. Prove: $\frac{x^{3}+y^{3}}{x^{2}+xy+y^{2}}+\frac{y^{3}+z^{3}}{y^{2}+yz+z^{2}}+$ $\frac{z^{3}+x^{3}}{z^{2}+zx+x^{2}} \geqslant 2$. | 5. Notice that $\frac{x^{2}-x y+y^{2}}{x^{2}+x y+y^{2}} \geqslant \frac{1}{3} \Leftrightarrow 3\left(x^{2}-x y+y^{2}\right) \geqslant x^{2}+x y+y^{2} \Leftrightarrow 2(x-$ $y)^{2} \geqslant 0$. Then $\frac{x^{3}+y^{3}}{x^{2}+x y+y^{2}}=\frac{x^{2}-x y+y^{2}}{x^{2}+x y+y^{2}}(x+y) \geqslant \frac{x+y}{3}$. Therefore, $\... | 2 | Inequalities | proof | Yes | Yes | inequalities | false | 738,137 |
Example 20 Let $n(n \geqslant 2)$ be an integer, $a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{R}^{+}$, prove that:
$$\left(a_{1}^{3}+1\right)\left(a_{2}^{3}+1\right) \cdots\left(a_{n}^{3}+1\right) \geqslant\left(a_{1}^{2} a_{2}+1\right)\left(a_{2}^{2} a_{3}+1\right) \cdots\left(a_{n}^{2} a_{1}+1\right) .$$ | Prove that for positive real numbers $x_{i}, y_{i}(i=1,2,3)$, we have
$$\prod\left(x_{i}^{3}+y_{i}^{3}\right) \geqslant\left(\prod x_{i}+\prod y_{i}\right)^{3}$$
In fact, by the AM-GM inequality, we get
$$\begin{array}{l}
\sqrt[3]{\frac{x_{1}^{3} x_{2}^{3} x_{3}^{3}}{\prod\left(x_{i}^{3}+y_{i}^{3}\right)}} \leqslant \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,138 |
Example 21 Let $a, b, c>0, a+b+c=1$. Prove: If positive real numbers $x_{1}, x_{2}, \cdots, x_{5}$ satisfy $x_{1} x_{2} \cdots x_{5}=1$, then
$$\prod_{i=1}^{5}\left(a x_{i}^{2}+b x_{i}+c\right) \geqslant 1$$ | Prove the generalization of the problem.
Consider the expression $\prod_{i=1}^{n}\left(a x_{i}^{2}+b x_{i}+c\right)$.
Then the term containing $a^{i} b^{j} c^{k}(i+j+k=n)$ is
$$a^{i} b^{j} c^{k}\left[\left(x_{1} x_{2} \cdots x_{i}\right)^{2}\left(x_{i+1} x_{i+2} \cdots x_{i+j}\right)+\cdots\right]$$
Therefore, there a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,139 |
Example 22 Given $x, y, z \in \mathbf{R}^{+} \cup\{0\}$, and $x+y+z=2$. | Prove: $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+x y z \leqslant 1$, and find the values of $x, y, z$ when the equality holds.
Proof Notice that
$$\begin{aligned}
& x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+x y z \\
= & \frac{1}{2}\left(2 x^{2} y^{2}+2 y^{2} z^{2}+2 z^{2} x^{2}+2 x y z\right) \\
= & \frac{1}{2}(x y \cdot 2 x y+y ... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,140 |
Example 23 Given that $a, b, c$ are positive real numbers. Prove:
$$\frac{a^{2} b(b-c)}{a+b}+\frac{b^{2} c(c-a)}{b+c}+\frac{c^{2} a(a-b)}{c+a} \geqslant 0 .$$ | $$\begin{aligned}
\text { Original } & \Leftrightarrow \frac{a^{2} b^{2}}{a+b}+\frac{b^{2} c^{2}}{b+c}+\frac{c^{2} a^{2}}{c+a} \geqslant a b c\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right) \\
& \Leftrightarrow \frac{a b}{c(a+b)}+\frac{b c}{a(b+c)}+\frac{a c}{b(c+a)} \geqslant \frac{a}{a+b}+\frac{b}{b+c}+\frac{c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,141 |
Example 25 Let $x, y, z \in(0,1)$, satisfying:
$$\sqrt{\frac{1-x}{y z}}+\sqrt{\frac{1-y}{z x}}+\sqrt{\frac{1-z}{x y}}=2,$$
Find the maximum value of $x y z$. | Let $u=\sqrt[6]{x y z}$, then by the condition and the AM-GM inequality, we have
$$\begin{aligned}
2 u^{3}= & 2 \sqrt{x y z}=\frac{1}{\sqrt{3}} \sum \sqrt{x(3-3 x)} \\
\leqslant & \frac{1}{\sqrt{3}} \sum \frac{x+(3-3 x)}{2} \\
& =\frac{3 \sqrt{3}}{2}-\frac{1}{\sqrt{3}}(x+y+z) \\
& \leqslant \frac{3 \sqrt{3}}{2}-\sqrt{3... | \frac{27}{64} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,143 |
Example 26 Let $x, y, z \in \mathbf{R}^{+}$, prove:
$$\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}>2 \sqrt[3]{x^{3}+y^{3}+z^{3}} .$$ | To prove the inequality is equivalent to
$$\begin{aligned}
& \left(\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}\right)^{3}>8\left(x^{3}+y^{3}+z^{3}\right) \\
\Leftrightarrow & \left(\frac{x y}{z}\right)^{3}+\left(\frac{y z}{x}\right)^{3}+\left(\frac{z x}{y}\right)^{3}+6 x y z+3 x^{3}\left(\frac{y}{z}+\frac{z}{y}\right) \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,144 |
Example 27 Let $x, y, z$ be non-negative real numbers, and $x+y+z=1$, prove that:
$$x y+y z+z x-2 x y z \leqslant \frac{7}{27} .$$ | Assume without loss of generality that $x \geqslant y \geqslant z$.
When $x \geqslant \frac{1}{2}$, then $y z-2 x y z \leqslant 0$, so
$$x y+y z+z x-2 x y z \leqslant x y+z x=x(1-x) \leqslant \frac{1}{4}<\frac{7}{27}$$
When $x<\frac{1}{2}$, then $y \leqslant \frac{1}{2}, z \leqslant \frac{1}{2}$.
$$(1-2 x)(1-2 y)(1-2 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,145 |
Example 28 Let $n$ be a positive integer, $\left(x_{1}, x_{2}, \cdots, x_{n}\right),\left(y_{1}, y_{2}, \cdots, y_{n}\right)$ be two sequences of positive numbers. Assume the sequence of positive real numbers $\left(z_{1}, z_{2}, \cdots, z_{2 n}\right)$ satisfies
$$z_{i+j}^{2} \geqslant x_{i} y_{j}, 1 \leqslant i, j \l... | Let $X=\max \left\{x_{1}, x_{2}, \cdots, x_{n}\right\}, Y=\max \left\{y_{1}, y_{2}, \cdots, y_{n}\right\}$. Without loss of generality, assume $X=Y=1$ (otherwise, use $a_{i}=\frac{x_{i}}{X}, b_{i}=\frac{y_{i}}{Y}, c_{i}=\frac{z_{i}}{\sqrt{X Y}}$ instead).
We will prove
$$M+z_{2}+z_{3}+\cdots+z_{2 n} \geqslant x_{1}+x_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,146 |
6 Let $a_{1}, a_{2}, \cdots, a_{n}>0$ and $a_{1}+a_{2}+\cdots+a_{n}=1$. Prove:
$$\left(\frac{1}{a_{1}^{2}}-1\right)\left(\frac{1}{a_{2}^{2}}-1\right) \cdots\left(\frac{1}{a_{n}^{2}}-1\right) \geqslant\left(n^{2}-1\right)^{n}$$ | 6. Since $a_{1}+a_{2}+\cdots+a_{n}=1$, by the AM-GM inequality we have $1+a_{i}=a_{1}+a_{2}+\cdots+a_{n}+a_{i} \geqslant(n+1)\left(a_{1} a_{2} \cdots a_{n} a_{i}\right)^{1 /(n+1)}, 1-a_{i}=a_{1}+a_{2}+\cdots+a_{n}-a_{i} \geqslant (n-1)\left(a_{1} a_{2} \cdots a_{n} / a_{i}\right)^{1 /(n-1)}$. Taking $i=1,2, \cdots, n$ ... | \left(\frac{1}{a_{1}^{2}}-1\right)\left(\frac{1}{a_{2}^{2}}-1\right) \cdots\left(\frac{1}{a_{n}^{2}}-1\right) \geqslant\left(n^{2}-1\right)^{n} | Inequalities | proof | Yes | Yes | inequalities | false | 738,148 |
Example 2 Let non-negative real numbers $a$ and $d$, and positive numbers $b$ and $c$, satisfy the condition $b+c \geqslant a+d$. Find the minimum value of $\frac{b}{c+d}+$ $\frac{c}{a+b}$. | Let's assume $a+b \geqslant c+d$. Since $\frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}-$ $c\left(\frac{1}{c+d}-\frac{1}{a+b}\right)$, note that $c \leqslant c+d$ and $b+c \geqslant a+d \Leftrightarrow b+c \geqslant \frac{1}{2}(a+b+$ $c+d)$. Therefore, we get
$$\begin{aligned}
\frac{b}{c+d}+\frac{c}{a+b} & \geqslant \frac... | \sqrt{2}-\frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,149 |
Example 3 Given $2 x>3 y>0$, find the minimum value of $\sqrt{2} x^{3}+\frac{3}{2 x y-3 y^{2}}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given that $2 x>3 y>0$, we have $2 x-3 y>0$. By the AM-GM inequality, we get
$$\begin{aligned}
2 x y-3 y^{2} & =y(2 x-3 y)=\frac{1}{3} \cdot 3 y(2 x-3 y) \\
& \leqslant \frac{1}{3} \cdot\left[\frac{3 y+(2 x-3 y)}{2}\right]^{2}=\frac{1}{3} x^{2}
\end{aligned}$$
Therefore,
$$\begin{aligned}
& \sqrt{2} x^{3}+\frac{3}{2 x... | 5 \sqrt[5]{\frac{27}{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,150 |
Example 4 If $x, y, z$ are positive real numbers, find the maximum value of $\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+18)}$, and prove your conclusion. | Given a fixed $y$,
$$\begin{aligned}
& \frac{x}{(1+5 x)(4 x+3 y)} \\
= & \frac{x}{20 x^{2}+(15 y+4) x+3 y} \\
= & \frac{1}{20 x+\frac{3 y}{x}+15 y+4} \\
\leqslant & \frac{1}{2 \sqrt{20 \times 3 y}+15 y+4} \\
= & \frac{1}{(\sqrt{15 y}+2)^{2}}
\end{aligned}$$
Equality holds if and only if $x=\sqrt{\frac{3 y}{20}}$.
Simi... | \frac{1}{5120} | Inequalities | proof | Yes | Yes | inequalities | false | 738,151 |
Example 5 If for any positive real numbers, $\frac{a^{2}}{\sqrt{a^{4}+3 b^{4}+3 c^{4}}}+\frac{k}{a^{3}} \cdot\left(\frac{c^{4}}{b}+\frac{b^{4}}{c}\right) \geqslant \frac{2 \sqrt{2}}{3}$. Always holds, find the minimum value of the real number $k$.
| Solve $\frac{a^{2}}{\sqrt{a^{4}+3 b^{4}+3 c^{4}}}=\frac{\sqrt{2} a^{4}}{\sqrt{2 a^{4}\left(a^{4}+3 b^{4}+3 c^{4}\right)}}$.
$$\begin{array}{l}
\geqslant \frac{\sqrt{2} a^{4}}{\frac{1}{2}\left[2 a^{4}+\left(a^{4}+3 b^{4}+3 c^{4}\right)\right]} \\
=\frac{2 \sqrt{2}}{3} \cdot \frac{a^{4}}{a^{4}+b^{4}+c^{4}}
\end{array}$$
... | \frac{1}{\sqrt[4]{24}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,152 |
Example 6 Given pairwise distinct positive integers $a, b, c, d, e, f, g, h, n$ satisfying
$$n=a b+c d=e f+g h$$
Find the minimum value of $n$. | If none of $a, b, c, d, e, f, g, h$ equals 1, then
$$\begin{aligned}
2 n & =a b+c d+e f+g h \\
& \geqslant 4 \sqrt[4]{a b c d e f g h} \\
& \geqslant 4 \sqrt[4]{2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9} \\
& =4 \sqrt[4]{2^{7} \times 3^{4} \times 5 \times 7} \\
& =4 \times 4 \times 3 \times \sqrt... | 31 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 738,153 |
Example 7 (1) If $a, b, c, d$ are real numbers, prove that:
$$a^{6}+b^{6}+c^{6}+d^{6}-6 a b c d \geqslant-2,$$
and determine when equality holds; | Prove (1) The given inequality is transformed into
$$a^{6}+b^{6}+c^{6}+d^{6}+1+1 \geqslant 6 a b c d$$
According to the arithmetic-geometric mean inequality, we have
$$\begin{aligned}
& \frac{a^{6}+b^{6}+c^{6}+d^{6}+1^{6}+1^{6}}{6} \\
\geqslant & \sqrt[6]{|a|^{6} \cdot|b|^{6} \cdot|c|^{6} \cdot|d|^{6} \cdot 1^{6} \cdo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,154 |
(2) For which positive integers $k$, does the inequality
$$a^{k}+b^{k}+c^{k}+d^{k}-k a b c d \geqslant M_{k}$$
hold for all real numbers $a, b, c, d$? Find the maximum possible value of $M_{k}$, and indicate when equality holds. | (2) Notice that, when $k$ is odd, choosing sufficiently large negative values for $a, b, c, d$ results in a sufficiently large negative value for $a^{k}+b^{k}+c^{k}+d^{k}-k a b c d$. Therefore, such a number $M_{k}$ does not exist.
When $k=2$, taking $a=b=c=d=r$, we get $a^{2}+b^{2}+c^{2}+d^{2}-2 a b c d=4 r^{2}-2 r^{... | 4-k | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,155 |
Example 8 Let $a, b, c \in \mathbf{R}^{+}$, satisfying $a+b+c=a b c$. Find the minimum value of $a^{7}(b c-1)+$ $b^{7}(a c-1)+c^{7}(a b-1)$. | Given that $a, b, c > 0$, and $a + b + c = abc$, we have $c(ab - 1) = a + b$.
Similarly, we get $b(ac - 1) = a + c$, and $a(bc - 1) = b + c$.
By the AM-GM inequality, we have
$$abc = a + b + c \geqslant 3 \sqrt[3]{abc},$$
which implies $abc \geqslant 3 \sqrt{3}$, with equality holding if and only if $a = b = c = \sqrt... | 162 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,156 |
Example 9 For positive real numbers $a, b, c$ satisfying $abc=1$, find the maximum value of
$$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right)$$ | Given that the expression is symmetric with respect to $a$, $b$, and $c$, when $a=b=c=1$, we have
$$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right)=1$$
Next, we prove that the maximum value is 1, i.e., for any real numbers $a$, $b$, and $c$ satisfying $a b c=1$, we have
$$\left(a-1... | 1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,157 |
Example 10 Let $a, b, c$ be positive real numbers, satisfying
$$a+b+c+3 \sqrt[3]{a b c} \geqslant k(\sqrt{a b}+\sqrt{b c}+\sqrt{c a}),$$
Find the maximum value of $k$. | Given that when $a=b=c$, from $6 \geqslant 3 k$, we get $k \leqslant 2$. Now we prove
$$\begin{array}{c}
a+b+c+3 \sqrt[3]{a b c} \geqslant 2(\sqrt{a b}+\sqrt{b c}+\sqrt{c a}) . \\
\text { Let } f(a, b, c)=a+b+c+3 \sqrt[3]{a b c}-2(\sqrt{a b}+\sqrt{b c}+\sqrt{c a}) .
\end{array}$$
Without loss of generality, assume $a ... | 2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,158 |
7 Let $a, b, c$ be positive numbers, and $a+b+c=3$. Prove:
$$\sqrt{a}+\sqrt{b}+\sqrt{c} \geqslant a b+b c+c a .$$ | 7. From the conditional equation, we have $(a+b+c)^{2}=9$. Thus, $a b+b c+c a=\frac{9-a^{2}-b^{-}-c}{2}$. To prove $2 \sqrt{a}+2 \sqrt{b}+2 \sqrt{c}+a^{2}+b^{2}+c^{2} \geqslant 9$, we first prove $2 \sqrt{a}+a^{2} \geqslant 3 a$. Indeed, $2 \sqrt{a}+a^{2}=\sqrt{a}+\sqrt{a}+a^{2} \geqslant 3 \sqrt[3]{a^{3}}=3 a$. Simila... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,159 |
Example 11 For $a, b, c \in \mathbf{R}^{+}$, find
$$\frac{(a+b)^{2}+(a+b+4 c)^{2}}{a b c}(a+b+c)$$
the minimum value. | Solve: By the AM-GM inequality, we have
$$\begin{aligned}
(a+b)^{2}+(a+b+4 c)^{2} & =(a+b)^{2}+[(a+2 c)+(b+2 c)]^{2} \\
& \geqslant(2 \sqrt{a b})^{2}+(2 \sqrt{2 a c}+2 \sqrt{2 b c})^{2} \\
& =4 a b+8 a c+8 b c+16 c \sqrt{a b}
\end{aligned}$$
Thus,
$$\begin{aligned}
& \frac{(a+b)^{2}+(a+b+4 c)^{2}}{a b c} \cdot(a+b+c) ... | 100 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,160 |
Example 1 For any $\triangle ABC$, let its area be $S$, perimeter be $l$, and $P, Q, R$ be the points where the incircle of $\triangle ABC$ touches sides $BC, CA, AB$ respectively. Prove:
$$\left(\frac{AB}{PQ}\right)^{3}+\left(\frac{BC}{QR}\right)^{3}+\left(\frac{CA}{RP}\right)^{3} \geqslant \frac{2}{\sqrt{3}} \cdot \f... | Let $B C=a, C A=b, A B=c, Q R=p, R P=q, P Q=r$. Let $A R=x, B P=y, C Q=z$. From $x+y=c, y+z=a, z+x=b$, we get
$$x=t-a, y=t-b, z=t-c\left(t=\frac{a+b+c}{2}\right) .$$
In $\triangle A B C$ and $\triangle A R Q$, by the Law of Cosines, we have respectively
$$\begin{array}{c}
a^{2}=b^{2}+c^{2}-2 b c \cos A=(b-c)^{2}+2 b c... | \frac{2}{\sqrt{3}} \cdot \frac{l^{2}}{S} | Inequalities | proof | Yes | Yes | inequalities | false | 738,161 |
Example 2 Let $a, b, c$ be the lengths of the three sides of a triangle. Let
$$\begin{array}{l}
A=\sum \frac{a^{2}+b c}{b+c} \\
B=\sum \frac{1}{\sqrt{(a+b-c)(b+c-a)}}
\end{array}$$
where, " $\sum$ " denotes the cyclic sum.
Prove: $A B \geqslant 9$. | Proof: Let $a=y+z, b=x+z, c=x+y$ (where $x, y, z$ are positive numbers). Then
$$\begin{aligned}
B & =\sum \frac{1}{2 \sqrt{x y}} \\
A & =\sum \frac{x^{2}+y^{2}+z^{2}+x y+z x+3 y z}{2 x+y+z} \\
A B & =\left(\sum \frac{1}{2 \sqrt{y z}}\right) \sum \frac{x^{2}+y^{2}+z^{2}+x y+3 y z+z x}{2 x+y+z} \\
& \geqslant\left(\sum \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,162 |
For example, the three sides of $\triangle ABC$ are $a, b, c$ and satisfy $a+b+c=1$. Prove:
$$5\left(a^{2}+b^{2}+c^{2}\right)+18 a b c \geqslant \frac{7}{3} .$$ | Prove that since $a^{2}+b^{2}+c^{2}$
$$\begin{array}{l}
=(a+b+c)^{2}-2(a b+b c+c a) \\
=1-2(a b+b c+c a)
\end{array}$$
Therefore, the inequality to be proved is equivalent to
$$\frac{5}{9}(a b+b c+c a)-a b c \leqslant \frac{4}{27}$$
Construct the function
$$f(x)=(x-a)(x-b)(x-c),$$
On one hand,
$$f(x)=x^{3}-(a+b+c) x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,163 |
Example 4 Let $P$ be any point inside an acute $\triangle ABC$, and let lines $AP, BP, CP$ intersect the circumcircles of $\triangle PBC, \triangle PCA, \triangle PAB$ at another point $A_1, B_1, C_1$ (different from $P$). Prove that:
$$\left(1+2 \cdot \frac{PA}{PA_1}\right)\left(1+2 \cdot \frac{PB}{PB_1}\right)\left(1... | Prove: As shown in the figure, connect $A_{1} B, A_{1} C, B_{1} C, B_{1} A, C_{1} A, C_{1} B$, and denote $\angle B A_{1} C=\angle C A B_{1}=\angle B A C_{1}=\alpha$, $\angle C B_{1} A=\angle A B C_{1}=\angle C B A_{1}=\beta, \angle A C_{1} B=$ $\angle B C A_{1}=\angle A C B_{1}=\gamma$
In quadrilateral $P B A_{1} C$,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,164 |
Example 5 Let $P$ be a point inside $\triangle A B C$, and let $D, E, F$ be the feet of the perpendiculars from $P$ to $B C, C A, A B$ respectively. Determine the point $P$ such that $P D \times P E \times P F$ is maximized. | Let the three interior angles of $\triangle ABC$ be $A, B, C$, and their opposite sides be $a, b, c$. Let the area of $\triangle ABC$ be $S$ (the notation will be the same in the following questions, and will not be repeated).
Let $PD=x, PE=y, PF=z$. Connect $AP, BP, CP$. It is easy to see that $S_{1}(\text{area of } ... | P \text{ is the centroid of } \triangle ABC | Geometry | math-word-problem | Yes | Yes | inequalities | false | 738,165 |
Example 6 Let $a, b, c$ be the lengths of the three sides of a triangle, and $\delta$ be the area. Prove that:
$$\delta \leqslant \frac{\sqrt{3}}{4}\left(\frac{a+b+c}{3}\right)^{2},$$
with equality if and only if $a=b=c$. | Prove that by Heron's formula, the original inequality is equivalent to
$$\sqrt{p(p-a)(p-b)(p-c)} \leqslant \frac{\sqrt{3}}{4}\left(\frac{2 p}{3}\right)^{2}=\sqrt{3}\left(\frac{p}{3}\right)^{2}$$
which is equivalent to
$$(p-a)(p-b)(p-c) \leqslant \frac{p^{3}}{27}$$
By the AM-GM inequality, we have
$$(p-a)(p-b)(p-c) \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,166 |
Example 7 Let $T_{a}, T_{b}, T_{c}$ be the lengths of the segments obtained by extending the angle bisectors of $\triangle A B C$ to intersect the circumcircle. Prove that:
$$a b c \leqslant \frac{3 \sqrt{3}}{8} T_{a} T_{b} T_{c} .$$ | Proof: Let $|A E|=T_{a}, D$ be the intersection of $A E$ and $B C$, then
$$B E^{2}=c^{2}+T_{a}^{2}-2 c T_{a} \cos \frac{A}{2}, C E^{2}=b^{2}+T_{a}^{2}-2 b T_{a} \cos \frac{A}{2}$$
Since $B E=C E$, we have
$$T_{a}=\frac{b+c}{2 \cos \frac{A}{2}}$$
By the arithmetic mean inequality, we get $T_{a} \geqslant \frac{\sqrt{b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,167 |
Example 8 Let $P$ be a point inside or on the boundary of $\triangle ABC$, and the distances from point $P$ to the three sides are $PD$, $PE$, and $PF$. Prove:
$$P A+P B+P C \geqslant 2(P D+P E+P F) .$$ | Proof: Let $P A=x, P B=y, P C=z, P D=p, P E=q, P F=r$, where $D, E, F$ are the projections of point $P$ on the three sides. Then, since $C, D, P, E$ are concyclic, we have
$$\begin{aligned}
D E & =\sqrt{p^{2}+q^{2}+2 p q \cos C} \\
& =\sqrt{(p \sin B+q \sin A)^{2}+(p \cos B-q \cos A)^{2}} \\
& \geqslant p \sin B+q \sin... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,168 |
8. Given $x, y, z \in \mathbf{R}^{+}$, and $x+y+z=1$. Prove: $\left(\frac{1}{x}-x\right)\left(\frac{1}{y}-y\right)\left(\frac{1}{z}-z\right) \geqslant$ $\left(\frac{8}{3}\right)^{3}$. | 8. Let $\frac{1}{x}=a, \frac{1}{y}=b, \frac{1}{z}=c$, substituting into the given equation we get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$, which simplifies to $a b c=a b+b c+c a$. By the AM-GM inequality, it is easy to see that $a b c \geqslant 27$, so,
$$\begin{array}{l}
\left(\frac{1}{x}-x\right)\left(\frac{1}{y}-y\r... | \left(\frac{8}{3}\right)^{3} | Inequalities | proof | Yes | Yes | inequalities | false | 738,170 |
Example 1 Let $a, b, c$ be positive real numbers, and satisfy $a^{2}+b^{2}+c^{2}=3$. Prove:
$$\frac{1}{1+2 a b}+\frac{1}{1+2 b c}+\frac{1}{1+2 c a} \geqslant 1$$ | Prove that from the arithmetic mean being greater than or equal to the geometric mean and the arithmetic mean being greater than or equal to the harmonic mean, we can derive:
$$\begin{aligned}
& \frac{1}{1+2 a b}+\frac{1}{1+2 b c}+\frac{1}{1+2 c a} \\
\geqslant & \frac{1}{1+a^{2}+b^{2}}+\frac{1}{1+b^{2}+c^{2}}+\frac{1}... | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 738,171 |
Example 2 Given positive real numbers $a, b, c$ satisfying
$$a b + b c + c a \leqslant 3 a b c .$$
Prove:
$$\begin{array}{l}
\quad \sqrt{\frac{a^{2}+b^{2}}{a+b}}+\sqrt{\frac{b^{2}+c^{2}}{b+c}}+\sqrt{\frac{c^{2}+a^{2}}{c+a}}+3 \\
\leqslant \sqrt{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})
\end{array}$$ | Prove that from $Q_{2} \geqslant A_{2}$ we get
$$\begin{aligned}
& \sqrt{2} \cdot \sqrt{a+b}=2 \sqrt{\frac{a b}{a+b}} \cdot \sqrt{\frac{1}{2}\left(2+\frac{a^{2}+b^{2}}{a b}\right)} \\
\geqslant & 2 \sqrt{\frac{a b}{a+b}} \cdot \frac{1}{2}\left(\sqrt{2}+\sqrt{\frac{a^{2}+b^{2}}{a b}}\right) \\
= & \sqrt{\frac{2 a b}{a+b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,172 |
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