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Example 4 A group of scouts, whose ages are integers from 7 to 13, come from 11 countries. Prove that there are at least 5 children such that for any one of these children, there are more children of the same age in the scout group than there are from the same country. (Canadian National Training Team Practice Question... | Proof Consider the weighted element relation table:
\begin{tabular}{|c|c|c|c|c|}
\hline & $A_{1}$ & $A_{2}$ & $\cdots$ & $A_{11}$ \\
\hline 7 & $a_{7,1}$ & $a_{7,2}$ & $\cdots$ & $a_{7,11}$ \\
\hline 8 & $a_{8,1}$ & $a_{8,2}$ & $\cdots$ & $a_{8,11}$ \\
\hline$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\
\hlin... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 737,956 |
Example 5 Let $A=\{1,2,3,4,5,6\}, B=\{7,8,9, \cdots, n\}$. Take 3 numbers from $A$ and 2 numbers from $B$ to form a set $A_{i}(i=1,2, \cdots, 20)$ with 5 elements, such that $\left|A_{i} \cap A_{j}\right| \leqslant 2,1 \leqslant i<j \leqslant 20$, find the minimum value of $n$. (2002 China National Training Team for IM... | We first prove: Each element in $B$ appears in the subsets $A_{i}(i=1,2, \cdots, 20)$ at most 4 times. If not, assume that an element $b$ in $B$ appears $k(k>4)$ times in the subsets $A_{i}(i=1,2, \cdots, 20)$. Consider the $k$ subsets containing $b$, they contain $3k > 12$ elements from $A$. Thus, by the pigeonhole pr... | 16 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,957 |
Example 6 Let $n$ be a given positive integer, $6 \mid n$. In an $n \times n$ chessboard, each square is filled with a positive integer, and the numbers in the squares of the $i$-th row from left to right are $(i-1) n+1,(i-1) n+2, \cdots,(i-1) n+n$. Now, take any 2 adjacent (sharing a common edge) squares, add 1 to one... | Let the sum of the numbers on the chessboard be denoted as \( S \).
Obviously, when all the numbers on the chessboard are equal, each number is at least \( n^2 \), so the sum of the numbers on the chessboard \( S' \geq n^2 \cdot n^2 = n^4 \). Each operation increases \( S \) by 3, and the initial sum of the numbers on ... | \frac{n^2(n^2 - 1)}{6} | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,958 |
Example 7 Let $n$ be a given positive integer, $X=\{1,2,3, \cdots, n\}, A$ be a subset of $X$, and for any $x<y<z, x, y, z \in A$, there exists a triangle with side lengths $x, y, z$. Let $|A|$ denote the number of elements in the set $A$. Find the maximum value of $|A|$. (Original problem) | Let $A=\left\{a_{1}, a_{2}, \cdots, a_{r}\right\}$ be a subset that meets the conditions, where $a_{1}2 k \geqslant z$, thus there exists a triangle with sides $x$, $y$, and $z$, and $|A|=k+1=\frac{n}{2}+1=\frac{n+2}{2}=\left[\frac{n+2}{2}\right]$.
If $n=2 k+1$, then let $A=\{k+1, k+2, \cdots, 2 k+1\}$, at this point,... | \left[\frac{n+2}{2}\right] | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,959 |
Example 8 Given a positive integer $n(n \geqslant 2)$, find the largest $\lambda$ such that: if there are $n$ bags, each containing some balls of integer powers of 2 grams, and the total weight of the balls in each bag is equal (there can be balls of equal weight in the same bag), then there must be a certain weight of... | Let the weight of the heaviest ball be 1, and let the total weight of the balls in each bag be $G$, then $G \geqslant 1$.
First, we prove that when $\lambda=\left[\frac{n}{2}\right]+1$ is valid, i.e., there must be a weight of balls whose total number is at least $\left[\frac{n}{2}\right]+1$.
Assume, for contradictio... | \left[\frac{n}{2}\right]+1 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,960 |
1 Let $A_{1}, A_{2}, \cdots, A_{29}$ be 29 different sets of positive integers. For $1 \leqslant i < j \leqslant 29$, the sum of any element from $A_{i}$ and any element from $A_{j}$ is greater than 200. (29th IMO Shortlist) | 1. First, transform the condition. Let $A_{i}^{\prime}=\left\{A_{i}\right.$ contain the numbers not greater than 1988 $\}=A_{i} \cap\{1,2, \cdots$, $1988\}$, then $N_{i}(1988)=\left|A_{i}^{\prime}\right|, N_{i j}(1988)=\left|A_{i}^{\prime} \cap A_{j}^{\prime}\right|$. Therefore, the essence of this problem is to prove ... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 737,961 |
2 Let $A_{i}$ be a subset of $M=\{1,2, \cdots, 10\}$, and $\left|A_{i}\right|=5(i=1,2, \cdots, k)$, $\left|A_{i} \cap A_{j}\right| \leqslant 2(1 \leqslant i<j \leqslant k)$. Find the maximum value of $k$. (1994 China National Training Team | 2. Since $\sum_{i=1}^{10} m_{i}=\sum_{i=1}^{k} 5=5 k$, by Cauchy's inequality, we have $2 \sum_{1 \leqslant i<j \leqslant k} \mid A_{i} \cap A_{j} \mid=2 \sum_{i=1}^{10} \mathrm{C}_{m_{i}}^{2}=\sum_{i=1}^{10} m_{i}^{2}-\sum_{i=1}^{10} m_{i} \geqslant \frac{\left(\sum_{i=1}^{10} m_{i}\right)^{2}}{\sum_{i=1}^{10} 1^{2}}-... | 6 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,962 |
3 Let $X$ be a finite set, $A_{1}, A_{2}, \cdots, A_{m}$ be subsets of $X$, and $\left|A_{i}\right|=r(1 \leqslant i \leqslant m)$. If for any $i \neq j$, $\left|A_{i} \cap A_{j}\right| \leqslant k$. Prove: $|X| \geqslant \frac{m r^{2}}{r+(m-1) k}$. | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Combinatorics | proof | Yes | Yes | inequalities | false | 737,963 |
Given $a>0, b>0, a+2b=1$. Prove: $\frac{1}{a}+\frac{2}{b} \geqslant 9$. | \[
\begin{array}{l}
\quad \text { 1. } 1=a+2 b=a+b+b \geqslant 3 \sqrt[3]{a b^{2}}, \text { so } \frac{1}{\sqrt[3]{a b^{2}}} \geqslant 3 \cdot \frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{1}{b}+ \\
\frac{1}{b} \geqslant 3 \sqrt[3]{\frac{1}{a b^{2}}} \geqslant 3 \times 3=9 .
\end{array}
\] | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,964 |
II Given that $a, b, c$ are positive real numbers, and $a b c=8$, prove:
$$\frac{a^{2}}{\sqrt{\left(1+a^{3}\right)\left(1+b^{3}\right)}}+\frac{b^{2}}{\sqrt{\left(1+b^{3}\right)\left(1+c^{3}\right)}}+\frac{c^{2}}{\sqrt{\left(1+c^{3}\right)\left(1+a^{3}\right)}} \geqslant \frac{4}{3} .$$ | 11. Proof: Notice that $\frac{a^{2}+2}{2}=\frac{\left(a^{2}-a+1\right)+(a+1)}{2} \geqslant \sqrt{\left(a^{2}-a+1\right) \cdot(a+1)}=$ $\sqrt{1+a^{3}}$. To prove the original inequality, it suffices to prove $\frac{a^{2}}{\left(a^{2}+2\right)\left(b^{2}+2\right)}+\frac{b^{2}}{\left(b^{2}+2\right)\left(c^{2}+2\right)}+$ ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,966 |
14 Let $a_{1}, a_{2}, a_{3} \in \mathbf{R}^{+}$, find
$$\frac{a_{1} a_{2}}{\left(a_{2}+a_{3}\right)\left(a_{3}+a_{1}\right)}+\frac{a_{2} a_{3}}{\left(a_{3}+a_{1}\right)\left(a_{1}+a_{2}\right)}+\frac{a_{3} a_{1}}{\left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right)}$$
the minimum value. | 14. The minimum value sought is $\frac{3}{4}$. When $a_{1}=a_{2}=a_{3}$, its value is $\frac{3}{4}$. The proof is as follows:
$$\begin{array}{l}
\frac{a_{1} a_{2}}{\left(a_{2}+a_{3}\right)\left(a_{3}+a_{1}\right)}+\frac{a_{2} a_{3}}{\left(a_{3}+a_{1}\right)\left(a_{1}+a_{2}\right)}+\frac{a_{3} a_{1}}{\left(a_{3}+a_{2}\... | \frac{3}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,967 |
15 Let positive real numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 2)$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$, find
$$\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}}$$
the minimum value. | 15. Let $b_{i}=2-a_{i} \geqslant 0(i=1,2, \cdots, n)$, then $\sum_{i=1}^{n} b_{i}=2 n-1$. By the inequality of arithmetic and geometric means, we get $\sum_{i=1}^{n} \frac{a_{i}}{b_{i}}=\sum_{i=1}^{n}\left(\frac{a_{i}}{b_{i}}-1\right)-n=2 \sum_{i=1}^{n} \frac{1}{b_{i}}-n \geqslant \frac{2 n}{\sqrt[n]{b_{1} \cdots b_{n}... | \frac{n}{2 n-1} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,968 |
16. Let $a>0, x_{1}, x_{2}, \cdots, x_{n} \in[0, a](n \geqslant 2)$ and satisfy
$$x_{1} x_{2} \cdots x_{n}=\left(a-x_{1}\right)^{2}\left(a-x_{2}\right)^{2} \cdots\left(a-x_{n}\right)^{2}$$
Find the maximum value of $x_{1} x_{2} \cdots x_{n}$. | 16. By the inequality of arithmetic and geometric means, we have $\left(x_{1} x_{2} \cdots x_{n}\right)^{\frac{1}{2 n}}=\left[\left(a-x_{1}\right)\left(a-x_{2}\right) \cdots(a-\right.$ $\left.\left.x_{n}\right)\right]^{\frac{1}{n}} \leqslant a-\frac{x_{1}+\cdots+x_{n}}{n} \leqslant a-\left(x_{1} \cdots x_{n}\right)^{\f... | \left(\frac{-1+\sqrt{4 a+1}}{2}\right)^{2 n} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,969 |
17 Let $n \geqslant 2, x_{1}, x_{2}, \cdots, x_{n}$ be real numbers, and $\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1}=1$, for each given positive integer $k, 1 \leqslant k \leqslant n$, find the maximum value of $\left|x_{k}\right|$. | 17. From the given conditions, we have $2 \sum_{i=1}^{n} x_{i}^{2} + 2 \sum_{i=1}^{n-1} x_{i} x_{i+1} = 2$. That is, $x_{1}^{2} + (x_{1} + x_{2})^{2} + (x_{2} + x_{3})^{2} + \cdots + (x_{n-2} + x_{n-1})^{2} + (x_{n-1} + x_{n})^{2} + x_{n}^{2} = 2$. For a given positive integer $k$, $1 \leqslant k \leqslant n$, by the m... | \sqrt{\frac{2 k (n+1-k)}{n+1}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,970 |
18 Prove: For any triangle with side lengths $a$, $b$, and $c$, and area $S$, we have
$$\frac{a b+b c+c a}{4 S} \geqslant \sqrt{3} \text {. }$$ | 18. As shown in the figure, in $\triangle ABC$, let $AB=c$, $AC=b$, $BC=a$, and $\angle BAC=\alpha$. Considering the arc $\overparen{BAC}$ on the circumcircle of $\triangle ABC$ opposite to side $BC$, since the midpoint $D$ of the arc is the farthest point from the chord $BC$, for the heights $AH=h$ and $DK$ of $\trian... | \sqrt{3} | Inequalities | proof | Yes | Yes | inequalities | false | 737,971 |
19 Prove: If $A D, B E$ and $C F$ are the angle bisectors of $\triangle A B C$, then the area of $\triangle D E F$ does not exceed one-fourth of the area of $\triangle A B C$. | 19. Let $a=BC, b=AC, c=AB, S=S_{\triangle ABC}, S_{0}=S_{\triangle DEF}$. By the property of the angle bisector of a triangle, we get $\frac{AF}{b}=\frac{BF}{a}=\frac{AF+BF}{b+a}=\frac{c}{a+b}$. Therefore, $AF=\frac{bc}{a+b}$, and similarly, $AE=\frac{bc}{a+c}$. Thus, $S_{\triangle AEF}=\frac{1}{2} AF \cdot AE \sin \an... | S_{0} \leqslant \frac{1}{4} S | Geometry | proof | Yes | Yes | inequalities | false | 737,972 |
20 Let the circumcircle $K$ of $\triangle ABC$ have radius $R$, and the internal angle bisectors intersect the circle $K$ at $A^{\prime}, B^{\prime}, C^{\prime}$. Prove: $16 Q^{3} \geqslant 27 R^{4} P$. Here, $Q$ and $P$ are the areas of $\triangle A^{\prime} B^{\prime} C^{\prime}$ and $\triangle ABC$, respectively. | 20. Let the three interior angles of $\triangle A B C$ be $\alpha, \beta, \gamma$, then $P=\frac{1}{2} R^{2}(\sin 2 \alpha+\sin 2 \beta+$ $\sin 2 \gamma)$. Since the interior angles of $\triangle A^{\prime} B^{\prime} C^{\prime}$ are $\frac{\beta+\gamma}{2}, \frac{\alpha+\gamma}{2}, \frac{\alpha+\beta}{2}$, we have $Q=... | proof | Geometry | proof | Yes | Yes | inequalities | false | 737,973 |
21 Let the three sides of $\triangle ABC$ be $a, b, c$. Now extend $AB, AC$ by $a$ units, extend $BC, BA$ by $b$ units, and extend $CA, CB$ by $c$ units. Let the convex polygon formed by the six endpoints have an area of $G$, and the area of $\triangle ABC$ be $F$. Prove:
$$\frac{G}{F} \geqslant 13$$ | 21. As shown in the figure, given $S_{\triangle A B_{2} G_{1}}=S_{\triangle B C_{2} A_{1}}=S_{\triangle C A_{2} B_{1}}=S_{\triangle A B C}$, so $\frac{G}{F}=$
$$\begin{array}{l}
\frac{S_{A B C_{2} C_{1}}+S_{B C A_{2} A_{2}}+S_{A C B_{1} B_{2}}+4 F}{F}=\frac{S_{\triangle A A_{1} A_{2}}+S_{\triangle B B_{1} B_{2}}+S_{\tr... | 13 | Geometry | proof | Yes | Yes | inequalities | false | 737,974 |
22 Let the maximum side length of an isosceles trapezoid be 13, and the perimeter be 28.
(1) If the area of the trapezoid is 27, find its side lengths;
(2) Can the area of such a trapezoid be 27.001? | 22. As shown in the figure, let $A D$ be the larger base, and $B H$ be the height of the given trapezoid $A B C D$. If $A B=C D=13$, then $A D+B C=2$, and $S_{\text {trapezoid } A B C D}=B H \cdot \frac{A D+B C}{2} \leqslant 13 \cdot \frac{2}{2}=$ $13<27$, which is impossible. Therefore, $A D=13$. Let $A B=x$, then $B ... | AB=BC=CD=5 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 737,975 |
23 Among all triangles with a given perimeter, find the triangle with the largest inradius.
Among all triangles with a given perimeter, find the triangle with the largest inradius. | 23. Let $a$, $b$, $c$ be the side lengths of a triangle with a fixed semiperimeter $p$, and let $S$ and $r$ be its area and the radius of its inscribed circle, respectively. Then, by the AM-GM inequality, we have $(r p)^{2}=S^{2}=p(p-a)(p-b)(p-c) \leqslant$ $p\left[\frac{(p-a)+(p-b)+(p-c)}{3}\right]^{3}=\frac{p^{4}}{27... | r \leqslant \frac{p}{\sqrt{27}} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 737,976 |
12. Let $a, b \in \mathbf{R}, \frac{1}{a}+\frac{1}{b}=1$. Prove that for all positive integers $n$, we have
$$(a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1} .$$ | 12. When $n=1$, it is obviously true. Assume that when $n=k$, we have $(a+b)^{k}-a^{k}-b^{k} \geqslant 2^{2 k}-2^{k+1}$. Then for $n=k+1$, given $\frac{1}{a}+\frac{1}{b}=1$, we have $a+b=a b$, thus $a b=a+b \geqslant 2 \sqrt{a b}$, which means
$$\begin{array}{l}
a b=a+b \geqslant 4 . \text { Therefore, }(a+b)^{k+1}-a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,977 |
24 In $\triangle A B C$, the three side lengths are $a, b, c$, and $a, b, c$ are rational numbers. Prove:
$$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c} \leqslant 1$$ | 24. Since $a, b, c$ are positive rational numbers, there exists $m \in \mathbf{N}$ such that $m a, m b, m c$ are positive integers. Also, since $a, b, c$ are the lengths of the sides of a triangle, we have $1+\frac{b-c}{a}>0, 1+\frac{c-a}{b}>0, 1+\frac{a-b}{c}>0$. By the AM-GM inequality, we get $\left[\left(1+\frac{b-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,978 |
25 Let $n \geqslant 2$, find the maximum and minimum value of the product $x_{1} x_{2} \cdots x_{n}$ under the conditions $x_{i} \geqslant \frac{1}{n}(i=1,2, \cdots, n)$ and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$. | 25. First, find the maximum value. By the AM-GM inequality, we have $\sqrt[n]{x_{1}^{2} x_{2}^{2} \cdots x_{n}^{2}} \leqslant \frac{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}{n}=\frac{1}{n}$. The equality holds when $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{\sqrt{n}}>\frac{1}{n}(n \geqslant 2)$. Therefore, the maximum value is $n^... | \frac{\sqrt{n^{2}-n+1}}{n^{n}} \text{ and } n^{-\frac{n}{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,979 |
26 Find the smallest positive number $\lambda$, such that for any triangle with side lengths $a, b, c$, if $a \geqslant \frac{b+c}{3}$, then $a c+b c-c^{2} \leqslant \lambda\left(a^{2}+b^{2}+3 c^{2}+2 a b-4 b c\right)$. | 26. It is known that $a^{2}+b^{2}+3 c^{2}+2 a b-4 b c=(a+b-c)^{2}+2 c^{2}+2 a c-2 b c=$ $(a+b-c)^{2}+2 c(a+c-b)$. Let $I=\frac{(a+b-c)^{2}+2 c(a+c-b)}{2 c(a+b-c)}=\frac{a+b-c}{2 c}$ $+\frac{a+c-b}{a+b-c}$, since $a \geqslant \frac{1}{3}(b+c)$, it follows that $a \geqslant \frac{1}{4}(a+b-c)+\frac{c}{2}$. Therefore, $a+... | \frac{2 \sqrt{2}+1}{7} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,980 |
27 For each positive integer $n$, prove:
$$\sum_{j=1}^{n} \frac{2 j+1}{j^{2}}>n\left[(n+1)^{\frac{2}{n}}-1\right] .$$ | 27. Clearly, $2 j+1=(j+1)^{2}-j^{2}$. By the AM-GM inequality, we get $\sum_{j=1} \frac{2 j+1}{j^{2}}=$
$$\begin{array}{l}
\sum_{j=1}^{n}\left[\frac{(j+1)^{2}}{j^{2}}-1\right]=\sum_{j=1}^{n} \frac{(j+1)^{2}}{j^{2}}-n \geqslant n\left[\frac{2^{2}}{1^{2}} \cdot \frac{3^{2}}{2^{2}} \cdot \cdots \cdot \frac{(n+1)^{2}}{n^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,981 |
28 Let $A$, $B$, $C$ be the three interior angles of a triangle, prove that:
$$\sin 3A + \sin 3B + \sin 3C \leqslant \frac{3}{2} \sqrt{3}$$ | 28. Let $A \geqslant 60^{\circ}$, then $B+C \leqslant 180^{\circ}-60^{\circ}=120^{\circ}$. $\sin 3 A+\sin 3 B+\sin 3 C=\sin 3 A+2 \sin \frac{3}{2}(B+C) \cos \frac{3}{2}(B-C) \leqslant \sin 3 A+2 \sin \frac{3}{2}(B+C)$. Let $\alpha=\frac{3}{2}(B+C)$, then $0 \leqslant \alpha \leqslant 180^{\circ}$, and $A=180^{\circ}-(B... | \frac{3}{2} \sqrt{3} | Inequalities | proof | Yes | Yes | inequalities | false | 737,982 |
29 Let $\alpha, \beta, \gamma$ be the three interior angles of a given triangle. Prove that:
$$\csc ^{2} \frac{\alpha}{2}+\csc ^{2} \frac{\beta}{2}+\csc ^{2} \frac{\gamma}{2} \geqslant 12$$
and find the condition for equality. | 29. By the AM-GM inequality, we have $\csc ^{2} \frac{\alpha}{2}+\csc ^{2} \frac{\beta}{2}+\csc ^{2} \frac{\gamma}{2} \geqslant$ $3\left(\csc \frac{\alpha}{2} \csc \frac{\beta}{2} \csc \frac{\gamma}{2}\right)^{\frac{2}{3}}$, with equality if and only if $\alpha=\beta=\gamma$. Using the AM-GM inequality and the properti... | 12 | Inequalities | proof | Yes | Yes | inequalities | false | 737,983 |
30 Let $x, y, z \geqslant 0$, and satisfy $y z+z x+x y=1$, prove:
$$x\left(1-y^{2}\right)\left(1-z^{2}\right)+y\left(1-z^{2}\right)\left(1-x^{2}\right)+z\left(1-x^{2}\right)\left(1-y^{2}\right) \leqslant \frac{4}{9} \sqrt{3} .$$ | 30. Let $x=\tan \frac{A}{2}, y=\tan \frac{B}{2}, z=\tan \frac{C}{2}$. Here $A, B, C \in [0, \pi)$. Since $\tan \left(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}\right)=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}-\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}-\tan \frac{A}{2... | x y z \leqslant \frac{1}{9} \sqrt{3} | Inequalities | proof | Yes | Yes | inequalities | false | 737,984 |
31 For $a_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n)$, prove: $\sum_{k=1}^{n} \sqrt[k]{a_{1} \cdots a_{k}} \leqslant \mathrm{e} \sum_{k=1}^{n} a_{k}$, where $\mathrm{e}=$ $\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}$ | 31. Given that $\left(1+\frac{1}{k}\right)^{k}$ is monotonically increasing and converges to $e$. For any $i \in \mathbf{N}$, we have $i\left(1+\frac{1}{i}\right)^{i} \leqslant i e$, and let $b_{i}=i\left(1+\frac{1}{i}\right)^{i}$. Then $\frac{b_{i}}{i} \leqslant e$. From $b_{1} b_{2} \cdots b_{k}=(1+k)^{k}$, we get $\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,985 |
32. Let $x_{i} \in \mathbf{R} (i=1,2, \cdots, n, n \geqslant 3)$. Let $p=\sum_{i=1}^{n} x_{i}, q=\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}$, prove:
(1) $\frac{n-1}{n} p^{2}-2 q \geqslant 0$;
(2) $\left|x_{i}-\frac{p}{n}\right| \leqslant \frac{n-1}{n} \sqrt{p^{2}-\frac{2 n}{n-1} q}, i=1,2, \cdots, n$. | 32. (1) Since $(n-1) p^{2}-2 n q=(n-1)\left(\sum_{i=1}^{n} x_{i}\right)^{2}-2 n \sum_{1 \leq i<j \leqslant n} x_{i} x_{j}=$ $(n-1) \sum_{i=1}^{n} x_{i}^{2}-2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}=\sum_{1 \leqslant i<j \leqslant n}\left(x_{i}-x_{j}\right)^{2}$, therefore $\frac{n-1}{n} p^{2}-2 q \geqslant 0$;
(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,986 |
33 Find the largest real number $\lambda$, such that for a real-coefficient polynomial $f(x)=x^{3}+a x^{2}+c$ with all roots being non-negative real numbers, if $x \geqslant 0$, then $f(x) \geqslant \lambda(x-a)^{3}$, and find the condition for equality. | 33. Let the three roots of $f(x)$ be $\alpha, \beta, \gamma$, and assume $0 \leqslant \alpha \leqslant \beta \leqslant \gamma$, then $x-a=x+\alpha+$ $\beta+\gamma, f(x)=(x-\alpha)(x-\beta)(x-\gamma)$. (1) When $0 \leqslant x \leqslant \alpha$, we have $-f(x)=$ $(\alpha-x)(\beta-x)(\gamma-x) \leqslant\left(\frac{\alpha+... | -\frac{1}{27} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,987 |
13 Let $a, b \in \mathbf{R}^{+}$, prove that $\sqrt{a}+1>\sqrt{b}$ holds if and only if for any $x>1$, $a x+\frac{x}{x-1}>b$. | 13. Given $a b>0, x-1>0$, then $a x+\frac{x}{x-1}=\left[a(x-1)+\frac{1}{x-1}\right]+a+1 \geqslant 2 \sqrt{a}+a+1=(\sqrt{a}+1)^{2}$. When and only when $a(x-1)=\frac{1}{x-1}$, i.e., $x=1+\frac{1}{\sqrt{a}}$, the minimum value of $a x+\frac{x}{x-1}$ is $(\sqrt{a}+1)^{2}$. Therefore, $a x+\frac{x}{x-1}>b$ holds for any $x... | \sqrt{a}+1>\sqrt{b} | Inequalities | proof | Yes | Yes | inequalities | false | 737,988 |
■Let $a, b, c \in \mathbf{R}^{+}$, prove: $a^{2 a} b^{2 b} c^{2 c} \geqslant a^{b+c} b^{c+a} c^{a+b}$. | 1. Due to the symmetry of $a, b, c$, without loss of generality, assume $a \geqslant b \geqslant c$, then $\frac{a^{2 a} b^{2 b} c^{2 c}}{a^{b+c} b^{a+c} c^{a+b}}=$ $\left(\frac{a}{b}\right)^{a-b}\left(\frac{b}{c}\right)^{b-c}\left(\frac{a}{c}\right)^{a-c} \geqslant 1$, so $a^{2 a} b^{2 b} c^{2 c} \geqslant a^{b+c} b^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,989 |
2 Let $a, b, c, d>0$ and $a+b+c+d=1$, prove:
$$\frac{1}{4 a+3 b+c}+\frac{1}{3 a+b+4 d}+\frac{1}{a+4 c+3 d}+\frac{1}{4 b+3 c+d} \geqslant 2 .$$ | 2. By Cauchy-Schwarz inequality, we have the left side of the original expression $\geqslant$
$$\frac{16}{(4 a+3 b+c)+(3 a+b+4 d)+(a+4 c+3 d)+(4 b+3 c+d)}=2 \text {. }$$ | 2 | Inequalities | proof | Yes | Yes | inequalities | false | 737,990 |
3 Given $a, b \in \mathbf{R}^{+}, n \geqslant 2, n \in \mathbf{N}^{+}$. Prove:
$$\sum_{i=1}^{n} \frac{1}{a+i b}<\frac{n}{\sqrt{a(a+n b)}} .$$ | 3. By the Cauchy-Schwarz inequality, we have $\left(\sum_{i=1}^{n} \frac{1}{a+i b}\right)^{2} \leqslant n \sum_{i=1}^{n}\left(\frac{1}{a+i b}\right)^{2}<$
$$\begin{array}{l}
n\left\{\frac{1}{a(a+b)}+\frac{1}{(a+b)(a+2 b)}+\cdots+\frac{1}{[a+(n-1) b](a+n b)}\right\}= \\
\frac{n}{b}\left[\left(\frac{1}{a}-\frac{1}{a+b}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,991 |
5 Let positive real numbers $a, b, c$ satisfy $ab + bc + ca = \frac{1}{3}$. Prove:
$$\frac{a}{a^{2}-bc+1}+\frac{b}{b^{2}-ca+1}+\frac{c}{c^{2}-ab+1} \geqslant \frac{1}{a+b+c} .$$ | 5. The denominator on the left side of the equation is clearly positive. By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\frac{a}{a^{2}-b c+1}+
\frac{b}{b^{2}-c a+1}+\frac{c}{c^{2}-a b+1}=\frac{a^{2}}{a^{3}-a b c+a}+\frac{b^{2}}{b^{3}-a b c+b}+\frac{c^{2}}{c^{3}-a b c+c} \geqslant \\
\frac{(a+b+c)^{2}}{a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,993 |
6 Given that $x, y, z$ are positive real numbers. Prove:
$$\frac{1+x y+x z}{(1+y+z)^{2}}+\frac{1+y z+y x}{(1+z+x)^{2}}+\frac{1+z x+z y}{(1+x+y)^{2}} \geqslant 1$$ | 6. By Cauchy-Schwarz inequality, $\left(1+\frac{y}{x}+\frac{z}{x}\right)(1+x y+x z) \geqslant(1+y+z)^{2} \Rightarrow$ $\frac{1+x y+x z}{(1+y+z)^{2}} \geqslant \frac{x}{x+y+z}$. Similarly, $\frac{1+y z+y x}{(1+z+x)^{2}} \geqslant \frac{y}{x+y+z} \cdot \frac{1+z x+z y}{(1+x+y)^{2}} \geqslant$ $\frac{z}{x+y+z}$. Adding th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,994 |
$7 x, y, z \in \mathbf{R}^{+}$, and $x y z \geqslant 1$. Prove:
$$\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \geqslant 0 .$$ | 7. The original inequality is equivalent to $\sum \frac{1}{x^{5}+y^{2}+z^{2}} \leqslant \frac{3}{x^{2}+y^{2}+z^{2}}$. Using $x y z \geqslant 1$ and the Cauchy-Schwarz inequality, we get $\left(x^{5}+y^{2}+z^{2}\right) \cdot\left(y z+y^{2}+z^{2}\right) \geqslant\left(\sum x^{2}\right)^{2}$. And $\sum\left(y z+y^{2}+z^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,995 |
8 Let $a, b, c \in \mathbf{R}^{+}$, and $a+b+c=3$. Prove:
$$\sum \frac{a^{4}}{b^{2}+c} \geqslant \frac{3}{2}$$
where, " $\sum$ " denotes the cyclic sum. | 8. By Cauchy-Schwarz inequality, we know $\left(b^{2}+c+c^{2}+a+a^{2}+b\right) \cdot$ $\left(\frac{a^{4}}{b^{2}+c}+\frac{b^{4}}{c^{2}+a}+\frac{c^{4}}{c^{2}+b}\right) \geqslant\left(a^{2}+b^{2}+c^{2}\right)^{2}$. Therefore, $\frac{a^{4}}{b^{2}+c}+\frac{b^{4}}{c^{2}+a}+\frac{c^{4}}{a^{2}+b} \geqslant$ $\frac{\left(a^{2}+... | \frac{a^{4}}{b^{2}+c}+\frac{b^{4}}{c^{2}+a}+\frac{c^{4}}{a^{2}+b} \geqslant \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,996 |
9 Let real numbers $a, b, c > 0$, and satisfy $a+b+c=3$. Prove:
$$\frac{a^{2}+3 b^{2}}{a b^{2}(4-a b)}+\frac{b^{2}+3 c^{2}}{b c^{2}(4-b c)}+\frac{c^{2}+3 a^{2}}{c a^{2}(4-c a)} \geqslant 4$$ | 9. Let $A=\frac{a^{2}}{a b^{2}(4-a b)}+\frac{b^{2}}{b c^{2}(4-b c)}+\frac{c^{2}}{c a^{2}(4-c a)}, B=\frac{b^{2}}{a b^{2}(4-a b)}$ $+\frac{c^{2}}{b c^{2}(4-b c)}+\frac{a^{2}}{c a^{2}(4-c a)}$. To prove the original inequality, it suffices to prove $A \geqslant 1, B \geqslant 1$. By the Cauchy-Schwarz inequality, we have... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,997 |
10 Let $a, b, c \in \mathbf{R}^{+}$, and $a b c=1$, prove:
$$\frac{1}{1+2 a}+\frac{1}{1+2 b}+\frac{1}{1+2 c} \geqslant 1$$ | 10. Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, then the original inequality is equivalent to $\frac{y}{2 x+y}+\frac{z}{2 y+z}+$ $\frac{x}{2 z+x} \geqslant 1$. By the Cauchy-Schwarz inequality, we have $[x(x+2 z)+y(y+2 x)+z(z+$ $2 y)]\left(\frac{x}{x+2 z}+\frac{y}{y+2 x}+\frac{z}{z+2 y}\right) \geqslant(x+y+z)^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,998 |
11 Let $a_{1}, a_{2}, \cdots, a_{n}$ be real numbers, prove:
$$\sqrt[3]{a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}} \leqslant \sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} .$$ | 11. From $\left(\sum_{i=1}^{n} a_{i}^{3}\right)^{2} \leqslant \sum_{i=1}^{n} a_{i}^{2} \sum_{i=1}^{n} a_{i}^{4} \leqslant \sum_{i=1}^{n} a_{i}^{2}\left(\sum_{i=1}^{n} a_{i}^{2}\right)^{2}=\left(\sum_{i=1}^{n} a_{i}^{2}\right)^{3}$, then $\left(\sum_{i=1}^{n} a_{i}^{3}\right)^{\frac{1}{3}} \leqslant\left(\sum_{i=1}^{n} ... | \left(\sum_{i=1}^{n} a_{i}^{3}\right)^{\frac{1}{3}} \leqslant \left(\sum_{i=1}^{n} a_{i}^{2}\right)^{\frac{1}{2}} | Inequalities | proof | Yes | Yes | inequalities | false | 738,000 |
12 Given that $a, b, c$ are positive real numbers, prove:
$$\frac{9}{a+b+c} \leqslant 2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right) .$$ | 12. By Cauchy-Schwarz inequality, we have $2(a+b+c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=[(a+b)+$ $(b+c)+(c+a)]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right) \geqslant 9$, hence the proposition is true. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,001 |
13 Let $a_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n)$, prove that:
$$\frac{1}{a_{1}}+\frac{2}{a_{1}+a_{2}}+\cdots+\frac{n}{a_{1}+\cdots+a_{n}}<2 \sum_{i=1}^{n} \frac{1}{a_{i}} .$$ | 13. By Cauchy's inequality, we have $\frac{k^{2}(k+1)^{2}}{4}=\left(\sum_{i=1}^{k} \frac{i}{\sqrt{a_{i}}} \cdot \sqrt{a_{i}}\right)^{2} \leqslant \sum_{i=1}^{k} \frac{i^{2}}{a_{i}} \sum_{i=1}^{k} a_{i}$. Therefore, $\frac{k}{\sum_{i=1}^{k} a_{i}} \leqslant \frac{4}{k(k+1)^{2}} \sum_{i=1}^{k} \frac{i^{2}}{a_{i}}$. Summi... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,002 |
14. Let $a_{i}, b_{i}, c_{i}, d_{i}$ be positive real numbers $(i=1,2, \cdots, n)$, prove that:
$$\left(\sum_{i=1}^{n} a_{i} b_{i} c_{i} d_{i}\right)^{4} \leqslant \sum_{i=1}^{n} a_{i}^{4} \sum_{i=1}^{n} b_{i}^{4} \sum_{i=1}^{n} c_{i}^{4} \sum_{i=1}^{n} d_{i}^{4} .$$ | 14. $\left(\sum_{i=1}^{n} a_{i} b_{i} c_{i} d_{i}\right)^{4} \leqslant\left[\sum_{i=1}^{n}\left(a_{i} b_{i}\right)^{2}\right]^{2}\left[\sum_{i=1}^{n}\left(c_{i} d_{i}\right)^{2}\right]^{2} \leqslant \sum_{i=1}^{n} a_{i}^{4} \sum_{i=1}^{n} b_{i}^{4} \sum_{i=1}^{n} c_{i}^{4} \sum_{i=1}^{n} d_{i}^{4}$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,003 |
15 Let $n(n \geqslant 2)$ be a positive integer, prove that:
$$\frac{4}{7}<1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n}<\frac{\sqrt{2}}{2}$$ | 15. $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}$. By the Cauchy-Schwarz inequality, we have $[(n+1)+(n+2)+\cdots+(2 n)]\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}\right)>$ $n^{2}$, so $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,004 |
16 Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, prove:
$$\frac{\left(\sum_{i=1}^{n} a_{i}\right)^{2}}{2 \sum_{i=1}^{n} a_{i}^{2}} \leqslant \frac{a_{1}}{a_{2}+a_{3}}+\frac{a_{2}}{a_{3}+a_{4}}+\cdots+\frac{a_{n}}{a_{1}+a_{2}}$$ | 16. Let $a_{n+1}=a_{1}, a_{n+2}=a_{2}$, then $\sum_{i=1}^{n} a_{i}\left(a_{i+1}+a_{i+2}\right) \sum_{i=1}^{n} \frac{a_{i}}{a_{i+1}+a_{i+2}} \geqslant$ $\left(\sum_{i=1}^{n} a_{i}\right)^{2}$. Therefore, $\sum_{i=1}^{n} \frac{a_{i}}{a_{i+1}+a_{i+2}} \geqslant \frac{\left(\sum_{i=1}^{n} a_{i}\right)^{2}}{\sum_{i=1}^{n} a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,005 |
17 Let $a, b, c, d$ be positive numbers, prove:
$$\sqrt{\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}} \geqslant \sqrt[3]{\frac{a b c+b c d+c d a+d a b}{4}}$$ | $\begin{array}{l}\text { 17. } \frac{1}{4}(a b c+b c d+c d a+d a b)=\frac{1}{4}[b c(a+d)+d a(b+c)] \leqslant \\ \frac{1}{4}\left[\left(\frac{b+c}{2}\right)^{2}(a+d)+\left(\frac{a+d}{2}\right)^{2}(b+c)\right]=\frac{1}{16}(b+c)(a+d)(a+b+c+d) \\ \leqslant \frac{1}{64}(a+b+c+d)^{3}=\left(\frac{a+b+c+d}{4}\right)^{3} \leqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,006 |
Example 1 Given $a, b, c \in \mathbf{R}^{+}$, and $a+b+c=1$, prove:
$$36 \leqslant \frac{1}{a}+\frac{4}{b}+\frac{9}{c} .$$ | Prove that by the Cauchy-Schwarz inequality,
$$\begin{aligned}
\frac{1}{a}+\frac{4}{b}+\frac{9}{c} & =\left(\frac{1}{a}+\frac{4}{b}+\frac{9}{c}\right) \cdot(a+b+c) \\
& \geqslant\left(\sqrt{a} \cdot \frac{1}{\sqrt{a}}+\sqrt{b} \cdot \frac{2}{\sqrt{b}}+\sqrt{c} \cdot \frac{3}{\sqrt{c}}\right)^{2}=36,
\end{aligned}$$
Th... | 36 | Inequalities | proof | Yes | Yes | inequalities | false | 738,008 |
Example 2 Let $a, b, c \in \mathbf{R}^{+}$, satisfying $a \cos ^{2} \alpha+b \sin ^{2} \alpha<c$, prove that:
$$\sqrt{a} \cos ^{2} \alpha+\sqrt{b} \sin ^{2} \alpha<\sqrt{c} .$$ | $$\begin{aligned}
\sqrt{a} \cos ^{2} \alpha+\sqrt{b} \sin ^{2} \alpha & =\sqrt{a} \cos \alpha \cdot \cos \alpha+\sqrt{b} \sin \alpha \cdot \sin \alpha \\
& \leqslant\left[(\sqrt{a} \cos \alpha)^{2}+(\sqrt{b} \sin \alpha)^{2}\right]^{\frac{1}{2}} \cdot\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)^{\frac{1}{2}} \\
& =\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,009 |
15 Let $a, b, c$ be positive real numbers, prove that:
$$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right) \geqslant 2\left(1+\frac{a+b+c}{\sqrt[3]{a b c}}\right)$$ | $\begin{array}{l}\text { 15. Since }\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=2+\left(\frac{a}{c}+\frac{c}{b}+\frac{b}{a}\right)+ \\ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)=2+\left(\frac{a}{c}+\frac{a}{b}+\frac{a}{a}\right)+\left(\frac{b}{a}+\frac{b}{c}+\frac{b}{b}\right)+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,010 |
Example 3 Let $a_{i}>0(i=1,2, \cdots, n)$ satisfy $\sum_{i=1}^{n} a_{i}=1$, prove that: $\frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n}^{2}}{a_{n}+a_{1}} \geqslant \frac{1}{2}$. | $$\begin{array}{l}
\left(\sum_{i=1}^{n} a_{i}\right)^{2}=\left(\sum_{i=1}^{n} \frac{a_{i}}{\sqrt{a_{i}+a_{i+1}}} \cdot \sqrt{a_{i}+a_{i+1}}\right)^{2} \\
\leqslant \sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+a_{i+1}} \cdot \sum_{i=1}^{n}\left(a_{i}+a_{i+1}\right) \\
=2 \sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+a_{i+1}} \cdot \su... | \frac{1}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 738,011 |
Example 4 Let $a, b, c$ be positive real numbers, and satisfy $a+b+c=1$. Prove:
$$\frac{a-b c}{a+b c}+\frac{b-c a}{b+c a}+\frac{c-a b}{c+a b} \leqslant \frac{3}{2}$$ | Prove that
$$1-\frac{a-b c}{a+b c}=\frac{2 b c}{a+b c}=\frac{2 b c}{1-b-c+b c}=\frac{2 b c}{(1-b)(1-c)} .$$
Similarly, $1-\frac{b-c a}{b+c a}=\frac{2 c a}{(1-c)(1-a)}, 1-\frac{c-a b}{c+a b}=\frac{2 a b}{(1-a)(1-b)}$. Therefore, the original inequality is equivalent to
$$\frac{2 b c}{(1-b)(1-c)}+\frac{2 c a}{(1-c)(1-a)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,012 |
Example 5 Let positive real numbers $a, b, c$ satisfy $a+b+c=3$. Prove:
$$\frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+\frac{1}{2+c^{2}+a^{2}} \leqslant \frac{3}{4} .$$ | Prove that the cyclic sum can be represented by the symbol $\sum$, i.e., prove:
$$\sum \frac{1}{2+a^{2}+b^{2}} \leqslant \frac{3}{4}$$
By the Cauchy-Schwarz inequality, we have
$$\left(\sum \frac{a^{2}+b^{2}}{2+a^{2}+b^{2}}\right) \sum\left(2+a^{2}+b^{2}\right) \geqslant\left(\sum \sqrt{a^{2}+b^{2}}\right)^{2}$$
Also... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,013 |
Example 6 Given $x, y, z>0$, and $x y z=1$. Prove:
$$\frac{(x+y-1)^{2}}{z}+\frac{(y+z-1)^{2}}{x}+\frac{(z+x-1)^{2}}{y} \geqslant 4(x+y+z)-12+\frac{9}{x+y+z} .$$ | Prove that because
$$(a-b)^{2}=a^{2}-2 a b+b^{2},$$
so
$$\begin{array}{l}
a^{2}=2 a b-b^{2}+(a-b)^{2}, \\
\frac{a^{2}}{b}=2 a-b+\frac{(a-b)^{2}}{b} .
\end{array}$$
When $b>0$, we have
Using the above formula and the Cauchy-Schwarz inequality, we know that
$$\begin{aligned}
& \frac{(x+y-1)^{2}}{z}+\frac{(y+z-1)^{2}}{x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,014 |
Example 7 Let non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ simultaneously satisfy the following conditions:
(1) $\sum_{i=1}^{n}\left(a_{i}+b_{i}\right)=1$;
(2) $\sum_{i=1}^{n} i\left(a_{i}-b_{i}\right)=0$;
(3) $\sum_{i=1}^{n} i^{2}\left(a_{i}+b_{i}\right)=10$.
Prove: For an... | Prove that for any $1 \leqslant k \leqslant n$, we have
$$\begin{aligned}
\left(k a_{k}\right)^{2} & \leqslant\left(\sum_{i=1}^{n} i a_{i}\right)^{2}=\left(\sum_{i=1}^{n} i b_{i}\right)^{2} \\
& \leqslant\left(\sum_{i=1}^{n} i^{2} b_{i}\right) \cdot\left(\sum_{i=1}^{n} b_{i}\right) \quad \text { (Cauchy-Schwarz inequal... | \max \left\{a_{k}, b_{k}\right\} \leqslant \frac{10}{10+k^{2}} | Algebra | proof | Yes | Yes | inequalities | false | 738,015 |
Example 8 Let $a_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n)$, if for any $x_{i} \geqslant 0$,
$$\sum_{i=1}^{n} r_{i}\left(x_{i}-a_{i}\right) \leqslant \sqrt{\sum_{i=1}^{n} x_{i}^{2}}-\sqrt{\sum_{i=1}^{n} a_{i}^{2}} .$$
Find $r_{i}(i=1,2, \cdots, n)$. | Let $x_{i}=0$, then $\sum_{i=1}^{n} r_{i} a_{i} \geqslant \sqrt{\sum_{i=1}^{n} a_{i}^{2}}$.
Let $x_{i}=2 a_{i}$, then $\sum_{i=1}^{n} r_{i} a_{i} \leqslant \sqrt{\sum_{i=1}^{n} a_{i}^{2}}$.
Thus, $\sum_{i=1}^{n} r_{i} a_{i}=\sqrt{\sum_{i=1}^{n} a_{i}^{2}}$.
Let $x_{i}=r_{i}$, then $\sum_{i=1}^{n} r_{i}\left(r_{i}-a_{i}... | r_{i}=\frac{a_{i}}{\sqrt{\sum_{i=1}^{n} a_{i}^{2}}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,016 |
Example 9 Let $x_{i}, y_{i}, \cdots, z_{i} \in \mathbf{R}(i=1,2, \cdots, n)$, prove:
$$\sum_{i=1}^{n} \sqrt{x_{i}^{2}+y_{i}^{2}+\cdots+z_{i}^{2}} \geqslant \sqrt{\left(\sum_{i=1}^{n} x_{i}\right)^{2}+\left(\sum_{i=1}^{n} y_{i}\right)^{2}+\cdots+\left(\sum_{i=1}^{n} z_{i}\right)^{2}}$$ | Let $a=\sum_{i=1}^{n} x_{i}, b=\sum_{i=1}^{n} y_{i}, \cdots, c=\sum_{i=1}^{n} z_{i}$. Without loss of generality, assume $a^{2}+b^{2}+\cdots+c^{2} \neq 0$. Then by the Cauchy-Schwarz inequality, we have
$$\left(a^{2}+b^{2}+\cdots+c^{2}\right)\left(x_{i}^{2}+y_{i}^{2}+\cdots+z_{i}^{2}\right) \geqslant\left(a x_{i}+b y_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,017 |
Example 10 Let $a_{i} \in \mathbf{R}^{+}, 1 \leqslant i \leqslant n$. Prove:
$$\frac{1}{\frac{1}{1+a_{1}}+\frac{1}{1+a_{2}}+\cdots+\frac{1}{1+a_{n}}}-\frac{1}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}} \geqslant \frac{1}{n}$$ | Prove that if $\sum_{i=1}^{n} \frac{1}{a_{i}}=a$, then $\sum_{i=1}^{n} \frac{1+a_{i}}{a_{i}}=n+a$. By the Cauchy-Schwarz inequality, we have
$$\sum_{i=1}^{n} \frac{a_{i}}{1+a_{i}} \cdot \sum_{i=1}^{n} \frac{1+a_{i}}{a_{i}} \geqslant n^{2}$$
Therefore, $\sum_{i=1}^{n} \frac{a_{i}}{a_{i}+1} \geqslant \frac{n^{2}}{n+a}$,... | \frac{1}{n} | Inequalities | proof | Yes | Yes | inequalities | false | 738,018 |
Example 11 Let $n$ be a positive integer, and $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$ be real numbers. Prove:
(1) $\left(\sum_{i, j=1}^{n}\left|x_{i}-x_{j}\right|\right)^{2} \leqslant \frac{2\left(n^{2}-1\right)}{3} \sum_{i, j=1}^{n}\left(x_{i}-x_{j}\right)^{2}$;
(2) The equality in (1) holds if and on... | Prove (1) Without loss of generality, we can assume $\sum_{i=1}^{n} x_{i}=0$, then
$$\sum_{i, j=1}^{n}\left|x_{i}-x_{j}\right|=2 \sum_{i<j}\left(x_{j}-x_{i}\right)=2 \sum_{i=1}^{n}(2 i-n-1) x_{i}$$
By the Cauchy-Schwarz inequality, we get
$$\begin{aligned}
\left(\sum_{i, j=1}^{n}\left|x_{i}-x_{j}\right|\right)^{2} & \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,019 |
Example 12 Prove: The integers satisfying the conditions
(1) $a_{1}+a_{2}+\cdots+a_{n} \geqslant n^{2}$;
(2) $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant n^{3}+1$
are only $\left(a_{1}, a_{2}, \cdots, a_{n}\right)=(n, n, \cdots, n)$. | Proof: Let $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ be an integer sequence satisfying the given conditions. By the Cauchy-Schwarz inequality, we have
$$a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant \frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2} \geqslant n^{3}.$$
Combining this with $a_{1}^{2}+a_{2}^{2}+\cdot... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,020 |
16 Let $x_{1}, x_{2}, x_{3} \in \mathbf{R}^{+}$, prove:
$$\frac{x_{2}}{x_{1}}+\frac{x_{3}}{x_{2}}+\frac{x_{1}}{x_{3}} \leqslant\left(\frac{x_{1}}{x_{2}}\right)^{2}+\left(\frac{x_{2}}{x_{3}}\right)^{2}+\left(\frac{x_{3}}{x_{1}}\right)^{2} .$$ | 16. By the AM-GM inequality, we have $\frac{x_{2}}{x_{1}}=\frac{x_{2}}{x_{3}} \cdot \frac{x_{3}}{x_{1}} \leqslant \frac{1}{2}\left[\left(\frac{x_{2}}{x_{3}}\right)^{2}+\left(\frac{x_{3}}{x_{1}}\right)^{2}\right], \frac{x_{3}}{x_{2}}=\frac{x_{3}}{x_{1}} \cdot \frac{x_{1}}{x_{2}} \leqslant \frac{1}{2}\left[\left(\frac{x_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,021 |
Example 13 Proof: The Ptolemy's inequality for two triangles
$$a^{2}\left(b_{1}^{2}+c_{1}^{2}-a_{1}^{2}\right)+b^{2}\left(c_{1}^{2}+a_{1}^{2}-b_{1}^{2}\right)+c^{2}\left(a_{1}^{2}+b_{1}^{2}-c_{1}^{2}\right) \geqslant 16 S S_{1},$$
where $a, b, c, S; a_{1}, b_{1}, c_{1}, S_{1}$ are the side lengths and areas of the two... | Prove that by the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& 16 S_{1}+2 a^{2} a_{1}^{2}+2 b^{2} b_{1}^{2}+2 c^{2} c_{1}^{2} \\
\leqslant & \left(16 S_{1}^{2}+2 a_{1}^{4}+2 b_{1}^{4}+2 c_{1}^{4}\right)^{\frac{1}{2}}\left(16 S^{2}+2 a^{4}+2 b^{4}+2 c^{4}\right)^{\frac{1}{2}} \\
= & \left(a_{1}^{2}+b_{1}^{2}+c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,022 |
Example 14 Let $a, b, c$ be the lengths of the three sides of a triangle, prove that:
$$a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \geqslant 0 .$$ | Prove that there obviously exist positive numbers $x, y, z$ such that $a=y+z, b=z+x, c=x+y$. Since
$$\begin{aligned}
& a^{2} b(a-b)=(y+z)^{2}(z+x)(y-x) \\
= & (y+z)(z+x)\left(y^{2}-z^{2}\right)+(y+z)^{2}\left(z^{2}-x^{2}\right)
\end{aligned}$$
Similarly handling $b^{2} c(b-c), c^{2} a(c-a)$, we have
$$\begin{aligned}
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,023 |
Example 15 Let $x_{i}>0, x_{i} y_{i}-z_{i}^{2}>0, i=1,2$, prove:
$$\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}}$$ | Notice that the right side of the inequality $\geqslant \frac{2}{\left[\left(x_{1} y_{1}-z_{1}^{2}\right)\left(x_{2} y_{2}-z_{2}^{2}\right)\right]^{\frac{1}{2}}}$, consider proving a stronger conclusion:
$$\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(z_{1}+z_{2}\right)^{2} \geqslant 4\left[\left(x_{1} y_{1}-z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,024 |
Example 16 Let $a_{i}, b_{i}, c_{i}, d_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n)$, prove that:
$$\left(\sum a_{i} b_{i} c_{i} d_{i}\right)^{4} \leqslant \sum a_{i}^{4} \cdot \sum b_{i}^{4} \cdot \sum c_{i}^{4} \cdot \sum d_{i}^{4}$$ | Prove that using the Cauchy-Schwarz inequality twice, we get
$$\begin{aligned}
\text { LHS } & =\left[\sum\left(a_{i} b_{i}\right)\left(c_{i} d_{i}\right)\right]^{4} \\
& \leqslant\left[\sum\left(a_{i} b_{i}\right)^{2}\right]^{2} \cdot\left[\sum\left(c_{i} d_{i}\right)^{2}\right]^{2}=\left[\sum a_{i}^{2} b_{i}^{2}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,025 |
Example 17 Let $t_{a}, t_{b}, t_{c}$ be the lengths of the angle bisectors of $\angle A, \angle B, \angle C$ in $\triangle ABC$, respectively. Prove:
$$\sum \frac{bc}{t_{a}^{2}} \geqslant 4$$ | It is not difficult to obtain $t_{a}^{2}=\frac{b c\left[(b+c)^{2}-a^{2}\right]}{(b+c)^{2}}$, then $\frac{b c}{t_{a}^{2}}=\frac{(b+c)^{2}}{(b+c)^{2}-a^{2}}$.
Similarly, we can get
$$\frac{a c}{t_{b}^{2}}=\frac{(a+c)^{2}}{(a+c)^{2}-b^{2}}, \frac{a b}{t_{c}^{2}}=\frac{(a+b)^{2}}{(a+b)^{2}-c^{2}}$$
Then
$$\sum \frac{b c}{... | 4 | Inequalities | proof | Yes | Yes | inequalities | false | 738,026 |
Example 18 Let $a, b, c, d$ be positive real numbers, satisfying $ab + cd = 1$. Points $P_{i}(x_{i}, y_{i}) (i=1, 2, 3, 4)$ are four points on the unit circle centered at the origin. Prove:
$$\begin{aligned}
& \left(a y_{1} + b y_{2} + c y_{3} + d y_{4}\right)^{2} + \left(a x_{4} + b x_{3} + c x_{2} + d x_{1}\right)^{2... | Prove that let $\alpha=a y_{1}+b y_{2}+c y_{3}+d y_{4}, \beta=a x_{4}+b x_{3}+c x_{2}+d x_{1}$, by the Cauchy-Schwarz inequality, we get
$$\begin{aligned}
\alpha^{2}= & \left(a y_{1}+b y_{2}+c y_{3}+d y_{4}\right)^{2} \\
\leqslant & {\left[\left(\sqrt{a d} y_{1}\right)^{2}+\left(\sqrt{b c} y_{2}\right)^{2}+\left(\sqrt{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,027 |
Example 19 Given a positive integer $n \geqslant 2$, let positive integers $a_{i}(i=1,2, \cdots, n)$ satisfy $a_{1}<$ $a_{2}<\cdots<a_{n}$ and $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$. Prove: For any real number $x$, we have
$$\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant \frac{1}{2} \cdot \fr... | Prove that when $x^{2} \geqslant a_{1}\left(a_{1}-1\right)$, due to $\sum \frac{1}{a_{i}} \leqslant 1$, we have
$$\begin{aligned}
\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} & \leqslant\left(\sum_{i=1}^{n} \frac{1}{2 a_{i}|x|}\right)^{2}=\frac{1}{4 x^{2}}\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right)^{2} \\
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,028 |
Example 20
$$\begin{array}{l}
\text { Let } x \in\left(0, \frac{\pi}{2}\right), n \in \mathbf{N} \text {, prove: } \\
\qquad\left(\frac{1-\sin ^{2 n} x}{\sin ^{2 n} x}\right)\left(\frac{1-\cos ^{2 n} x}{\cos ^{2 n} x}\right) \geqslant\left(2^{n}-1\right)^{2} .
\end{array}$$ | Prove that since
$$\begin{aligned}
1-\sin ^{2 n} x & =\left(1-\sin ^{2} x\right)\left(1+\sin ^{2} x+\sin ^{4} x+\cdots+\sin ^{2(n-1)} x\right) \\
& =\cos ^{2} x\left(1+\sin ^{2} x+\sin ^{4} x+\cdots+\sin ^{2(n-1)} x\right), \\
1-\cos ^{2 n} x & =\left(1-\cos ^{2} x\right)\left(1+\cos ^{2} x+\cos ^{4} x+\cdots+\cos ^{2(... | \left(2^{n}-1\right)^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 738,029 |
Example 21 Let $a_{1}, a_{2}, \cdots, a_{n}$ be an infinite sequence of real numbers, such that for all positive integers $i$, there exists a real number $c$, satisfying $0 \leqslant a_{i} \leqslant c$, and $\left|a_{i}-a_{j}\right| \geqslant \frac{1}{i+j}$ for all positive integers $i, j$ $(i \neq j)$. Prove: $c \geqs... | For $n \geqslant 2$, let $\sigma(1), \sigma(2), \cdots, \sigma(n)$ be a permutation of $1,2, \cdots, n$, and satisfy
then
$$\begin{array}{c}
0 \leqslant a_{\sigma(1)}<a_{\sigma(2)}<\cdots<a_{\sigma(n)} \leqslant c \\
c \geqslant a_{\sigma(n)}-a_{\sigma(1)} \\
\left(a_{\sigma(n)}-a_{\sigma(n-1)}\right)+\left(a_{\sigma(... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,030 |
Example 22 Let $a, b, c$ be positive real numbers. Prove that:
$$\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(a+2 b+c)^{2}}{2 b^{2}+(a+c)^{2}}+\frac{(a+b+2 c)^{2}}{2 c^{2}+(b+a)^{2}} \leqslant 8$$ | Proof In Chapter 2, we used two different methods to prove this inequality. Here, we provide another new proof using the Cauchy-Schwarz inequality.
By the Cauchy-Schwarz inequality, we have
$$\begin{array}{c}
\sqrt{\frac{2 a^{2}+\frac{(b+c)^{2}}{2}+\frac{(b+c)^{2}}{2}}{3}} \geqslant \frac{\sqrt{2} a+\frac{\sqrt{2}}{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,031 |
177 Let $a$, $b$, $c$ be positive real numbers, and $a+b+c=1$. Prove:
$$(1+a)(1+b)(1+c) \geqslant 8(1-a)(1-b)(1-c) .$$ | 17. Since $1+a=2-b-c=1-b+1-c \geqslant 2 \sqrt{(1-b)(1-c)}$, similarly we get $1+b \geqslant 2 \sqrt{(1-a)(1-c)}, 1+c \geqslant 2 \sqrt{(1-a)(1-b)}$. Multiplying the above three inequalities will do. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,032 |
Example 23 Let $a_{i}>0$, and $\sum_{i=1}^{n} a_{i}=k$, prove that:
$$\sum_{i=1}^{n}\left(a_{i}+\frac{1}{a_{i}}\right)^{2} \geqslant n\left(\frac{n^{2}+k^{2}}{n k}\right)^{2}$$ | Prove that by the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(1^{2}+1^{2}+\cdots+1^{2}\right) \sum_{i=1}^{n}\left(a_{i}+\frac{1}{a_{i}}\right)^{2} \\
\geqslant & {\left[\sum_{i=1}^{n}\left(a_{i}+\frac{1}{a_{i}}\right)\right]^{2}=\left(k+\sum_{i=1}^{n} \frac{1}{a_{i}}\right)^{2} } \\
\geqslant & \left(k... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,033 |
Example 24 Prove: For positive real numbers $a, b, c$, we have
$$\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1 .$$ | Prove by the transformed Cauchy inequality,
$$\text { LHS }=\sum \frac{a}{\sqrt{a^{2}+8 b c}}=\sum \frac{a^{\frac{3}{2}}}{\sqrt{a^{3}+8 a b c}} \geqslant \frac{\left(\sum a\right)^{\frac{3}{2}}}{\left[\sum\left(a^{3}+8 a b c\right)\right]^{\frac{1}{2}}} \text {. }$$
To prove the original inequality, it suffices to pro... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,034 |
Example 1 Solve the system of equations
$$\left\{\begin{array}{l}
a^{2}=\frac{\sqrt{b c} \sqrt[3]{b c d}}{(b+c)(b+c+d)} \\
b^{2}=\frac{\sqrt{c d} \sqrt[3]{c d a}}{(c+d)(c+d+a)} \\
c^{2}=\frac{\sqrt{d a} \sqrt[3]{d a b}}{(d+a)(d+a+b)} \\
d^{2}=\frac{\sqrt{a b} \sqrt[3]{a b c}}{(a+b)(a+b+c)}
\end{array}\right.$$
for rea... | First, note that none of the variables equals zero. Without loss of generality, assume $b=0$, from (31) we get $a=0$, from (34) we get $d=0$, and from (33) we get $c=0$, which means all values are zero, but this is impossible because the denominator would be zero.
Second, note that the square roots of $b c, c d, d a, ... | a=b=c=d=\frac{\sqrt{6}}{6} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,035 |
Example 2 Given real numbers $x, y, z > 3$, find all real solutions $(x, y, z)$ of the equation
$$\frac{(x+2)^{2}}{y+z-2}+\frac{(y+4)^{2}}{z+x-4}+\frac{(z+6)^{2}}{x+y-6}=36$$ | Given $x, y, z > 3$, we know
$$y+z-2>0, z+x-4>0, x+y-6>0$$
By the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& {\left[\frac{(x+2)^{2}}{y+z-2}+\frac{(y+4)^{2}}{x+z-4}+\frac{(z+6)^{2}}{x+y-6}\right] } \\
& {[(y+z-2)+(x+z-4)+(x+y-6)] } \\
\geqslant & (x+y+z+12)^{2} \\
\Leftrightarrow & \frac{(x+2)^{2}}{y+z-2}+\... | (10,8,6) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,036 |
Example 3 $n$ is a positive integer, $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$ are $2 n$ positive real numbers, satisfying $a_{1}+a_{2}+\cdots+a_{n}=1, b_{1}+b_{2}+\cdots+b_{n}=1$, find the minimum value of $\frac{a_{1}^{2}}{a_{1}+b_{1}}+\frac{a_{2}^{2}}{a_{2}+b_{2}}+\cdots+\frac{a_{n}^{2}}{a_{n}+b_{n}... | By Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(a_{1}+a_{2}+\cdots+a_{n}+b_{1}+b_{2}+\cdots+b_{n}\right)\left(\frac{a_{1}^{2}}{a_{1}+b_{1}}+\frac{a_{2}^{2}}{a_{2}+b_{2}}+\cdots+\frac{a_{n}^{2}}{a_{n}+b_{n}}\right) \\
\geqslant & \left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}=1
\end{aligned}$$
and given that... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,037 |
Example 4 Given that $x, y, z$ are real numbers, and satisfy
$$x+y+z=xy+yz+zx \text{.}$$
Find the minimum value of $\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1}$. | Let $x=1, y=z=-1$. Then
$$\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1}=-\frac{1}{2}$$
Conjecture the minimum value is $-\frac{1}{2}$.
It suffices to prove:
$$\begin{aligned}
& \frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1} \geqslant-\frac{1}{2} \\
\Leftrightarrow & \frac{(x+1)^{2}}{x^{2}+1}+\frac{(y+1)... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,038 |
Example 5 Let $a, b, c, x, y, z$ be real numbers, and
$$a^{2}+b^{2}+c^{2}=25, x^{2}+y^{2}+z^{2}=36, a x+b y+c z=30 .$$
Find the value of $\frac{a+b+c}{x+y+z}$. | By Cauchy-Schwarz inequality, we have
$$25 \times 36=\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right) \geqslant(a x+b y+c z)^{2}=30^{2} .$$
For the equality to hold, we get
$$\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=k$$
Thus, \( k^{2}\left(x^{2}+y^{2}+z^{2}\right)=25 \), so \( k= \pm \frac{5}{6} \) (the negati... | \frac{5}{6} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,039 |
Example 6 Let real numbers $a, b, c, d, e$ satisfy
$$a+b+c+d+e=8, a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16,$$
Find the maximum value of $e$. | Solve: Rewrite the conditions as
$$8-e=a+b+c+d, 16-e^{2}=a^{2}+b^{2}+c^{2}+d^{2},$$
From this, we get an inequality involving $e$. By the Cauchy-Schwarz inequality, we have
$$a+b+c+d \leqslant(1+1+1+1)^{\frac{1}{2}}\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{\frac{1}{2}} .$$
Substitute the conditions and square both sides,... | \frac{16}{5} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,040 |
Example 7 Let $x \geqslant 0, y \geqslant 0, z \geqslant 0, a, b, c, l, m, n$ be given positive numbers, and $a x+b y+c z=\delta$ be a constant, find
$$w=\frac{l}{x}+\frac{m}{y}+\frac{n}{z}$$
the minimum value. | By the Cauchy-Schwarz inequality, we have
So
$$\begin{array}{l}
w \cdot \delta= {\left[\left(\sqrt{\frac{l}{x}}\right)^{2}+\left(\sqrt{\frac{m}{y}}\right)^{2}+\left(\sqrt{\frac{n}{z}}\right)^{2}\right] } \\
\cdot\left[(\sqrt{a x})^{2}+(\sqrt{b y})^{2}+(\sqrt{c z})^{2}\right] \\
\geqslant(\sqrt{a l}+\sqrt{b m}+\sqrt{c... | \frac{(\sqrt{a l}+\sqrt{b m}+\sqrt{c n})^{2}}{\delta} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,041 |
Example 8 For positive real numbers $a, b$ that satisfy $a+b=1$, find
$$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$$
the minimum value. | When $a=b=\frac{1}{2}$, we have
$$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}=\frac{25}{2}$$
Below we prove
$$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2} \geqslant \frac{25}{2}$$
Thus, the minimum value is $\frac{25}{2}$.
Let $x=a+\frac{1}{a}, y=b+\frac{1}{b}$, by
$$\frac{x^{2}+y^{... | \frac{25}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,042 |
18 Let $x, y, z$ be positive real numbers, and $x \geqslant y \geqslant z$. Prove:
$$\frac{x^{2} y}{z}+\frac{y^{2} z}{x}+\frac{z^{2} x}{y} \geqslant x^{2}+y^{2}+z^{2} .$$ | $\begin{array}{l}\text { 18. } \frac{x^{2} y}{z}+\frac{y^{2} z}{x}+\frac{z^{2} x}{y}-x^{2}-y^{2}-z^{2}=\frac{x^{2}}{z}(y-z)+\frac{y^{2} z}{x}+\frac{z^{2} x}{y}-y^{2}-z^{2} \geqslant \\ \frac{y^{2}}{z}(y-z)+2 z \sqrt{y z}-y^{2}-z^{2}=\frac{y-z}{z}\left(y^{2}-y z+z^{2}-\frac{2 z^{2} \sqrt{y}}{\sqrt{y}+\sqrt{z}}\right)= \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,043 |
Example 9 Let $n$ and $k$ be given positive integers $(k<n)$, and let $a_{1}, a_{2}, \cdots, a_{k}$ be given positive real numbers. Try to find positive real numbers $a_{k+1}, a_{k+2}, \cdots, a_{n}$ such that the sum
takes the minimum value.
$$M=\sum_{i \neq j} \frac{a_{i}}{a_{j}}$$ | By calculating for $n=1,2,3$, we get
$$M=\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)-n .$$
Let $a=a_{1}+a_{2}+\cdots+a_{k}, b=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{k}}$, then by the assumption $a, b$ are given constants. Therefore, by the Cauchy... | a_{k+1}=\cdots=a_{n}=\sqrt{\frac{a}{b}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,044 |
Example 10 Let $2 n$ real numbers $a_{1}, a_{2}, \cdots, a_{2 n}$ satisfy $\sum_{i=1}^{2 n-1}\left(a_{i+1}-a_{i}\right)^{2}=1$, find
$$\left(a_{n+1}+a_{n+2}+\cdots+a_{2 n}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$$
the maximum value. | When $n=1$, $\left(a_{2}-a_{1}\right)^{2}=1$, then $a_{2}-a_{1}= \pm 1$, the maximum value is 1.
When $n \geqslant 2$, let $x_{1}=a_{1}, x_{i+1}=a_{i+1}-a_{i}, i=1,2, \cdots, 2 n-1$. Then $\sum_{i=2}^{2 n} x_{i}^{2}=$ 1, and $a_{k}=x_{1}+x_{2}+\cdots+x_{k}, k=1,2, \cdots, 2 n$.
By the Cauchy-Schwarz inequality, we get... | \sqrt{\frac{n\left(2 n^{2}+1\right)}{3}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,045 |
Example 11 Let $x_{i} \geqslant 0, i=1,2, \cdots, n$, satisfy
$$\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leq j<k \leqslant n} \sqrt{\frac{j}{k}} x_{j} x_{k}=1 .$$
Find the minimum and maximum values of $x_{1}+x_{2}+\cdots+x_{n}$. | Solve: From
$$\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2}=\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant j<k \leqslant n} x_{j} x_{k} \geqslant 1$$
Take $x_{1}=1, x_{2}=\cdots=x_{n}=0$, then the minimum value of $x_{1}+x_{2}+\cdots+x_{n}$ is 1. Let $y_{i}=$ $\frac{x_{i}}{\sqrt{i}}$, then the condition becomes
$$\sum_{i=1}... | \sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,046 |
Example 12 Let $x, y, z$ be real numbers greater than -1. Find the minimum value of
$$\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}}$$ | Given $x, y, z > -1$, then $\frac{1+x^{2}}{1+y+z^{2}}, \frac{1+y^{2}}{1+z+x^{2}}, \frac{1+z^{2}}{1+x+y^{2}}$ have positive numerators and denominators, so
$$\begin{aligned}
& \frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \\
\geqslant & \frac{1+x^{2}}{1+z^{2}+\frac{1+y^{2}}{2}}+\frac{1+y^... | 2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,047 |
Example 14 Let $n>3$ be a given positive integer, and real numbers $x_{1}, x_{2}, \cdots, x_{n+1}, x_{n+2}$ satisfy $0<$ $x_{1}<x_{2}<\cdots<x_{n+1}<x_{n+2}$. Find
$$\frac{\left(\sum_{i=1}^{n} \frac{x_{i+1}}{x_{i}}\right)\left(\sum_{j=1}^{n} \frac{x_{j+2}}{x_{j+1}}\right)}{\sum_{k=1}^{n} \frac{x_{k+1} x_{k+2}}{x_{k+1}^... | Let $t_{i}=\frac{x_{i+1}}{x_{i}}(1 \leqslant i \leqslant n+1)$, then the original expression equals
$$\frac{\sum_{i=1}^{n} t_{i} \sum_{i=1}^{n} t_{i+1}}{\sum_{i=1}^{n} \frac{t_{i} t_{i+1}}{t_{i}+t_{i+1}} \sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)}$$
By the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \sum_{i=1... | 1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,049 |
Example 1 Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive integers, prove that:
$$\frac{a_{1}^{2}}{a_{2}}+\frac{a_{2}^{2}}{a_{3}}+\cdots+\frac{a_{n}^{2}}{a_{1}} \geqslant a_{1}+a_{2}+\cdots+a_{n} .$$ | $$\begin{aligned}
& \left(\frac{a_{1}^{2}}{a_{2}}+\frac{a_{2}^{2}}{a_{3}}+\cdots+\frac{a_{n}^{2}}{a_{1}}\right)\left(a_{2}+a_{3}+\cdots+a_{1}\right) \\
\geqslant & \left(\frac{a_{1}}{\sqrt{a_{2}}} \cdot \sqrt{a_{2}}+\frac{a_{2}}{\sqrt{a_{3}}} \cdot \sqrt{a_{3}}+\cdots+\frac{a_{n}}{\sqrt{a_{1}}} \cdot \sqrt{a_{1}}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,050 |
Example 2 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 2)$ satisfying $\sum_{i=1}^{n} a_{i}=1$, prove:
$$\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}} \geqslant \frac{n}{2 n-1}$$ | Prove that
$$\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}}=\sum_{i=1}^{n}\left(\frac{2}{2-a_{i}}-1\right)=\sum_{i=1}^{n} \frac{2}{2-a_{i}}-n .$$
By the Cauchy-Schwarz inequality, we have
$$\left(\sum_{i=1}^{n} \frac{1}{2-a_{i}}\right)\left[\sum_{i=1}^{n}\left(2-a_{i}\right)\right] \geqslant n^{2}$$
Therefore,
$$\begin{array}{... | \frac{n}{2 n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 738,051 |
Example 3 Let $a_{i}, b_{i}, i \geqslant 1$ be positive numbers, satisfying $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}$. Prove that,
$$\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}} \geqslant \frac{1}{2}\left(\sum_{i=1}^{n} a_{i}\right)$$ | Prove that by Cauchy's inequality,
$$\left(\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}}\right) \sum_{i=1}^{n}\left(a_{i}+b_{i}\right) \geqslant\left(\sum_{i=1}^{n} a_{i}\right)^{2} .$$
Since $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}$, the above inequality becomes
$$2\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,052 |
Example 4 Let $P_{1}, P_{2}, \cdots, P_{n}$ be any permutation of $1,2, \cdots, n$. Prove that:
$$\frac{1}{P_{1}+P_{2}}+\frac{1}{P_{2}+P_{3}}+\cdots+\frac{1}{P_{n-2}+P_{n-1}}+\frac{1}{P_{n-1}+P_{n}}>\frac{n-1}{n+2}$$ | Prove that by Cauchy-Schwarz inequality,
$$\begin{array}{c}
{\left[\left(P_{1}+P_{2}\right)+\left(P_{2}+P_{3}\right)+\cdots+\left(P_{n-1}+P_{n}\right)\right] \cdot} \\
\left(\frac{1}{P_{1}+P_{2}}+\frac{1}{P_{2}+P_{3}}+\cdots+\frac{1}{P_{n-2}+P_{n-1}}+\frac{1}{P_{n-1}+P_{n}}\right) \geqslant(n-1)^{2}
\end{array}$$
Ther... | \frac{n-1}{n+2} | Inequalities | proof | Yes | Yes | inequalities | false | 738,053 |
19 Let $a, b, c$ be positive real numbers, satisfying $a^{2}+b^{2}+c^{2}=1$. Prove:
$$\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b} \geqslant \sqrt{3} .$$ | $$\begin{array}{l}
\text { 19. Since }\left(\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b}\right)^{2}=\frac{a^{2} b^{2}}{c^{2}}+\frac{b^{2} c^{2}}{a^{2}}+\frac{c^{2} a^{2}}{b^{2}}+2\left(a^{2}+b^{2}+c^{2}\right)= \\
\frac{1}{2}\left(\frac{a^{2} b^{2}}{c^{2}}+\frac{c^{2} a^{2}}{b^{2}}\right)+\frac{1}{2}\left(\frac{b^{2} c^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,054 |
Example 5 Let positive numbers $x_{i}$ satisfy $\sum_{i=1}^{n} x_{i}=1$, prove:
$$\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1-x_{i}}} \geqslant \frac{1}{\sqrt{n-1}} \sum_{i=1}^{n} \sqrt{x_{i}} .$$ | Prove that by the Cauchy-Schwarz inequality,
$$\sum_{i=1}^{n} \frac{1}{\sqrt{1-x_{i}}} \cdot \sum_{i=1}^{n} \sqrt{1-x_{i}} \geqslant n^{2}$$
and
$$\sum_{i=1}^{n} \sqrt{1-x_{i}} \leqslant \sqrt{\sum_{i=1}^{n} 1 \cdot \sum_{i=1}^{n}\left(1-x_{i}\right)}=\sqrt{n(n-1)},$$
so $\square$
$$\begin{aligned}
\sum_{i=1}^{n} \fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,055 |
Example 6 Let $a, b, c$ be real numbers greater than -1, prove:
$$\frac{1+a^{2}}{1+b+c^{2}}+\frac{1+b^{2}}{1+c+a^{2}}+\frac{1+c^{2}}{1+a+b^{2}} \geqslant 2$$ | Proof: Given the assumption that we have $1+a^{2}, 1+b^{2}, 1+c^{2}, 1+b+c^{2}, 1+c+a^{2}, 1+a+b^{2}$ are all greater than zero.
By the Cauchy-Schwarz inequality, we get
$$\begin{aligned}
& \left(\frac{1+a^{2}}{1+b+c^{2}}+\frac{1+b^{2}}{1+c+a^{2}}+\frac{1+c^{2}}{1+a+b^{2}}\right) \cdot\left[\left(1+a^{2}\right)\left(1... | 2 | Inequalities | proof | Yes | Yes | inequalities | false | 738,056 |
Example 7 Positive numbers $a, b, c$ satisfy $a b c=1, n$ is a positive integer, prove:
(a) $\frac{1}{1+2 a}+\frac{1}{1+2 b}+\frac{1}{1+2 c} \geqslant 1$; | Proof (a) First, let's prove
$$\begin{aligned}
& \frac{1}{1+2 a} \geqslant \frac{a^{-\frac{2}{3}}}{a^{-\frac{2}{3}}+b^{-\frac{2}{3}}+c^{-\frac{2}{3}}} \\
\Leftrightarrow & a^{-\frac{2}{3}}+b^{-\frac{2}{3}}+c^{-\frac{2}{3}} \geqslant a^{-\frac{2}{3}}+2 a^{\frac{1}{3}} \\
\Leftrightarrow & b^{-\frac{2}{3}}+c^{-\frac{2}{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,057 |
(b) $\frac{c^{n}}{a+b}+\frac{b^{n}}{c+a}+\frac{a^{n}}{b+c} \geqslant \frac{3}{2}$. | (b) Suppose $a \geqslant b \geqslant c$, then $a^{n-1} \geqslant b^{n-1} \geqslant c^{n-1}, \frac{a}{b+c} \geqslant \frac{b}{c+a} \geqslant \frac{c}{a+b}$.
By the rearrangement inequality, we get
$$\begin{array}{l}
\frac{c^{n}}{a+b}+\frac{b^{n}}{c+a}+\frac{a^{n}}{b+c} \geqslant \frac{c a^{n-1}}{a+b}+\frac{b c^{n-1}}{c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,058 |
Example 8 Proof: For any real numbers $x, y, z$ satisfying $x+y+z=0$, we have
$$\frac{x(x+2)}{2 x^{2}+1}+\frac{y(y+2)}{2 y^{2}+1}+\frac{z(z+2)}{2 z^{2}+1} \geqslant 0 \text {. }$$ | Notice that $\frac{x(x+2)}{2 x^{2}+1}=\frac{(2 x+1)^{2}}{2\left(2 x^{2}+1\right)}-\frac{1}{2}$, so the original inequality is equivalent to
$$\frac{(2 x+1)^{2}}{2 x^{2}+1}+\frac{(2 y+1)^{2}}{2 y^{2}+1}+\frac{(2 z+1)^{2}}{2 z^{2}+1} \geqslant 3 \text {, }$$
By the Cauchy-Schwarz inequality, we have
$$2 x^{2}=\frac{4}{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,059 |
Example 9 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n}(n>2)$ satisfying $a_{1}+a_{2}+\cdots+a_{n}=1$. Prove: $\frac{a_{2} a_{3} \cdots a_{n}}{a_{1}+n-2}+\frac{a_{1} a_{3} \cdots a_{n}}{a_{2}+n-2}+\cdots+\frac{a_{1} a_{2} \cdots a_{n-1}}{a_{n}+n-2} \leqslant \frac{1}{(n-1)^{2}}$. | Prove that by the Cauchy-Schwarz inequality, for positive numbers \( x_{1}, x_{2}, \cdots, x_{n} \), we have
\[
\frac{1}{\sum_{i=1}^{n} x_{i}} \leqslant \frac{1}{n^{2}} \sum_{i=1}^{n} \frac{1}{x_{i}}
\]
Given \( a_{1} + a_{2} + \cdots + a_{n} = 1 \) (where \( n > 2 \)), then
\[
\begin{aligned}
\sum_{i=1}^{n} & \frac{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,060 |
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