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Example 11. In $\triangle A B C$, prove that:
$$\frac{\sin A}{h_{a}}+\frac{\sin B}{h_{b}}+\frac{\sin C}{h_{c}} \geqslant \frac{\sqrt{3}}{R} .$$ | Prove that the left side of the above equation is the ordered sum, denoted as $M$. By Chebyshev's inequality, we have
$$M \geqslant \frac{1}{3}\left(\frac{1}{h_{c}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}\right)(\sin A+\sin B+\sin C) .$$
By the inequality $\frac{1}{h_{d}}+\frac{1}{h_{b}}+\frac{1}{h_{c}} \geqslant \frac{2}{\sq... | \frac{\sqrt{3}}{R} | Inequalities | proof | Yes | Yes | inequalities | false | 736,703 |
5. Given the function $f(x)=x^{2}+(a+1)^{2}+|x+a+1|$ has a minimum value $y^{\prime}>3$. Find the range of the real number $a$.
保留了原文的换行和格式。 | (Solution: $a \leq$
$$\frac{-1-\sqrt{6}}{2} \text { or } a>\frac{\sqrt{2}}{2} \text { ) }$$ | a \leq \frac{-1-\sqrt{6}}{2} \text { or } a>\frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,707 |
6. Let $x^{2}+y^{2}=a^{2}, a^{2}<1$. Try to find the maximum and minimum values of $S=\sqrt{1-x^{2}}+$ $\sqrt{1-y^{2}}$. | (Answer;
$$\left.S_{\text {large }}=\sqrt{4-3 a^{2}}, S_{\text {small }}=1+\sqrt{1-a^{2}}\right)$$
) | S_{\text {large }}=\sqrt{4-3 a^{2}}, S_{\text {small }}=1+\sqrt{1-a^{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,708 |
7. If the equal numbers $x, y$ satisfy the condition $2 x^{2}-6 x+y^{2}=0$, then what is the maximum value of $x^{2}+y^{2}+2 x$?
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | (Plan: ! !
The text above has been translated into English, preserving the original text's line breaks and format. | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,709 |
Example 1. Let $x, y, z$ be positive numbers. Prove that:
$$\frac{z^{2}-x^{2}}{x+y}+\frac{x^{2}-y^{2}}{y+z}+\frac{y^{2}-z^{2}}{z+x} \geqslant 0 .$$
(W・Janous Conjecture) | To prove the inequality is equivalent to
$$\begin{array}{l}
\frac{z^{2}}{x+y}+\frac{y^{2}}{x+z}+\frac{x^{2}}{y+z} \\
\geqslant \frac{x^{2}}{x+y}+\frac{y^{2}}{y+z}+\frac{z^{2}}{z+x}
\end{array}$$
Without loss of generality, assume \( x \leqslant y \leqslant z \), then
$$\begin{array}{l}
x^{2} \leqslant y^{2} \leqslant ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,711 |
Example 2. Given $x_{i}, y_{i}(i=1,2, \cdots, n)$ are real numbers, and $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}, y_{1} \geqslant y_{2} \geqslant \cdots \geqslant y_{n}$, and $z_{1}$, $z_{2}, \cdots, z_{n}$ is any permutation of $y_{1}, y_{2}, \cdots, y_{n}$. Prove: $\sum_{i=1}^{n}\left(x_{i}-y_{i}\right... | Proof $\because \sum_{i=1}^{n} y_{i}^{2}=\sum_{i=1}^{n} z_{i}^{2}$,
thus the original inequality is equivalent to
$$\sum_{i=1}^{n} x_{i} y_{i} \geqslant \sum_{i=1}^{n} x_{i} z_{i}$$
The left side of this equation is the sum in order, and the right side is the sum in disorder. According to the rearrangement inequality,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,712 |
Example 1: A class of students is holding a New Year's riddle competition and needs to purchase 4, 5, and 6 prizes of different prices. Now, they are selecting prizes with unit prices of 1 yuan, 2 yuan, and 3 yuan from a supermarket. The minimum and maximum amount of money spent should be $(\quad)$
A. 28 yuan and 32 yu... | Analysis: According to the rearrangement inequality, the least amount of money spent should be the reverse order sum: $4 \times 3 + 5 \times 2 + 6 \times 1 = 28$ yuan, and the most amount of money spent should be the direct order sum: $4 \times 1 + 5 \times 2 + 6 \times 3 = 32$ yuan. Therefore, the answer is $A$. | A | Logic and Puzzles | MCQ | Yes | Yes | inequalities | false | 736,713 |
Example 2 If $a>0, b>0, c>0$, compare $a^{3}+b^{3}+c^{3}$ and $a^{2} b+b^{2} c+c^{2} a$. | Analysis: Although the problem does not specify the size relationship of the three numbers, the expressions being compared have symmetry. Therefore, we can assume the order of the three numbers without loss of generality.
Solution: Without loss of generality, let $a \geq b \geq c>0$, then $a^{2} \geq b^{2} \geq c^{2}>... | a^{3}+b^{3}+c^{3} \geq a^{2} b+b^{2} c+c^{2} a | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,714 |
Variation: Let $c_{1}, c_{2}, \cdots, c_{n}$ be a permutation of the positive integers $a_{1}, a_{2}, \cdots, a_{n}$, then $\frac{a_{1}}{c_{1}}+\frac{a_{2}}{c_{2}}+\cdots+\frac{a_{n}}{c_{n}} \geq n$. | Proof: Without loss of generality, let $a_{1} \leq a_{2} \leq \cdots \leq a_{n}$, then $\frac{1}{a_{1}} \geq \frac{1}{a_{2}} \geq \cdots \geq \frac{1}{a_{n}}$, and $\frac{a_{1}}{c_{1}}+\frac{a_{2}}{c_{2}}+\cdots+\frac{a_{n}}{c_{n}}$ is a disordered sum, which must be greater than the reverse ordered sum $\frac{a_{1}}{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,716 |
Example 3 In $\triangle A B C$, $a, b, c$ are the sides opposite to $A, B, C$ respectively, then $\frac{a A+b B+c C}{a+b+c}-\frac{\pi}{3}$ (fill in $\left.\geq 、 \leq 、=\right)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Analysis: The numerator $a A+b B+c C$ of the left side of this problem is easily reminiscent of the rearrangement inequality, and when $a \leq b \leq c$, we have $A \leq B \leq C$, so $a A+b B+c C$ is the sum in the same order. If we notice that $\pi=A+B+C$, the problem becomes easy to solve.
Solution: Without loss of... | \geq | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,717 |
1. Let $a, b, c$ be positive numbers, prove using the rearrangement inequality that
$$2\left(a^{3}+b^{3}+c^{3}\right) \geq a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) .$$ | Assume without loss of generality that $a \leq b \leq c$, then $a^{2} \leq b^{2} \leq c^{2}$. By the rearrangement inequality, we have
$$\begin{array}{l}
a^{2} \cdot a+b^{2} \cdot b+c^{2} \cdot c \geq a^{2} \cdot b+b^{2} \cdot c+c^{2} \cdot a \\
a^{2} \cdot a+b^{2} \cdot b+c^{2} \cdot c \geq a^{2} \cdot c+b^{2} \cdot a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,718 |
2. Let $a, b, c$ be positive numbers, then the minimum value of $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$ is $\qquad$ . | Let's assume $a \leq b \leq c$,
then $a+b \leq a+c \leq b+c, \frac{1}{a+b} \geq \frac{1}{a+c} \geq \frac{1}{b+c}$,
so $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$ is a sequence sum,
$$\begin{array}{l}
\therefore \frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a} \geq \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \\
\frac{c}{a+b}+\fr... | \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,719 |
Example 4 Let $a, b, c$ be positive numbers, prove that
$$\frac{c^{2}-a^{2}}{a+b}+\frac{a^{2}-b^{2}}{b+c}+\frac{b^{2}-c^{2}}{c+a} \geq 0$$ | Analysis: This problem requires appropriately transforming the inequality to be proven, making the forms on both sides symmetrical, to create conditions for using the rearrangement inequality.
Proof: The inequality to be proven is equivalent to
$$\frac{c^{2}}{a+b}+\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a} \geq \frac{a^{2}}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,720 |
Variation: Let $a, b, c$ be positive real numbers, prove that:
$$\frac{a^{3}}{b c}+\frac{b^{3}}{a c}+\frac{c^{3}}{b a} \geq a+b+c .$$ | $$\begin{array}{c}
\text { Proof: Without loss of generality, assume } 0<a \leq b \leq c, \therefore a b \leq c a \leq b c, \\
a^{3} \leq b^{3} \leq c^{3}, a^{2} \leq b^{2} \leq c^{2}, \therefore \frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}, \frac{1}{a b} \geq \frac{1}{c a} \geq \frac{1}{b c} \text {. }
\end{array}$$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,721 |
Question 1 (Example 1 by Li Wen, 1963 Mordica Mathematical Competition) Let $a, b, c \in R^{+}$,
prove that $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geqslant \frac{3}{2}$. | Given $a \geqslant b \geqslant c>0$,
then $0<b+c \leqslant c+a \leqslant a+b$, thus we have
$$\frac{1}{b+c} \geqslant \frac{1}{c+a} \geqslant \frac{1}{a+b}$$
Applying the rearrangement inequality to (1) and (2), we get
$$\begin{array}{l}
\frac{c}{a+b}+\frac{b}{c+a}+\frac{a}{b+c} \geqslant \frac{a}{a+b}+\frac{c}{c+a}+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,722 |
Question 2 (Example 3 by Li Wen, 36th IMO Problem)
Let $a, b, c$ be positive real numbers, and satisfy $abc=1$,
prove that $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}$. | Prove $\because a b c=1, \therefore$ the inequality to be proved is equivalent to $\frac{(a b c)^{2}}{a^{3}(b+c)}+\frac{(a b c)^{2}}{b^{3}(c+a)}+\frac{(a b c)^{2}}{c^{3}(a+b)} \geqslant \frac{3}{2}$. That is, $\frac{b c}{a(b+c)} \cdot b c+\frac{a c}{b(c+a)} \cdot a c+\frac{a b}{c(a+b)} \cdot a b \geqslant \frac{3}{2}$.... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,723 |
Question 3 (Example 5 by Li Wen, 2nd "Friendship Forest" Spring Question) Let $a$, $b$, $c$ be positive numbers,
prove that $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{a+b+c}{2}$. | Given $a \geqslant b \geqslant c>0$
(1)
$$\begin{array}{l}
\text { then } a+b \geqslant a+c \geqslant b+c>0, \\
\frac{1}{b+c} \geqslant \frac{1}{a+c} \geqslant \frac{1}{a+b}>0 \\
\frac{a}{b+c} \geqslant \frac{b}{a+c} \geqslant \frac{c}{a+b}>0
\end{array}$$
Applying the rearrangement inequality to (1) and (2), we get
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,724 |
Question 4 (Example 8 by Li Wen, 31st IMO Preliminary Question)
Let $a, b, c, d$ be non-negative real numbers satisfying $a b+b c+c d+d a=1$. Prove that:
$$\frac{a^{3}}{b+c+d}+\frac{b^{3}}{a+c+d}+\frac{c^{3}}{a+b+d}+\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3}$$ | Given $a \geqslant b \geqslant c \geqslant d \geqslant 0$, then
\[ a+b+c \geqslant a+b+d \geqslant a+c+d \geqslant b+c+d > 0, \]
we have
\[ \frac{a^{2}}{b+c+d} \geqslant \frac{b^{2}}{a+c+d} \geqslant \frac{c^{2}}{a+b+d} \geqslant \frac{d^{2}}{a+b+c} \geqslant 0. \]
Let
\[ S = \frac{a^{3}}{b+c+d} + \frac{b^{3}}{a+c+d} +... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,725 |
Question 5 (Example 9 by Li Wen) Let $a_{i}, b_{i} \in \mathbb{R}^{+}(i=1,2, \cdots, n)$, and $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}$,
Prove: $\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}} \geqslant \frac{1}{2} \sum_{i=1}^{n} a_{i}$. | $$\begin{array}{l}
\left(\sum_{i=1}^{n} a_{i}\right)^{2}=\left(\sum_{i=1}^{n} \frac{a_{i}}{\sqrt{a_{i}+b_{i}}} \cdot \sqrt{a_{i}+b_{i}}\right)^{2} \\
\leqslant \sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}} \cdot \sum_{i=1}^{n}\left(a_{i}+b_{i}\right) \\
=\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}} \cdot 2 \sum_{i=1}^{n... | \sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}} \geqslant \frac{1}{2} \sum_{i=1}^{n} a_{i} | Inequalities | proof | Yes | Yes | inequalities | false | 736,726 |
【Example 1】Prove: $1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots+\frac{1}{n!}<\frac{7}{4}(n$ $\left.\in N_{.}\right)$. | 【Analysis】The conventional enlargement and reduction is $\frac{1}{n!}=\frac{1}{n(n-1)(n-2) \cdots 2 \times 1}$
$$<\frac{1}{2 \times 2 \times \cdots 2 \times 1}=\frac{1}{2^{n-1}}(n \geq 3)$$
Then: the left side $<1+\frac{1}{2!}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{n-1}}=2-\frac{1}{2^{n}}$,
it cannot be le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,729 |
【Example 2】Prove that $\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+\cdots+\frac{1}{(2 n+1)^{2}}<\frac{17}{72}$ $\left(n \in N^{+}\right)$ | $$\begin{array}{c}
\text { [Analysis] } \frac{1}{(2 n+1)^{2}}=\frac{1}{(2 n+1)(2 n+1)}\frac{17}{72}$, the inequality chain is broken, the amplification fails.
Reduce the number of terms to be amplified, retain the previous term, and amplify from the second term, we get
Left side $<\frac{1}{3^{2}}+\frac{1}{4}\left(\fr... | \frac{17}{72} | Inequalities | proof | Yes | Yes | inequalities | false | 736,730 |
【Example 3】Given the general term formula of the sequence as $a_{n}=3^{2}-(-2)^{n}$.
(1) Prove that when $k$ is an odd number, $\frac{1}{a_{k}}+\frac{1}{a_{k+1}}<\frac{4}{3^{k+1}}$;
(2) Prove that $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<\frac{1}{2}\left(n \in N^{+}\right)$. | (1) When $n$ is even, we have
$$\begin{array}{l}
\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}\right)+\left(\frac{1}{a_{3}}+\frac{1}{a_{4}}\right)+\cdots+\left(\frac{1}{a_{n-1}}+\frac{1}{a_{n}}\right)0$, then
$$\begin{array}{l}
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,731 |
【Example 4】Prove $\frac{1}{n+1}\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\right.$
$$\left.\frac{1}{2 n-1}\right)>\frac{1}{n}\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2^{n}}\right)(n \geq 2, n \in N) .$$ | 【Analysis】The conventional enlargement and reduction is:
$$\begin{array}{l}
1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2 n-1}>\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}, \\
\quad \text { then } \frac{1}{n+1}\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2 n-1}\right)>\frac{1}{n+1} \\
\left(\frac{1}{2}+\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,732 |
【Example 5】Prove that $\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2 n}<\frac{\sqrt{2}}{2}$ $\left(n \in N_{+}\right)$. | 【Analysis】Using the Cauchy inequality to avoid "loss of control", we have
$$\begin{array}{l}
\left(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2 n}\right)^{2} \\
\leq\left(1^{2}+1^{2}+\cdots+1^{2}\right)\left[\frac{1}{(n+1)^{2}}+\frac{1}{(n+1)^{2}}+\cdots+ \\
\frac{1}{(2 n)^{2}}\right] \\
< n\left[\frac{1... | \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2 n} < \frac{\sqrt{2}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,733 |
Lemma 1 Let $x, y \in \overline{\mathbf{R}}_{-}$, and $x y=1$, then
$$\frac{1}{\sqrt{2 x+1}}+\frac{1}{\sqrt{2 y+1}} \leqslant \frac{2}{\sqrt{3}}$$
Equality holds in (3) if and only if $x=y=1$. | Prove (3) $\Leftrightarrow \sqrt{3}(\sqrt{2 x+1}+$
$$\begin{array}{l}
\sqrt{2 y+1}) \leqslant 2 \sqrt{(2 x+1)(2 y+1)} \\
\Leftrightarrow \quad 3(\sqrt{2 x+1}+\sqrt{2 y+1})^{2} \\
\quad \leqslant 4(2 x+1)(2 y+1) \\
\Leftrightarrow \quad 3 \sqrt{(2 x+1)(2 y+1)} \leqslant x+y+7 \\
\Leftrightarrow \quad 9(2 x+1)(2 y+1) \le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,736 |
Example 1 Let $a, b, c > 0$, and $ab + bc + ca = 1$.
Prove: $\frac{b^{2}+c^{2}}{\sqrt{a^{2}+1}}+\frac{c^{2}+a^{2}}{\sqrt{b^{2}+1}}+\frac{a^{2}+b^{2}}{\sqrt{c^{2}+1}} \geqslant \sqrt{3}$. | Proof: Since $ab + bc + ca = 1$, we have:
$$\begin{array}{l}
\frac{b^{2} + c^{2}}{\sqrt{a^{2} + 1}} = \frac{b^{2} + c^{2}}{\sqrt{a^{2} + ab + bc + ca}} \\
= \frac{2(b^{2} + c^{2})}{2 \sqrt{(a + b)(a + c)}}
\end{array}$$
Since $\frac{(b+c)^{2}}{(a+b) + (a+c)} = \frac{(b+c)^{2}}{2a + b + c}$, we get:
$$\begin{array}{l}
... | \sqrt{3} | Inequalities | proof | Yes | Yes | inequalities | false | 736,737 |
Example 2 Let $a, b, c > 0$, and $ab + bc + ca = 1$. Prove:
$$\sqrt{a^{3}+a}+\sqrt{b^{3}+b}+\sqrt{c^{3}+c} \geqslant 2 \sqrt{a+b+c} .$$ | Proof: Notice that
$$\begin{array}{l}
a^{3}+a=a\left(a^{2}+1\right) \\
=a\left(a^{2}+ab+bc+ca\right) \\
=a(a+b)(a+c).
\end{array}$$
Thus, the original inequality is equivalent to
$$\begin{array}{l}
\sum \sqrt{a(a+b)(a+c)} \\
\quad \geqslant 2 \sqrt{(a+b+c)(ab+bc+ca)} \\
\Leftrightarrow \sum a^{3}+2 \sum[\sqrt{a(a+b)(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,738 |
Example 3 Let $a, b, c > 0$, and $ab + bc + ca = 1$.
Prove: $\sum \frac{1}{\left(1+a^{2}\right)\left(1+b^{2}\right)} \leqslant \frac{1}{8abc} \sum \frac{1}{a+b}$. | Prove: Notice that
$$\begin{array}{l}
\left(1+a^{2}\right)\left(1+b^{2}\right) \\
=\left(a b+b c+c a+a^{2}\right)\left(a b+b c+c a+b^{2}\right) \\
=(a+b)(a+c)(b+a)(b+c) .
\end{array}$$
Therefore, $\sum \frac{1}{\left(1+a^{2}\right)\left(1+b^{2}\right)}$
$$\begin{array}{l}
=\sum \frac{1}{(a+b)(a+c)(b+a)(b+c)} \\
=\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,739 |
Example 4 Let $a, b, c \geqslant 0$, and $ab + bc + ca = 1$. Prove: $\frac{a^{3}}{1+b^{2}}+\frac{b^{3}}{1+c^{2}}+\frac{c^{3}}{1+a^{2}} \geqslant \frac{\sqrt{3}}{4}$. | Proof: Notice that
$$\begin{array}{l}
\frac{a^{3}}{1+b^{2}}=\frac{a^{3}}{a b+b c+c a+b^{2}}=\frac{a^{3}}{(b+a)(b+c)} \\
=\left[\frac{a^{3}}{(b+a)(b+c)}+\frac{b+a}{8}+\frac{b+c}{8}\right]-\frac{b}{4}-\frac{a}{8}-\frac{c}{8}
\end{array}$$
equals $\sqrt[3]{\frac{a^{3}}{64}}-\frac{b}{4}-\frac{a}{8}-\frac{c}{8}=\frac{5 a}{... | \frac{\sqrt{3}}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 736,740 |
Example 6 Let $a, b, c > 0, a + b + c = abc$. Prove:
$$\sum \frac{1}{\sqrt{1+a^{2}}} \leqslant \frac{3}{2} .$$ | Proof: Let $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$.
From $a+b+c=abc$, we get $xy+yz+zx=1$.
Thus, the original inequality is equivalent to
$$\begin{array}{l}
\sum \frac{x}{\sqrt{x^{2}+1}} \leqslant \frac{3}{2} \\
\Leftrightarrow \sum \frac{2 x}{\sqrt{x^{2}+xy+yz+zx}} \leqslant 3 \\
\Leftrightarrow \sum \frac{2 x}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,741 |
Example 1 Given $a, b, c \in \mathrm{R}^{+}$, prove that $\frac{a}{b+c}+$
$$\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2} .$$ | Prove: The inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2} \Leftrightarrow$
$$\begin{array}{l}
(a+b+c)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) \geqslant \frac{9}{2} \Leftrightarrow \\
{[(a+b)+(b+c)+(c+a)] \cdot} \\
\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) \geqslan... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,742 |
Example 2 Let $a, b, c \in \mathrm{R}_{+}, \lambda \geqslant 0$, prove:
$$\frac{\sqrt{a^{2}+\lambda b^{2}}}{a}+\frac{\sqrt{b^{2}+\lambda c^{2}}}{b}+\frac{\sqrt{c^{2}+\lambda a^{2}}}{c} \geqslant$$
$3 \sqrt{1+\lambda}$. (Mathematical Problems 1613, Mathematical Bulletin) | Proof: By the binary Cauchy inequality, we get $(1 + \lambda) \left(1 + \lambda \cdot \frac{b^2}{a^2}\right) \geqslant \left(1 + \lambda \cdot \frac{b}{a}\right)^2$. Taking the square root on both sides, we get $\sqrt{1 + \lambda} \cdot \sqrt{1 + \lambda \cdot \frac{b^2}{a^2}} \geqslant 1 + \lambda \cdot \frac{b}{a}$. ... | 3 \sqrt{1 + \lambda} | Inequalities | proof | Yes | Yes | inequalities | false | 736,743 |
Example 3 Let $x, y, z>0$ and $4x+3y+5z=1$, find the minimum value of $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}$, and indicate the corresponding values of $x, y, z$. (Mathematical Problems 1625, Mathematics Bulletin) | Solution: $\because 4 x+3 y+5 z=(x+y)+2(y+$ $z)+3(z+x)=1$, by Cauchy-Schwarz inequality, we get $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=(4 x+3 y+$
$5z) \left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=[(x+y)+2$ $(y+z)+3(z+x)]$. $\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right) \geqslant(1+\sqrt{2}+\sqrt{... | (1+\sqrt{2}+\sqrt{3})^{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,744 |
Example 3 Given $a, b, c \in \mathbf{R}$ and $a+b+c=1$, prove: $\sqrt{\frac{7}{5}} \leqslant \sqrt{a+b}+\sqrt{2 b+c}+\sqrt{3 c+a} \leqslant$ $\sqrt{\frac{119}{10}}$. (Mathematical Problems 1464, Mathematics Bulletin) | \begin{array}{l}\text { Prove: Let } x=\sqrt{a+b}, y=\sqrt{2 b+c}, z \\ =\sqrt{3 c+a}, \text { then } x, y, z \geqslant 0 \text { and } 5 x^{2}+y^{2}+2 z^{2} \\ =7, \text { by the Cauchy-Schwarz inequality, we have }(x+y+z)^{2}= \\ \left(\frac{1}{\sqrt{5}} \cdot \sqrt{5} x+y+\frac{1}{\sqrt{2}} \cdot \sqrt{2} z\right)^{... | \sqrt{\frac{7}{5}} \leqslant \sqrt{a+b}+\sqrt{2 b+c}+\sqrt{3 c+a} \leqslant \sqrt{\frac{119}{10}} | Inequalities | proof | Yes | Yes | inequalities | false | 736,745 |
Example 4 When point $P$ moves along the line $y=2 x+6$, and point $Q$ moves on the ellipse $\frac{x^{2}}{6}+\frac{y^{2}}{4}=1$, the minimum length of segment $P Q$ is $\qquad$ | Solution: Let $Q\left(x_{0}, y_{0}\right)$ be the point whose distance to the line $y=2 x+6$ is $d$. Since point $Q$ lies on the ellipse $\frac{x^{2}}{6}+\frac{y^{2}}{4}=1$, we have $\frac{x_{0}{ }^{2}}{6}+\frac{y_{0}^{2}}{4}=1$. Also, $|P Q| \geqslant d=\frac{\left|2 x_{0}-y_{0}+6\right|}{\sqrt{5}}$. By the Cauchy-Sch... | \frac{6 \sqrt{5}-2 \sqrt{35}}{5} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,746 |
For example, $1 a, b, c \in \mathrm{R}^{+}, a+b+c=1$. Prove:
$$\sqrt{13 a+1}+\sqrt{13 b+1}+\sqrt{13 c+1} \leqslant 4 \sqrt{3}$$ | Proof: Since $a, b, c \in \mathrm{R}^{+}, a+b+c=1$, by transformation (1),
so $\sqrt{13 a+1}+\sqrt{13 b+1}+\sqrt{13 c+1} \leqslant (13 a+1+13 b+1+13 c+1)^{\frac{1}{2}} \cdot \sqrt{3}=4 \sqrt{3}$
Similarly, we can succinctly prove several inequalities in [1].
(1) $a, b, c, d \in \mathbf{R}^{+}, a+b+c+d=1$.
Prove: $\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,747 |
Example 2 Let $P$ be a point inside $\triangle ABC$, and let $D$, $E$, $F$ be the feet of the perpendiculars from $P$ to $BC$, $CA$, $AB$ respectively. Prove that: $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF} \geqslant \frac{2p^2}{S}$ (22nd MD problem)
where $p$ is the semiperimeter of $\triangle ABC$, and $S$ is th... | $$\begin{aligned}
\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F} & \geqslant \frac{(B C+C A+A B)^{2}}{B C \cdot P D+C A \cdot P E+A B \cdot P F} \\
& =\frac{(2 p)^{2}}{2 S}=\frac{2 p^{2}}{S}
\end{aligned}$$
When and only when $P D=P E=P F$, that is, $P$ is the incenter of $\triangle A B C$, the equality of the origin... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,748 |
Example 4 Given: $a, b, c$ are positive numbers, and $a b c=1$.
Prove: $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+$ $\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2},(M O-36)$ | Proof: Since $a b c=1, \frac{1}{a^{3}(b+c)}=$ $\frac{a^{2} b^{2} c^{2}}{a^{3}(b+c)}=\frac{b^{2} c^{2}}{a(b+c)}$ by transformation (3)
$$\begin{array}{l}
\text { Therefore } \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \\
=\frac{b^{2} c^{2}}{a(b+c)}+\frac{a^{2} c^{2}}{b(c+a)}+\frac{a^{2} b^{2}}{c(a+b)}... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,749 |
Example 5 Given that $a, b$ are positive real numbers, and $\frac{1}{a}+\frac{1}{b}=1$, prove that for every $n \in \mathrm{N}$, we have:
$$(a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1}(1998 \text { year }$$
National High School League Question) | Prove: Since $\frac{1}{a}+\frac{1}{b}=1$, then $a+b=ab$ $(a-1) \cdot(b-1)=1$, and because $\frac{1}{a}+\frac{1}{b} \geqslant 2 \sqrt{\frac{1}{ab}}$, so $ab \geqslant 4$.
Thus: $(a+b)^{n}-a^{n}-b^{n}+1=a^{n} b^{n}-a^{n}-b^{n}+1=\left(a^{n}-1\right) \cdot\left(b^{n}-1\right)=$ $(a-1) \cdot(b-1) \cdot\left(a^{n-1}+a^{n-2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,750 |
Example 1 Let $a, b \in \mathbf{R}_{+}$, and $a+b=2$, prove: $\frac{a^{2}}{2-a}+\frac{b^{2}}{2-b} \geqslant 2$. | Analysis Before using the Cauchy inequality, it is necessary to observe the structural characteristics of the inequality. In this problem, we are looking for the minimum value of $\frac{a^{2}}{2-a}+\frac{b^{2}}{2-b}$, so we need to form the structure $\left(a^{2}+\right.$ $\left.b^{2}\right)\left(c^{2}+d^{2}\right)$. W... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,752 |
Example 2 Let $a, b, c$ all be positive numbers, and $a+b+c=1$, prove:
$$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}+\left(c+\frac{1}{c}\right)^{2} \geqslant \frac{100}{3} .$$ | Prove $\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}+\left(c+\frac{1}{c}\right)^{2}$
$$\begin{array}{l}
=\frac{1}{3}\left(1^{2}+1^{2}+1^{2}\right)\left[\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}+\left(c+\frac{1}{c}\right)^{2}\right] \\
\geqslant \frac{1}{3}\left[1 \cdot\left(a+\frac{1... | \frac{100}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 736,753 |
Example 11 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 the Cauchy-Schwarz inequality, we have
$$25 \times 36=\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right) \geqslant(a x+b y+c z)^{2}=30^{2} .$$
From the equality condition of the above inequality, we get \(a=k x, b=k y, c=k z\).
Thus, \(25=a^{2}+b^{2}+c^{2}=k^{2} x^{2}+k^{2} y^{2}+k^{2} z^{2}\)
$$=k^{2}\le... | \frac{5}{6} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,754 |
Example 12 Given $a, b \in \mathbf{R}$ and $a \sqrt{2-2 b^{2}}+b \sqrt{1-2 a^{2}}=1$, find the value of $2 a^{2}+b^{2}$. | By Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(2 a^{2}+b^{2}\right) \cdot\left[\left(1-b^{2}\right)+\left(1-2 a^{2}\right)\right] \\
=\left[(\sqrt{2} a)^{2}+b^{2}\right] \cdot\left[\left(\sqrt{1-b^{2}}\right)^{2}+\left(\sqrt{1-2 a^{2}}\right)^{2}\right] \\
\geqslant\left(\sqrt{2} a \cdot \sqrt{1-b^{2}}+... | 2 a^{2}+b^{2}=1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,755 |
Example 13 Given real numbers $m, n>0$.
(1) Prove: $\frac{a^{2}}{m}+\frac{b^{2}}{n} \geqslant \frac{(a+b)^{2}}{m+n}$;
(2) Find the minimum value of the function $y=\frac{2}{x}+\frac{9}{1-2 x}\left[x \in\left(0, \frac{1}{2}\right)\right]$. | Proof (1) $\because m, n>0$, applying the Cauchy-Schwarz inequality,
we get $(m+n)\left(\frac{a^{2}}{m}+\frac{b^{2}}{n}\right) \geqslant(a+b)^{2}$.
$$\therefore \frac{a^{2}}{m}+\frac{b^{2}}{n} \geqslant \frac{(a+b)^{2}}{m+n} \text {. }$$
(2) From (1), we know that the function
$$y=\frac{2}{x}+\frac{9}{1-2 x}=\frac{2^{2... | 25 | Inequalities | proof | Yes | Yes | inequalities | false | 736,756 |
Example 14 Let the three sides of $\triangle ABC$ be $a, b, c$ with corresponding altitudes $h_{a}$, $h_{b}$, $h_{c}$, and the radius of the incircle of $\triangle ABC$ be $r=2$. If $h_{a}+h_{b}+h_{c}=18$, find the area of $\triangle ABC$. | Given that $h_{a}+h_{b}+h_{c}=18$, we have $(a+b+c)\left(h_{a}+h_{b}+\right.$ $\left.h_{c}\right)=18(a+b+c)=9(a+b+c) r$.
By the Cauchy-Schwarz inequality, we get
$(a+b+c)\left(h_{a}+h_{b}+h_{c}\right) \geqslant\left(\sqrt{a h_{a}}+\sqrt{b h_{b}}+\sqrt{c h_{c}}\right)^{2}$.
Also, the area of $\triangle A B C$ is $S=\fr... | 12\sqrt{3} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,757 |
Example 15 Given a positive integer $n$ and a positive number $M$, for all arithmetic sequences $a_{1}, a_{2}, a_{3}, \cdots$ satisfying the condition $a_{1}^{2}+$ $a_{n+1}^{2} \leqslant M$, find the maximum value of $S=a_{n+1}+a_{n+2}+\cdots+a_{2 n+1}$. | According to the problem, we can obtain \( S = \frac{(n+1)\left(a_{n+1} + a_{2n+1}\right)}{2} \),
\[
\because a_{1} + a_{2n+1} = 2a_{n+1},
\]
Substituting into the above equation, we get \( S = \frac{(n+1)\left(3a_{n+1} - a_{1}\right)}{2} \).
By the Cauchy-Schwarz inequality, we have
\[
S \leqslant \frac{(n+1)}{2} \sq... | \frac{(n+1)}{2} \sqrt{10M} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,758 |
Example 16 (1) Let three 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 that $a, b, c$ must be the lengths of the three sides of a triangle;
(2) Let $n$ positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the inequality
$$\left(a_{1}^{2}+a_... | Proof (1) From the problem, we get
$$\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)^{2}-2\left(a^{4}+b^{4}+c^{4}\right)>0 \text {, i.e., } \\
(a+b+c)(a+b-c)(a-b+c)(b+c-a)>0,
\end{array}$$
Since \(a, b, c > 0\), at least three of the factors on the left side of the inequality are positive. Since the product of the fou... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,759 |
Example 3 $P$ is a point inside $\triangle A B C$, $x, y, z$ are the distances from $P$ to the three sides $a, b, c$, $R$ is the radius of the circumcircle of $\triangle A B C$, prove that: $\sqrt{x}+\sqrt{y}+\sqrt{z} \leqslant$ $\frac{1}{\sqrt{2 R}} \cdot \sqrt{a^{2}+b^{2}+c^{2}}$. | Let $S$ be the area of $\triangle ABC$, then $a x+b y+c z=2 S=\frac{a b c}{2 R}$. By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{a x} \cdot \sqrt{\frac{1}{a}}+\sqrt{b y} \cdot \sqrt{\frac{1}{b}}+\sqrt{c z} \cdot \sqrt{\frac{1}{c}} \\
\leqslant \sqrt{(\sqrt{a x})^{2}+(\sqrt... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,760 |
Example 4 If $n$ is a positive integer no less than 2, try to prove:
$$\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} .$$ | Prove $\because 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n}$
$$\begin{array}{l}
=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2 n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right) \\
=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}
\end{array}$$
$\therefore$ The inequ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,761 |
Example 5 Let $a, b, c$ all be positive numbers, $a+b+4c^2=1$, find the maximum value of $\sqrt{a}+\sqrt{b}+\sqrt{2}c$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given that $a, b, c$ are all positive numbers,
we have $a+b+4 c^{2}=(\sqrt{a})^{2}+(\sqrt{b})^{2}+(2 c)^{2}$.
By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(a+b+4 c^{2}\right) \cdot\left[1^{2}+1^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}\right] \\
=\left[(\sqrt{a})^{2}+(\sqrt{b})^{2}+(2 c)^{2}\right] \... | \frac{\sqrt{10}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,762 |
Example 6 As shown in Figure 1, in the isosceles right triangle $AOB$ with one leg being 1, a point $P$ is randomly taken within this triangle. Parallel lines to the three sides are drawn through $P$, forming three triangles (shaded areas in the figure) with $P$ as a vertex. Find the minimum value of the sum of the are... | Parse: Taking $O A, O B$ as the $x$-axis and $y$-axis respectively to establish a rectangular coordinate system as shown in Figure 2, the equation of the line $A B$ is $x+y=1$. Let the coordinates of point $P$ be $P\left(x_{p}, y_{P}\right)$, then the sum of the areas of the three triangles sharing $P$ as a common vert... | \frac{1}{6} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,763 |
Example 7 Solve the equation $\sqrt{4 x+3}+2 \sqrt{1-2 x}=\sqrt{15}$. | Solve $15=\left(\sqrt{2} \cdot \sqrt{2 x+\frac{3}{2}}+2 \cdot \sqrt{1-2 x}\right)^{2}$
$$\begin{array}{l}
\leqslant\left[(\sqrt{2})^{2}+2^{2}\right] \cdot\left[\left(\sqrt{2 x+\frac{3}{2}}\right)^{2}+(\sqrt{1-2 x})^{2}\right] \\
\leqslant 6\left(2 x+\frac{3}{2}+1-2 x\right)=6 \times \frac{5}{2}=15
\end{array}$$
The ne... | x=-\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,764 |
Example 8 Solve the system of equations in the set of real numbers
$$\left\{\begin{array}{l}
2 x+3 y+z=13 \\
4 x^{2}+9 y^{2}+z^{2}-2 x+15 y+3 z=82
\end{array}\right.$$ | By adding the left and right sides of the two equations and transforming, we get
$$(2 x)^{2}+(3 y+3)^{2}+(z+2)^{2}=108$$
From the equation $2 x+3 y+z=13$ transformed, we get
$$2 x+(3 y+3)+(z+2)=18,$$
Thus, by the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
18^{2} & =[1 \times(2 x)+1 \times(3 y+3)+1 \times(z+... | x=3, y=1, z=4 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,765 |
Example 9 Given real numbers $a, b, c, d$ satisfy $a+b+c+d=3, a^{2}+2 b^{2}+$ $3 c^{2}+6 d^{2}=5$, try to find the range of values for $a$.
untranslated text remains in its original format and lines. | Given that $b+c+d=3-a, 2 b^{2}+3 c^{2}+6 d^{2}=5-a^{2}$. By the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(2 b^{2}+3 c^{2}+6 d^{2}\right) \cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right) \\
= & {\left[(\sqrt{2} b)^{2}+(\sqrt{3} c)^{2}+(\sqrt{6} d)^{2}\right] } \\
& \cdot\left[\left(\frac{1}{\sqrt... | [1,2] | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,766 |
Example 10 Given positive numbers $x, y, z$ satisfy $x+y+z=x y z$, and the inequality $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \leqslant \lambda$ always holds, find the range of $\lambda$. | Parse $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \leqslant \frac{1}{2 \sqrt{x y}}+\frac{1}{2 \sqrt{y z}}+\frac{1}{2 \sqrt{z x}}$ $=\frac{1}{2}\left(1 \times \sqrt{\frac{z}{x+y+z}}+1 \times \sqrt{\frac{x}{x+y+z}}+1 \times \sqrt{\frac{y}{x+y+z}}\right)$ $\leqslant \frac{1}{2}\left[\left(1^{2}+1^{2}+1^{2}\right)\left(\fra... | \left[\frac{\sqrt{3}}{2},+\infty\right) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,767 |
Example 1 Let $a, b, c$ all be positive numbers, prove:
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{a+b+c}{2}$$ | Proof: According to the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\quad[(b+c)+(c+a)+(a+b)]\left(\frac{a^{2}}{b+c}+ \\
\left.\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\right) \geqslant(a+b+c)^{2} .
\end{array}$$
Thus, $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{a+b+c}{2}$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,768 |
Example 2 Let $a, b, c$ be positive real numbers, and satisfy $abc=1$, prove that: $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}$. | Prove: Since $abc=1$, the original inequality is equivalent to
$$\frac{b^{2} c^{2}}{a b+a c}+\frac{c^{2} a^{2}}{b c+b a}+\frac{a^{2} b^{2}}{a c+b c} \geqslant \frac{3}{2}$$
According to the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\quad[(a b+a c)+(b c+b a)+(a c+b c)] \\
\left(\frac{b^{2} c^{2}}{a b+a c}+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,769 |
Example 3 Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}^{+}$, and $\sum_{i=1}^{n} x_{1}=1$, prove: $\frac{x_{1}^{2}}{1-x_{1}}+\frac{x_{2}^{2}}{1-x_{2}}+\frac{x_{n}^{2}}{1-x_{n}} \geqslant \frac{1}{n-1}$. | $$\begin{aligned}
& {\left[\left(1-x_{1}\right)+\left(1-x_{2}\right)+\cdots+\left(1-x_{n}\right)\right]\left(\frac{x_{1}^{2}}{1-x_{1}}\right.} \\
+ & \left.\frac{x_{2}^{2}}{1-x_{2}}+\cdots+\frac{x_{n}^{2}}{1-x_{n}}\right) \geqslant\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} \\
& \text { Given } \sum_{i=1}^{n} x_{1}=1, \t... | \frac{1}{n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 736,770 |
Example 4 If $0<a_{1}, a_{2}, \cdots, a_{n}<1$, and $\sum_{i=1}^{n} a i=a$, prove: $\frac{a_{1}}{1-a_{1}}+\frac{a_{2}}{1-a_{2}}+\cdots+\frac{a_{n}}{1-a_{n}} \geqslant \frac{n a}{n-a}$. | Prove that the original inequality is equivalent to
$$\begin{array}{c}
\frac{a_{1}}{1-a_{1}}+1+\frac{a_{2}}{1-a_{2}}+1+\cdots+\frac{a_{n}}{1-a_{n}}+1 \geqslant \\
\frac{n a}{n-a}+n \Leftrightarrow \frac{1}{1-a_{1}}+\frac{1}{1-a_{2}}+\cdots+\frac{1}{1-a_{n}} \geqslant \frac{n^{2}}{n-a}
\end{array}$$
According to the Ca... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,771 |
Example 1 (The Third Mathematical Competition of Yerevan, USSR) If $a, b, c \in \mathbf{R}^{+}$, and $a+b+c=1$, prove:
$$a^{2}+b^{2}+c^{2} \geqslant \frac{1}{3}$$ | $$\begin{array}{l}
a^{2}+b^{2}+c^{2}=\frac{a^{2}}{1}+\frac{b^{2}}{1}+\frac{c^{2}}{1} \geqslant \frac{(a+b+c)^{2}}{1+1+1} \\
=\frac{1}{3} \text { (when and only when } a=b=c \text { equality holds). }
\end{array}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,773 |
Example 2 (1984 Leningrad Mathematical Olympiad) If $a, b, c \in \mathbf{R}^{+}$, and $a+b+c=1$, prove that:
$$a^{3} b+b^{3} c+c a^{3} \geqslant a b c .$$ | Prove that the original inequality is equivalent to $\frac{a^{2}}{c}+\frac{b^{2}}{a}+\frac{c^{2}}{b} \geqslant 1$.
Since $a, b, c \in \mathbf{R}^{+}$, and $a+b+c=1$, by the corollary of the Cauchy-Schwarz inequality, we have
$$\frac{a^{2}}{c}+\frac{b^{2}}{a}+\frac{c^{2}}{b} \geqslant \frac{(a+b+c)^{2}}{a+b+c}=a+b+c=1$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,774 |
Example 3 (24th Soviet Mathematical Olympiad) Let $a_{1}$, $a_{2}, \cdots, a_{n}$ be positive numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1$. Prove that:
$$\frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n}^{2}}{a_{n}+a_{1}} \geqslant \frac{1}{2} .$$ | Prove that since $a_{1}, a_{2}, \cdots, a_{n}$ are positive numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1$, by the corollary of the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots \cdots+\frac{a_{n}^{2}}{a_{n}+a_{1}} \\
\geqslant \frac{\left(a_{1}+a_{2... | \frac{1}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,775 |
Example 4 (Mathematical Problem 668 from "Mathematical Bulletin") Let $\mu, \lambda \in \mathbf{R}^{+}, 0<x, y, z<\frac{\lambda}{\mu}$, and $x+y+z=1$. Prove that:
$$\frac{x}{\lambda-\mu x}+\frac{y}{\lambda-\mu y}+\frac{z}{\lambda-\mu z} \geqslant \frac{3}{3 \lambda-\mu} .$$ | Prove that since $\mu, \lambda \in \mathbf{R}^{+}, 0<x, y, z<\frac{\lambda}{\mu}$, and $x + y + z = 1$, by the corollary of the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
x^{2}+y^{2}+z^{2}=\frac{x^{2}}{1}+\frac{y^{2}}{1}+\frac{z^{2}}{1} \\
\geqslant \frac{(x+y+z)^{2}}{1+1+1}=\frac{1}{3}
\end{array}$$
i.e., ... | \frac{3}{3 \lambda-\mu} | Inequalities | proof | Yes | Yes | inequalities | false | 736,776 |
Example 5 (1984 Balkan Mathematical Olympiad) Let $a_{i} \in$ $\mathbf{R}^{+}, i=1,2, \cdots, n, n \geqslant 2, n \in \mathbf{N}$, and $\sum_{i=1}^{n} a_{i}=1$, prove that:
$$\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}} \geqslant \frac{n}{2 n-1}$$ | Prove that because $a_{i} \in \mathbf{R}^{+}, i=1,2, \cdots, n, n \geqslant 2, n \in$ N, and $\sum_{i=1}^{n} a_{i}=1$, so
$$\begin{array}{l}
\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}} \\
=\frac{a_{1}^{2}}{2 a_{1}-a_{1}^{2}}+\frac{a_{2}^{2}}{2 a_{2}-a_{2}^{2}}+\cdots+\frac{a_{n}^{2}}{2 a_{n}-a_{n}^{2}} \\
\geqslant \frac{\lef... | \frac{n}{2 n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 736,777 |
Example 6 Given $\sin \alpha \cos \alpha \sin \beta \cos \beta \neq 0$, prove:
$$\frac{1}{\sin ^{2} \alpha}+\frac{1}{\cos ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta} \geqslant 9 .$$ | $$\begin{array}{l}
\text { Prove that the left side of the equation }=\frac{1}{\sin ^{2} \alpha}+\frac{\sin ^{2} \beta+\cos ^{2} \beta}{\cos ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta} \\
=\frac{1}{\sin ^{2} \alpha}+\frac{1}{\cos ^{2} \alpha \cos ^{2} \beta}+\frac{1}{\cos ^{2} \alpha \sin ^{2} \beta} \\
\geqslant \fra... | 9 | Inequalities | proof | Yes | Yes | inequalities | false | 736,778 |
Example 7 (Problem from the 36th IMO) Let $a, b, c$ be positive numbers and $abc=1$. Prove that:
$$\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}.$$ | $$\begin{array}{l}
=\frac{b^{2} c^{2}}{a^{3} b^{2} c^{2}(b+c)}+\frac{a^{2} c^{2}}{a^{2} b^{3} c^{2}(a+c)}+\frac{a^{2} b^{2}}{a^{2} b^{2} c^{3}(a+b)} \\
=\frac{b^{2} c^{2}}{a b+a c}+\frac{a^{2} c^{2}}{a b+b c}+\frac{a^{2} b^{2}}{a c+b c} \\
\geqslant \frac{(a b+b c+a c)^{2}}{2(a b+b c+a c)} \\
=\frac{1}{2}(a b+b c+a c) ... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,779 |
Example 8 (1990 Japanese Spring Selection Problem) Let $x, y, z > 0$, and $x+y+z=1$, find the minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$. | Since $x, y, z > 0$ and $x + y + z = 1$, we have
$$\frac{1}{x} + \frac{4}{y} + \frac{9}{z} \geqslant \frac{(1+2+3)^{2}}{x+y+z} = 36$$
Equality holds if and only if $\frac{1}{x} = \frac{2}{y} = \frac{3}{z}$, i.e., $x = \frac{1}{6}, y = \frac{1}{3}, z = \frac{1}{2}$. Therefore, the minimum value is 36. | 36 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,780 |
Example 9 (2004 Croatian Mathematical Competition Problem) Prove the inequality $\frac{a^{2}}{(a+b)(a+c)}+\frac{b^{2}}{(b+c)(b+a)}+$ $\frac{c^{2}}{(c+a)(c+b)} \geqslant \frac{3}{4}$ for all positive real numbers $a, b, c$. | Prove the left side of the given inequality:
$$\begin{aligned}
\geqslant & \frac{(a+b+c)^{2}}{(a+b)(a+c)+(b+c)(b+a)+(c+a)(c+b)} \\
= & \frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}+3(ab+bc+ac)} \\
= & \frac{(a+b+c)^{2}}{(a+b+c)^{2}+(ab+bc+ac)} \\
& \because(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc \\
& \geqslant(ab+ac+bc)+2ab+2... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 736,781 |
Example 1 Find the maximum value of the function $y=\frac{\sin x}{2+\cos x}(0<x<\pi)$. | Solution: From the analysis, we have $4 y^{2}=[1 \cdot \sin x+(-y) \cdot \cos x]^{2}$ $\leqslant\left[1^{2}+(-y)^{2}\right]\left(\sin ^{2} x+\cos ^{2} x\right)=1+y^{2}$, which means $y^{2} \leqslant$ $\frac{1}{3}$.
Also, $y>0, \therefore 0<y \leqslant \frac{\sqrt{3}}{3}$. The equality holds when $\tan x=-\sqrt{3}$, i.... | \frac{\sqrt{3}}{3} | Calculus | math-word-problem | Yes | Yes | inequalities | false | 736,782 |
Example 2 Given $a, b, c \in R^{+}$, and $a+b+c=1$, find the maximum value of $\sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1}$. | Solution: By Cauchy-Schwarz inequality, we have ( $1 \cdot \sqrt{4 a+1}+1$.
$$\begin{array}{l}
\sqrt{4 b+1}+1 \cdot \sqrt{4 c+1})^{2} \leqslant\left(1^{2}+1^{2}+1^{2}\right)(4 a+1 \\
+4 b+1+4 c+1)=12(a+b+c)+9=21, \text { when and only when }
\end{array}$$
$a=b=c=\frac{1}{3}$, the equality holds, the maximum value soug... | \sqrt{21} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,783 |
Example 3 Find the maximum value of the function $y=6 \sqrt{x-5}+8 \sqrt{6-x}$. | Analysis: Observing that $(x-5)+(6-x)=1$, we can construct two sets of numbers $\sqrt{x-5}, \sqrt{6-x}; 6,8$, and use the Cauchy-Schwarz inequality to solve.
$$y=6 \sqrt{x-5}+8 \sqrt{6-x} \leqslant(x-5+6-x)^{\frac{1}{2}}$$
- $\left(6^{2}+8^{2}\right)^{\frac{1}{2}}=10, \therefore$ the maximum value sought is 10. | 10 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,784 |
Example 4 Let $a>b>c, n \in N^{+}$, and $\frac{1}{a-b}+\frac{1}{b-c}+$ $\frac{n}{c-a} \geqslant 0$ always holds, find the maximum value of $n$.
untranslated text remains in its original format and line breaks are preserved. | Solution: Transposing terms, we get $\frac{n}{a-c} \leqslant \frac{1}{a-b}+\frac{1}{b-c}, n \leqslant(a-$
c) $\left(\frac{1}{a-b}+\frac{1}{b-c}\right), n \leqslant[(a-b)+(b-c)]\left(\frac{1}{a-b}\right.$ $\left.+\frac{1}{b-c}\right), n \leqslant(1+1)^{2}=4, \therefore$ the maximum value of $n$ is 4. | 4 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,785 |
Example 6 Given $7 \sin \alpha + 24 \cos \alpha = 25$, find the value of $\sin \alpha$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Example 6 Given $7 \sin \alpha + 24 \cos \alpha = 25$, find the value of $\sin \alpha$. | Analysis: Observing that $7.24$ and $25$ form a Pythagorean triple, we square both sides of $7 \sin \alpha + 24 \cos \alpha = 25$, obtaining $25^{2} = (7 \sin \alpha + 24 \cos \alpha)^{2} \leqslant (\sin ^{2} \alpha + \cos ^{2} \alpha)(7^{2} + 24^{2}) = 25^{2}$. According to the condition for equality in the Cauchy-Sch... | \sin \alpha = \frac{7}{25} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,787 |
Example 7 Solve the equation $\sqrt{x-\frac{1}{x}}+\sqrt{\frac{x-1}{x}}=x$. | Analysis: The general method is to eliminate the square roots by squaring both sides twice, which is quite cumbersome. Observing that $x>1$. Construct two sets of numbers $\sqrt{x-\frac{1}{x}}$, $\frac{1}{\sqrt{x}} ; 1, \sqrt{x-1}$, by the Cauchy-Schwarz inequality, $\left(\sqrt{x-\frac{1}{x}}+\right.$ $\left.\sqrt{\fr... | x=\frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,788 |
Example 8 Solve the system of equations in the set of real numbers
$$\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=\frac{9}{4} \\
-8 x+6 y-24 z=39 .
\end{array}\right.$$ | Solution: By the Cauchy-Schwarz inequality, we have $\left(x^{2}+y^{2}+z^{2}\right)\left[(-8)^{2}+6^{2}+(-24)^{2}\right] \geqslant(-8 x+6 y-24 z)^{2}$. Note that the left side of the above equation $=\frac{9}{4} \times(64+36+4 \times 144)=39^{2}$, and $(-8 x+6 y-24 y)^{2}=39^{2}$. By the condition for equality in the C... | x=-\frac{6}{13}, y=\frac{9}{26}, z=-\frac{18}{13} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,789 |
Example 9 Let $a, b, c \in R^{+}$ and $a b c=1$, prove that $\frac{1}{1+2 a}+$ $\frac{1}{1+2 b}+\frac{1}{1+2 c} \geqslant 1$ | Prove: Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, the original inequality is equivalent to $\frac{y}{2 x+y}+\frac{z}{2 y+z}+\frac{x}{2 z+x} \geqslant 1$. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
{[x(2 z+x)+y(2 x+y)+z(2 y+z)] \cdot\left[\frac{x}{2 z+x}+\right.} \\
\left.\frac{y}{2 x+y}+\frac{z}{2 y+z}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,790 |
Example 10 Given $a, b, c \in R$ and $a+b+c=1$. Prove:
$$\sqrt{\frac{7}{5}} \leqslant \sqrt{a+b}+\sqrt{2 b+c}+\sqrt{3 c+a} \leqslant \sqrt{\frac{119}{10}} .$$ | Proof: Let $x=\sqrt{a+b}, y=\sqrt{2b+c}, z=\sqrt{3c+a}$, then $x, y, z \geqslant 0$ and $5x^2 + y^2 + 2z^2 = 7$. By the Cauchy-Schwarz inequality, we have $(x+y+z)^2 = \left(\frac{1}{\sqrt{5}} \sqrt{5} x + y + \frac{1}{\sqrt{2}} \sqrt{2} z\right)^2$ $\leqslant \left(\frac{1}{5} + 1 + \frac{1}{2}\right)\left(5x^2 + y^2 ... | \sqrt{\frac{7}{5}} \leqslant x + y + z \leqslant \sqrt{\frac{119}{10}} | Inequalities | proof | Yes | Yes | inequalities | false | 736,791 |
Example 1 Find the distance from point $P\left(x_{0}, y_{0}\right)$ to the line $l: A x+B y+C=$ $0\left(A^{2}+B^{2} \neq 0\right)$. | Let point $Q(x, y)$ be any point on the line, then the problem is transformed into finding the minimum value of
$$t=\sqrt{\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}}$$
under the condition $A x+B y+C=0$.
(1) can be transformed into
$$\frac{\left(A x-A x_{0}\right)^{2}}{(A t)^{2}}+\frac{\left(B y-B y_{0}\right)^{2... | d=\frac{\left|A x_{0}+B y_{0}+C\right|}{\sqrt{A^{2}+B^{2}}} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,792 |
Example 2 Given that the line $l$ passing through point $P(1,2)$ intersects the positive half-axes of the $x$-axis and $y$-axis at points $A, B$ respectively, find the minimum value of $|O A|+|O B|$. | Let the equation of line $l$ be
$$\frac{x}{a}+\frac{y}{b}=1(a>0, b>0),$$
then $\frac{1}{a}+\frac{2}{b}=1$, i.e., $\frac{1^{2}}{a}+\frac{(\sqrt{2})^{2}}{b}=1$.
By Theorem 1, we get
$$|O A|+|O B|=a+b \geqslant(1+\sqrt{2})^{2}=3+2 \sqrt{2}$$ | 3+2\sqrt{2} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,793 |
Example 3 Given $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, find the maximum value of $\frac{x+y-1}{y+2}$.
untranslated text is retained in its original format and directly output the translation result. | Let $t=\frac{x+y-1}{y+2}$, then $x+y-t y=2 t+1$. The original
equation can be transformed into
$$\frac{x^{2}}{9}+\frac{(1-t)^{2} y^{2}}{4(1-t)^{2}}=1$$
By Theorem 1, we get
$$9+4(1-t)^{2} \geqslant(x+y-t y)^{2}=(2 t+1)^{2}$$
Solving this, we get $t \leqslant 1$, hence the maximum value of $\frac{x+y-1}{y+2}$ is 1. | 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,794 |
Example 4 Find the maximum of $y=\sin ^{2} x+2 \sin x \cos x+3 \cos ^{2} x$. | Solve: The original function can be transformed into $y=2+2 \sin 2 x+\cos 2 x$. Since $\sin ^{2} 2 x+\cos ^{2} 2 x=1$, by the corollary, we get $2 \geqslant(y-2)^{2}$, so $|y-2| \leqslant \sqrt{2}$, thus $y_{\max }=2+\sqrt{2}, y_{\min }=2-\sqrt{2}$. | y_{\max }=2+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,795 |
Example 5 Given $a, \beta \in\left(0, \frac{\pi}{2}\right)$, and $\cos \alpha+\cos \beta-$ $\cos (\alpha+\beta)=\frac{3}{2}$. Find the values of $\alpha, \beta$. | Solving the original condition can be transformed into
$$\sin \alpha \sin \beta+(1-\cos \alpha) \cos \beta=\frac{3}{2}-\cos \alpha$$
From $\cos ^{2} \beta+\sin ^{2} \beta=1$, we can get
$$\frac{\sin ^{2} \alpha \sin ^{2} \beta}{\sin ^{2} \alpha}+\frac{(1-\cos \alpha)^{2} \cos ^{2} \beta}{(1-\cos \alpha)^{2}}=1,$$
The... | \alpha=\frac{\pi}{3}, \beta=\frac{\pi}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,796 |
Example 6 Solve the system of equations
$$\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=\frac{9}{4} \\
-8 x+6 y-24 z=39
\end{array}\right.$$ | From (1), we get $x^{2}+z^{2}=\frac{9}{4}-y^{2}$, which can be transformed into
$$\frac{(8 x)^{2}}{8^{2}\left(\frac{9}{4}-y^{2}\right)}+\frac{(24 z)^{2}}{24^{2}\left(\frac{9}{4}-y^{2}\right)}=1$$
By Theorem 1, we have
$$64\left(\frac{9}{4}-y^{2}\right)+576\left(\frac{9}{4}-y^{2}\right) \geqslant(6 y-39)^{2},$$
Simpli... | x=-\frac{6}{13}, z=-\frac{8}{13}, y=\frac{9}{26} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,797 |
Example 7 If $a, b \in \mathbf{R}$, and $a \sqrt{1-b^{2}}+b \sqrt{1-a^{2}}$ $=1$. Prove: $a^{2}+b^{2}=1$. | Let $a^{2}+b^{2}=t$, then we have
$$\frac{\left(a \sqrt{1-b^{2}}\right)^{2}}{t\left(\sqrt{1-b^{2}}\right)^{2}}+\frac{\left(b \sqrt{1-a^{2}}\right)^{2}}{t\left(\sqrt{1-a^{2}}\right)^{2}}=1$$
By Theorem 1, we get
$$t\left(1-b^{2}+1-a^{2}\right) \geqslant\left(a \sqrt{1-b^{2}}+b \sqrt{1-a^{2}}\right)^{2}=1,$$
Thus, $t^{... | a^{2}+b^{2}=1 | Algebra | proof | Yes | Yes | inequalities | false | 736,798 |
Example 8 (Question from the 7th Canadian Junior High School Mathematics Competition) Determine the largest real number $z$ such that the real numbers $x, y$ satisfy:
$$\left\{\begin{array}{l}
x+y+z=5 \\
x y+y z+x z=3 .
\end{array}\right.$$ | Solve: From $x+y+z=5$, we get
$$x^{2}+y^{2}+z^{2}+2(xy+yz+xz)=25$$
Also, $xy+yz+xz=3$, substituting into (1) we get
$$x^{2}+y^{2}+z^{2}=19 \text{.}$$
Thus, $x^{2}+y^{2}=19-z^{2}$ can be seen as a circle in terms of $x, y$. By the corollary, we get
$$2(19-z^{2}) \geqslant (5-z)^{2}.$$
Solving this, we get $3z^{2}-10z... | \frac{13}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,799 |
Example 9 (1986 National High School Mathematics League Test Question) If $a$, $b, c \in \mathbf{R}$, and $a, b, c$ satisfy the system of equations
$$\left\{\begin{array}{l}
a^{2}-b c-8 a+7=0 \\
b^{2}+c^{2}+b c-6 a+6=0
\end{array}\right.$$
Try to find the range of values for $a$. | Solve: From (1) + (2), we get $b^{2}+c^{2}=-a^{2}+14 a-13$. From (2) - (1), we get $b+c= \pm(a-1)$, so by the corollary, $2(-$ $\left.a^{2}+14 a-13\right) \geqslant[ \pm(a-1)]^{2}$, solving this yields $1 \leqslant a \leqslant 9$, hence the range of values for $a$ is $1 \leqslant a \leqslant 9$. | 1 \leqslant a \leqslant 9 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,800 |
Example 2 Let $a, b, c \in \mathrm{R}_{+}, \lambda \geqslant 0$, prove that
$$\frac{\sqrt{a^{2}+\lambda b^{2}}}{a}+\frac{\sqrt{b^{2}+\lambda c^{2}}}{b}+\frac{\sqrt{c^{2}+\lambda a^{2}}}{c} \geqslant$$
$3 \sqrt{1+\lambda}$. (Mathematical Problems 1613, Mathematical Bulletin) | Proof: By the binary Cauchy inequality, we get $(1+\lambda) \left(1+\lambda \cdot \frac{b^{2}}{a^{2}}\right) \geqslant \left(1+\lambda \cdot \frac{b}{a}\right)^{2}$. Taking the square root on both sides, we get $\sqrt{1+\lambda} \cdot \sqrt{1+\lambda \cdot \frac{b^{2}}{a^{2}}} \geqslant 1+\lambda \cdot \frac{b}{a}$. Si... | 3 \sqrt{1+\lambda} | Inequalities | proof | Yes | Yes | inequalities | false | 736,801 |
Example 1 Given $a, b, c \in \mathbf{R}^{+}$, prove:
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{a+b+c}{2} .$$ | Analysis: Based on the structural characteristics of the problem, we can use the Cauchy-Schwarz inequality to explore. Thinking of the larger side as the "sum of squares and products," we can consider $\frac{a^{2}}{b+c} + \frac{b^{2}}{c+a} + \frac{c^{2}}{a+b}$ as a "sum of squares." Therefore, we need another "sum of s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,802 |
Example 2 Given $x, y, z \in \mathbf{R}$, and satisfying $x+y+z=2$, find the minimum value of $T=x^{2}-2 x+2 y^{2}+3 z^{2}$. | Parse $T+1=(x-1)^{2}+2 y^{2}+3 z^{2}$ as a "sum of squares," and then multiply it by another "sum of squares," which results in the form of the product of "sum of squares" on the left side of the Cauchy-Schwarz inequality. Thus, $\frac{11}{6}(T+1)$
$$\begin{array}{l}
=\left[1^{2}+\left(\frac{\sqrt{2}}{2}\right)^{2}+\le... | -\frac{5}{11} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,803 |
Example 11 Let $a b c \neq 0$, if $(a+2 b+3 c)^{2}=14\left(a^{2}+\right.$ $b^{2}+c^{2}$ ), then the value of $\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}$ is $\qquad$. | Considering the Cauchy-Schwarz inequality, we have
$$\left(1^{2}+2^{2}+3^{2}\right)\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+2 b+3 c)^{2}$$
The equality "=" holds when $a: b: c=1: 2: 3$, hence
$$\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=8 \text {. }$$ | 8 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,804 |
Example 1 Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x$ $\geqslant 0$, $f(x)=x^{2}$. If for any $x \in[t, t+2]$, the inequality $f(x+t) \geqslant 2 f(x)$ always holds, then the range of the real number $t$ is ( ).
A. $[\sqrt{2},+\infty)$
B. $[2,+\infty)$
C. $(0,2]$
D. $[-\sqrt{2},-1] \cup[\sqrt{2}... | Solution: According to the problem, the function is increasing on $\mathbf{R}$, and the inequality $f(x+t) \geqslant 2 f(x)$ can be transformed into $f(x+t) \geqslant f(\sqrt{2} x)$,
so $x+t \geqslant \sqrt{2} x$, which means $t \geqslant(\sqrt{2}-1) x$.
Since the function $g(x)=(\sqrt{2}-1) x$ is increasing on $[t, t+... | A | Algebra | MCQ | Yes | Yes | inequalities | false | 736,805 |
Example 2 If $x \in(-\infty, 1]$ , the inequality $\left(a-a^{2}\right) 4^{x}+$ $2^{x}+1>0$ always holds, then the range of real number $a$ is $(\quad)$.
A. $\left(-\infty, \frac{1}{4}\right)$
B. $\left(-\frac{1}{2}, \frac{3}{2}\right)$
C. $\left(-2, \frac{1}{4}\right)$
D. $(-\infty, 6)$ | Solve: From $\left(a-a^{2}\right) 4^{x}+2^{x}+1>0$, separating the parameter yields $a^{2}-a<\left(\frac{1}{2}\right)^{x}+\left(\frac{1}{4}\right)^{x}$. | not found | Inequalities | MCQ | Yes | Yes | inequalities | false | 736,806 |
Example 12 If $a, b \in \mathbf{R}$, and $a \sqrt{1-b^{2}}+b \sqrt{1-a^{2}}=$ 1 , prove: $a^{2}+b^{2}=1$. | Parse the construction of "sum of products" and "product of sums" from the conditions,
$$\begin{array}{l}
\quad 1=\left(a \sqrt{1-b^{2}}+b \sqrt{1-a^{2}}\right)^{2} \\
\quad \leqslant\left(a^{2}+b^{2}\right)\left(1-b^{2}+1-a^{2}\right)=\left(a^{2}+b^{2}\right)[2- \\
\left.\left(a^{2}+b^{2}\right)\right]
\end{array}$$
... | a^{2}+b^{2}=1 | Algebra | proof | Yes | Yes | inequalities | false | 736,807 |
Example 13 Let the three sides of $\triangle ABC$ be $a, b, c$ with corresponding altitudes $h_{a}$, $h_{b}$, $h_{c}$, and the radius of the incircle of $\triangle ABC$ be $r=2$. If $h_{a}+h_{b}+h_{c}=$ 18, find the area of $\triangle ABC$. | Given that $h_{a}+h_{b}+h_{c}=18$, we have $(a+b+c)$ $\left(h_{a}+h_{b}+h_{c}\right)=18(a+b+c)=9(a+b+c) r$.
By the Cauchy-Schwarz inequality, we get $(a+b+c)\left(h_{a}+h_{b}+h_{c}\right) \geqslant$ $\left(\sqrt{a h_{a}}+\sqrt{b h_{b}}+\sqrt{c h_{c}}\right)^{2}$.
Also, the area of $\triangle A B C$ is $S=\frac{1}{2} ... | 12 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,808 |
Example 3 Given that the line $l$ passing through point $P(1,2)$ intersects the positive half-axes of the $x$-axis and $y$-axis at points $A$ and $B$ respectively, find the minimum value of $|O A|+|O B|$. | Let the equation of line $l$ be $\frac{x}{a}+\frac{y}{b}=1(a>0$, $b>0)$, then $\frac{1}{a}+\frac{2}{b}=1$. By associating with the Cauchy-Schwarz inequality to construct the product of sums, we know that $|O A|+|O B|=a+b=(a+b)\left(\frac{1}{a}+\frac{2}{b}\right)$ $\geqslant(1+\sqrt{2})^{2}=3+2 \sqrt{2}$. Therefore, the... | 3+2\sqrt{2} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,809 |
Example 4 Given $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, find the minimum value of $\frac{x+y-1}{y+2}$.
untranslated text is retained in its original format and directly output the translation result. | Parse $\frac{\text { Let } \frac{x+y-1}{y \pm 2}}{y}=t$, rearranging gives $x-y-t y=2 t +1$. The given equation is $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$. Using the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
9+4(1-t)^{2}=\left[9+4(1-t)^{2}\right]\left[\frac{x^{2}}{9}+\frac{y^{2}}{4}\right] \\
\geqq \underline{(... | 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,810 |
Example 6 Given a positive integer $n$ and a positive number $M$, for all arithmetic sequences $a_{1}, a_{2}, a_{3}, \cdots$ satisfying the condition $a_{1}^{2} + a_{n+1}^{2} \leqslant M$, find the maximum value of $S=a_{n+1} + a_{n+2}+\cdots+a_{2 n+1}$. | Given the problem, $S=\frac{1}{2}(n+1)\left(a_{n+1}+a_{2 n+1}\right)$.
$\because a_{1}+a_{2 n+1}=2 a_{n+1}$, i.e., $a_{2 n+1}=2 a_{n+1}-a_{1}$,
$$\therefore S=\frac{1}{2}(n+1)\left(3 a_{n+1}-a_{1}\right) \text {. }$$
By the Cauchy-Schwarz inequality,
$$\begin{array}{l}
S \leqslant \frac{1}{2}(n+1) \cdot \sqrt{\left[3^... | \frac{1}{2}(n+1) \sqrt{10 M} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,812 |
Example 7 Given $x \in\left(0, \frac{\pi}{2}\right)$, try to find the maximum value of the function $f(x)=3 \cos x + 4 \sqrt{1+\sin ^{2} x}$, and the corresponding $x$ value. | Given $x \in\left(0, \frac{\pi}{2}\right), \cos x>0$, it is easy to see that $f(x)>0$.
$$\begin{array}{l}
\therefore f^{2}(x)=\left(3 \cos x+4 \sqrt{1+\sin ^{2} x}\right)^{2} \\
\leqslant\left(3^{2}+4^{2}\right)\left[\cos ^{2} x+\left(1+\sin ^{2} x\right)\right]=50 . \\
\therefore f(x) \leqslant 5 \sqrt{2}
\end{array}$... | 5 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,813 |
Example 8 Suppose the equation $\sin ^{2} \theta+\sin 2 \theta=2 \cos ^{2} \theta+m$ has real solutions, try to find the range of real number $m$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The note about translati... | The original equation can be transformed into $2 \sin 2 \theta-3 \cos 2 \theta=2 m+1$. By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
(2 m+1)^{2}=(2 \sin 2 \theta-3 \cos 2 \theta)^{2} \\
\leqslant\left[2^{2}+(-3)^{2}\right]\left(\sin ^{2} 2 \theta+\cos ^{2} 2 \theta\right)=13
\end{array}$$
Solving this, ... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,814 |
Example 10 If $\cos \alpha+2 \sin \alpha=-\sqrt{5}$, then $\tan \alpha=$ （）.
A. $\frac{1}{2}$
B. 2
C. $-\frac{1}{2}$
D. -2 | From the given conditions and Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
5=(1 \times \cos \alpha+2 \times \sin \alpha)^{2} \\
\leqslant\left(1^{2}+2^{2}\right)\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)=5
\end{array}$$
Equality holds if and only if $\frac{\cos \alpha}{1}=\frac{\sin \alpha}{2}$, i.e., $\t... | B | Algebra | MCQ | Yes | Yes | inequalities | false | 736,816 |
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