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Example 1 The maximum value of the function $y=3 \sqrt{x-1}+4 \sqrt{5-x}$ is $\square$ $\qquad$ . | According to the problem, $y>0$ and $x \in[1,5]$, so by the Cauchy-Schwarz inequality, we have: $\square$
$$y^{2}=(3 \sqrt{x-1}+4 \sqrt{5-x})^{2} \leqslant\left(3^{2}+\right.$$
$\left.4^{2}\right)\left[(\sqrt{x-1})^{2}+(\sqrt{5-x})\right]=100$, equality holds if and only if $\frac{\sqrt{x-1}}{3}=\frac{\sqrt{5-x}}{4}$, ... | 10 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,817 |
Variation 1. The maximum value of the function $y=3 \sqrt{x-1}+\sqrt{35-7 x}$ is $\qquad$ . | According to the problem, we know that $y=3 \sqrt{x-1}+\sqrt{7} \cdot \sqrt{5-x}>$ 0 and $x \in[1,5]$, so by the Cauchy-Schwarz inequality we get:
$$\begin{array}{c}
y^{2}=[3 \sqrt{x-1}+\sqrt{7} \cdot \sqrt{5-x}]^{2} \leqslant\left[3^{2}+\right. \\
\left.(\sqrt{7})^{2}\right]\left[(\sqrt{x-1})^{2}+(\sqrt{5-x})^{2}\righ... | 8 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,818 |
Variant 2. (2008 National Exam I for Science) If the line $\frac{x}{a}+\frac{y}{b}$ $=1$ passes through the point $M(\cos \alpha, \sin \alpha)$, then $(\quad)$
A. $a^{2}+b^{2} \leqslant 1$
B. $a^{2}+b^{2} \geqslant 1$
C. $\frac{1}{a^{2}}+\frac{1}{b^{2}} \leqslant 1$
D. $\frac{1}{a^{2}}+\frac{1}{b^{2}} \geqslant 1$ | Parse: Substituting the coordinates of point $M$ into the line $\frac{x}{a}+\frac{y}{b}=1$ yields:
$$\begin{aligned}
& 1=\left(\frac{\cos \alpha}{a}+\frac{\sin \alpha}{b}\right)^{2}=\left(\cos \alpha \cdot \frac{1}{a}+\sin \alpha \cdot \frac{1}{b}\right)^{2} \\
\leqslant & \left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)... | D | Geometry | MCQ | Yes | Yes | inequalities | false | 736,819 |
Example 2 (2007 Guangzhou First Mock Exam for Science) Given $a, b>0$, and $a+b=1$, then the minimum value of $1 / 2a + 1 / b$ is $\qquad$ | Given: $\because a, b>0$, and $a+b=1$, by Cauchy-Schwarz inequality we know:
$$\frac{1}{2 a}+\frac{1}{b}=\frac{(\sqrt{2} / 2)^{2}}{a}+\frac{1^{2}}{b} \geqslant \frac{(\sqrt{2} / 2)^{2}}{a+b}=\frac{3}{2}+\sqrt{2} \text {, }$$
with equality if and only if $\frac{\sqrt{2} / 2}{a}=\frac{1}{b}$, i.e., $a=\sqrt{2}-1, b=2-\s... | \frac{3}{2}+\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,820 |
Variation 1. (Guangzhou, 2009 $\in \boldsymbol{R}$, and $a+b+c=2, a^{2}+2 b^{2}+3 c^{2}=4$, then the range of values for $a$ is $\qquad$ . | According to the problem, we have: $b+c=2-a, 2 b^{2}+3 c^{2}=4-$ $a^{2}$. By the Cauchy-Schwarz inequality, we get: $4-a^{2}=2 b^{2}+3 c^{2}=\frac{b^{2}}{1 / 2}+\frac{c^{2}}{1 / 3}$ $\geqslant \frac{(b+c)^{2}}{1 / 2+1 / 3}=\frac{6}{5}(2-a)^{2}$, so $5\left(4-a^{2}\right) \geqslant 6(2-$ $a)^{2}$,
Simplifying, we get: ... | [2 / 11, 2] | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,821 |
Variation 2 (Excerpt from the 2008 National Exam II, Science Question 21) Let the center of the ellipse be at the origin of the coordinate system, with $A(2,0), B(0,1)$ being two of its vertices. The line $y=k x(k>0)$ intersects $A B$ at point $D$, and intersects the ellipse at points $E$ and $F$. Find the maximum valu... | For the quadrilateral $A O B F$ doubled, let point $F(x, y)(x>0)$, so the area of quadrilateral $A E B F$ is $S=2\left(S_{\triangle O B F}+S_{\triangle O A F}\right)=x+2 y$. Also, since $1=\frac{x^{2}}{4}+y^{2}=\frac{x^{2}}{4}+\frac{(2 y)^{2}}{4} \geqslant \frac{(x+2 y)^{2}}{4+4}=\frac{(x+2 y)^{2}}{8}$, thus $x+2 y \le... | 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,822 |
Lemma 2 Let $x_{i}>0, i=1,2, \cdots, n$, then when $0<p<1, n>1$ or $p>1, n>1$, we have $\sum_{i=1}^{n} x_{i}^{p}<\left(\sum_{i=1}^{n} x_{i}\right)^{p}$. | Prove that when $p=1$, (6) obviously holds. Next, we prove that when $0<p<1$, we have $p-1<0$. By the given conditions $x_{i}>0, i=1,2, \cdots, n, n>1$, we get
$$\sum_{i=1}^{n} x_{i}^{p}=\left(\sum_{i=1}^{n} x_{i}\right)^{p}-\sum_{i=1}^{n} x_{i}\left[\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{p-1}-x_{i}^{p-1}\right]<\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,825 |
Theorem 2 Let $a_{i}, b_{i} \geqslant 0(i=1,2, \cdots, n)$, then $\sum_{i=1}^{n} a_{i} b_{i} \leqslant\left(\sum_{i=1}^{n} a_{i}^{\frac{3}{2}}\right)^{\frac{2}{3}}\left(\sum_{i=1}^{n} b_{i}^{3}\right)^{\frac{1}{3}}$. | Prove that in the inequality $a b \leqslant \frac{2}{3} a^{\frac{3}{2}}+\frac{1}{3} b^{3}$, taking $a=\frac{a_{i}}{\left(\sum_{i=1}^{n} a_{i}^{\frac{3}{2}}\right)^{\frac{2}{3}}}, b=\frac{b_{i}}{\left(\sum_{i=1}^{n} b_{i}^{3}\right)^{\frac{1}{3}}} \quad(i=1,2, \cdots, n)$,
we obtain
$$\frac{a_{i} b_{i}}{\left(\sum_{i=1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,829 |
Theorem 7 (Young's Inequality) (1) For any $a, b \geqslant 0$, rational numbers $0 < r < 1$, the Young's inequality holds: $x y \leqslant \frac{1}{r} x^{r}+\left(1-\frac{1}{r}\right) y^{\frac{r}{r-1}}$;
(3) For any $n(n \geqslant 2)$ non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$, rational numbers $p_{1}, p_{2... | Prove (1) There exist positive integers $m, n$, such that $1 \leqslant m<n, s=\frac{m}{n}$; using the arithmetic mean-geometric mean inequality, we get $a^{\frac{m}{n}} b^{\frac{n-m}{n}} \leqslant \frac{m}{n} a+\frac{n-m}{n} b$, i.e.,
$$a^{\prime} b^{1-1} \leqslant s a+(1-s) b$$
(2) There exist positive integers $m, n$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,830 |
Theorem 12 ${ }^{[2,4]}$ Let positive integer $n \geqslant 2$, real numbers $a_{i} \geqslant 0(i=1,2, \cdots, n)$, then we have
(1) $\left(1+\left(a_{1} a_{2} \cdots a_{n}\right)^{\frac{1}{n}}\right)^{n} \leqslant\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)$ (Chrystal Inequality);
(2) $\left(1+a_... | Prove (1) Directly using the corollary of Hölder's inequality, Theorem 11, we get
$$\left(1+\left(a_{1} a_{2} \cdots a_{n}\right)^{\frac{1}{n}}\right)=1 \cdot 1 \cdots \cdot 1+\left(a_{1}^{\frac{1}{n}}\right)\left(a_{2}^{\frac{1}{n}}\right) \cdots\left(a_{n}^{\frac{1}{n}}\right) \leqslant\left(1+a_{1}\right)^{\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,835 |
$\begin{array}{c}\text { Theorem } 13^{[2,4]} \text { (Newman's Inequality) Let } n \geqslant 2 \text { be a positive integer, real numbers } a_{i}>0(i=1,2, \cdots, n), \text { and } \sum_{i=1}^{n} a_{i}=1, \text { then } \\ \prod_{i=1}^{n}\left(\frac{1}{a_{i}}-1\right) \geqslant(n-1)^{n} .\end{array}$ | Prove that
$$\frac{1}{a_{i}}-1=\frac{\sum_{k=1}^{n} a_{k}}{a_{i}}-1=\frac{\sum_{k \neq i} a_{k}}{a_{i}} \geqslant \frac{(n-1)\left(\prod_{k \neq i} a_{k}\right)^{\frac{1}{(n-1)}}}{a_{i}}$$
Therefore,
$$\prod_{i=1}^{n}\left(\frac{1}{a_{i}}-1\right) \geqslant(n-1)^{n}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,836 |
Theorem 14 $1^{[2,4]}$ (Klamkin Inequality) Let $n \geqslant 2$ be a positive integer, and let $a_{i}>0(i=1,2, \cdots, n)$ be real numbers such that $\sum_{i=1}^{n} a_{i}=1$. Then we have (1) $\prod_{i=1}^{n}\left(1+\frac{1}{a_{i}}\right) \geqslant(n+1)^{n} ;\left(2 \prod_{i=1}^{n}\left(\frac{1+a_{i}}{1-a_{i}}\right) \... | Prove (1) Using Chrystal's inequality in Theorem 12, we get
Since
$$\begin{array}{c}
\prod_{i=1}^{n}\left(1+\frac{1}{a_{i}}\right) \geqslant\left(1+\left(\prod_{i=1}^{n} \frac{1}{a_{i}}\right)^{\frac{1}{n}}\right)^{n}, \\
\left(\prod_{i=1}^{n} \frac{1}{a_{i}}\right)^{\frac{1}{n}} \geqslant \frac{n}{\sum_{i=1}^{n} a_{i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,837 |
Example 1 Let $x, y \geqslant 0$, and $x^{3}+y^{3}=1$, find the range of $x+y$.
| Using the result of Theorem 2, we get $x+y \leqslant\left(x^{3}+y^{3}\right)^{\frac{1}{3}}(1+1)^{\frac{2}{3}}=2^{\frac{2}{3}}$, and by $\left(x^{3}+y^{3}\right)^{\frac{1}{3}} \leqslant x+y$, we obtain $1 \leqslant x+y \leqslant 2^{\frac{2}{3}}$. | 1 \leqslant x+y \leqslant 2^{\frac{2}{3}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,844 |
Example $2^{[2]}$ Let $0 \leqslant x \leqslant \frac{\pi}{2}$, find the range of $(\sin x)^{3}+(\cos x)^{3}$.
| Solve: From
$$1=(\sin x)^{2}+(\cos x)^{2} \leqslant\left[(\sin x)^{3}+(\cos x)^{3}\right]^{\frac{2}{3}}(1+1)^{\frac{1}{3}},$$
we get
Then from this, we have
$$\begin{array}{c}
\frac{1}{\sqrt{2}} \leqslant(\sin x)^{3}+(\cos x)^{3} \\
{\left[(\sin x)^{3}+(\cos x)^{3}\right]^{\frac{1}{3}} \leqslant\left[(\sin x)^{2}+(\c... | \frac{1}{\sqrt{2}} \leqslant (\sin x)^{3} + (\cos x)^{3} \leqslant 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,845 |
Example $5^{(2,4]}$ Let $a_{1}, a_{2}>0$, and $a_{1}+a_{2}=1$, then it holds that $\frac{1+a_{1}}{1-a_{1}} \cdot \frac{1+a_{2}}{1-a_{2}} \geqslant 3^{2}$. | Prove that by using the condition and the three-variable geometric-arithmetic mean inequality, we get
$$\frac{1+a_{1}}{1-a_{1}} \cdot \frac{1+a_{2}}{1-a_{2}}=\frac{a_{1}+a_{2}+a_{1}}{a_{2}} \cdot \frac{a_{1}+a_{2}+a_{2}}{a_{1}} \geqslant \frac{3 \sqrt[3]{a_{1}^{2} a_{2}} \cdot 3 \sqrt[3]{a_{1} a_{2}^{2}}}{a_{1} a_{2}}=... | 3^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,846 |
Example $6^{[12.4]}$ Let $a_{1}, a_{2}, a_{3}>0$, and $a_{1}+a_{2}+a_{3}=1$, then it holds that $\frac{1+a_{1}}{1-a_{1}} \cdot \frac{1+a_{2}}{1-a_{2}} \cdot \frac{1+a_{3}}{1-a_{3}} \geqslant 2^{3}$. | Prove that
$$\frac{1+a_{1}}{1-a_{1}} \cdot \frac{1+a_{2}}{1-a_{2}} \cdot \frac{1+a_{3}}{1-a_{3}}=\frac{\left(a_{1}+a_{2}\right)+\left(a_{1}+a_{3}\right)}{a_{2}+a_{3}} \cdot \frac{\left(a_{1}+a_{2}\right)+\left(a_{2}+a_{3}\right)}{a_{1}+a_{3}} \cdot \frac{\left(a_{1}+a_{3}\right)+\left(a_{2}+a_{3}\right)}{a_{1}+a_{2}} ... | 2^3 | Inequalities | proof | Yes | Yes | inequalities | false | 736,847 |
Example 2 Let $x, y, z \geqslant 0$, and $x+y+z=1$, prove:
$$0 \leqslant y z+z x+x y-2 x y z \leqslant \frac{7}{27}$$ | By the transformation II of Schur's inequality, we get
$$\left(\sum x\right)^{3}-4 \sum x \sum y z+9 x y z \geqslant 0$$
Given the condition $\sum x=1$, we have
$$\begin{array}{c}
1-4 \sum y z+9 x y z \geqslant 0 \\
\sum y z-2 x y z \leqslant \frac{1}{4}+\frac{1}{4} x y z \\
\leqslant \frac{1}{4}+\frac{1}{4}\left(\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,851 |
Example 3 Let $x, y, z \in \mathbf{R}^{+}$, and $x+y+z=$ $x y z$, prove:
$$x^{2}+y^{2}+z^{2}-2(x y+y z+z x)+9 \geqslant 0$$ | Prove that because $x+y+z=x y z$, (3) is equivalent to
$$\begin{array}{l}
{\left[x^{2}+y^{2}+z^{2}-2(x y+y z+z x)\right](x+y} \\
+z)+9 x y z \geqslant 0 \Leftrightarrow x^{3}+y^{3}+z^{3}-\left(x^{2} y+y^{2} z\right. \\
+\left.z^{2} x+x y^{2}+y z^{2}+z x^{2}\right)+3 x y z \geqslant 0
\end{array}$$
That is, $\sum x^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,852 |
Example 4 Let $a, b, c \in \mathbf{R}^{+}$, prove:
$$\begin{aligned}
& \sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c})+(a+b+c)^{2} \\
\geqslant & 4 \sqrt{3 a b c(a+b+c)} .
\end{aligned}$$ | Prove that by Schur's inequality (in (2), let $r=$ 2)
$$\sum x^{2}(x-y)(x-z) \geqslant 0, x, y, z \in \mathbf{R}^{+},$$
Therefore, $\sum x^{4}+x y z \sum x \geqslant \sum x^{3}(y+z)$.
Also, $\sum x^{3}(y+z)=2 \sum y^{2} z^{2}+\sum y z(y-$ $z)^{2} \geqslant 2 \sum y^{2} z^{2}$
Thus, $\sum x^{4}+x y z \sum x \geqslant ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,853 |
Let $w_{1}, w_{2}, \cdots, w_{n}$ be
positive real numbers, $w_{1}+w_{2}+\cdots+w_{n}=1$, for any positive real number $a_{ij}$ we have
$$\begin{array}{l}
\left(a_{11}+a_{12}+\cdots+a_{1 m}\right)^{w_{1}}\left(a_{21}+a_{22}+\cdots\right. \\
\left.+a_{2 m}\right)^{w_{2}} \cdots\left(a_{n 1}+a_{n 2}+\cdots+a_{n m}\right)... | Let $A_{a}=\sum_{j=1}^{m} a_{a j}(\alpha=1,2, \cdots, n)$, then
(5) is
$$\left(A_{1}{ }^{w_{1}} A_{2}{ }^{w_{2}} \cdots A_{n}{ }^{w_{n}}\right)^{-1} \sum_{j=1}^{m} a_{1 j}{ }^{w_{1}} a_{2 j}{ }^{w_{2}} \cdots a_{n j}{ }^{w_{n}}$$
$\leqslant 1$,
i.e., $\sum_{j=1}^{m}\left(\frac{a_{1 j}}{A_{1}}\right)^{w_{1}}\left(\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,854 |
Example $\mathbf{5}$ Let $a, b \in \mathbf{R}^{+}$.
1) Find the minimum value of $S=\frac{(a+1)^{2}}{b}+\frac{(b+3)^{2}}{a}$;
2) Find the minimum value of $T=\frac{(a+1)^{3}}{b^{2}}+\frac{(b+3)^{3}}{a^{2}}$. | 1) By Cauchy-Schwarz inequality, we have
$$S \cdot (b+a) \geqslant (a+1+b+3)^{2},$$
Thus, $S \geqslant \frac{(a+b+4)^{2}}{a+b} = (a+b) + \frac{16}{a+b} + 8 \geqslant 2 \sqrt{16} + 8 = 16,$
Equality holds when $a=\frac{7}{3}, b=\frac{5}{3}$. Therefore, the minimum value of $S$ is 16.
2) By (8) (Hölder's inequality), w... | 27 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,855 |
Example 6 Let $a, b, c \in \mathbf{R}^{+}$, prove:
$$\begin{array}{c}
\frac{a+b+c}{3} \geqslant \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}} \\
\geqslant \frac{\sqrt{a b}+\sqrt{b c}+\sqrt{c a}}{3}
\end{array}$$ | Prove that by the AM-GM inequality, we have
$$\begin{aligned}
& \frac{(a+b)(b+c)(c+a)}{3} \\
\geqslant & \sqrt[3]{(a+b)(b+c)(c+a)}
\end{aligned}$$
Therefore, $\frac{a+b+c}{3}$
$$\geqslant \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}$$
By (7) (Hölder's inequality), we have
$$\begin{aligned}
& \frac{(a+b)(b+c)(c+a)}{8} \\
= & \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,856 |
Example 7 Let $a, b, c$ be positive real numbers, prove that:
$$\begin{aligned}
& \left(a^{5}-a^{2}+3\right)\left(b^{5}-b^{2}+3\right)\left(c^{5}-c^{2}+3\right) \\
\geqslant & (a+b+c)^{3} .
\end{aligned}$$ | For $x \in \mathbf{R}^{+}, x^{2}-1$ and $x^{3}-1$ have the same sign, so
$$\left(x^{2}-1\right)\left(x^{3}-1\right) \geqslant 0$$
which implies $x^{5}-x^{2}+3 \geqslant x^{3}+2$.
Thus, $\left(a^{5}-a^{2}+3\right)\left(b^{5}-b^{2}+3\right)\left(c^{5}-c^{2}+3\right)$
$$\geqslant\left(a^{3}+2\right)\left(b^{3}+2\right)\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,857 |
Example 8 (Power Mean Inequality) Let $a_{1}, a_{2}, \cdots$, $a_{n}$ be positive real numbers, $\alpha>\beta>0$, then
$$\left(\frac{1}{n} \sum_{i=1}^{n} a_{i}^{\beta}\right)^{\frac{1}{\beta}} \leqslant\left(\frac{1}{n} \sum_{i=1}^{n} a_{i}^{\sigma}\right)^{\frac{1}{\sigma}} .$$ | To prove: In (10) (Hölder's inequality), let $x_{i}=$ $1, i=1,2, \cdots, n$, then we have
$$\sum_{i=1}^{n} y_{i} \leqslant n^{\frac{1}{p}}\left(\sum_{i=1}^{n} y_{i}^{q}\right)^{\frac{1}{q}}$$
Since $\frac{1}{p}=1-\frac{1}{q}$, the above inequality can be written as
$$\frac{1}{n} \sum_{i=1}^{n} y_{i} \leqslant\left(\fr... | \left(\frac{1}{n} \sum_{i=1}^{n} a_{i}^{\beta}\right)^{\frac{1}{\beta}} \leqslant\left(\frac{1}{n} \sum_{i=1}^{n} a_{i}^{\alpha}\right)^{\frac{1}{\alpha}} | Inequalities | proof | Yes | Yes | inequalities | false | 736,858 |
For example, let positive real numbers $x, y, z$ satisfy $x y z \geqslant 1$, prove:
$$\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \geqslant 0$$ | Prove: The original inequality is equivalent to:
$$\begin{array}{l}
\quad\left(\frac{x^{2}-x^{5}}{x^{5}+y^{2}+z^{2}}+1\right)+\left(\frac{y^{2}-y^{5}}{y^{5}+z^{2}+x^{2}}+1\right)+ \\
\left(\frac{z^{2}-z^{5}}{z^{5}+x^{2}+y^{2}}+1\right) \leqslant 3 \\
\quad \text { i.e., } \frac{x^{2}+y^{2}+z^{2}}{x^{5}+y^{2}+z^{2}}+\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,859 |
Example 2 Given $a, b, c>0$, and $\frac{1}{a+b+1}+\frac{1}{b+c+1}$ $+\frac{1}{c+a+1} \geqslant 1$, prove: $a+b+c \geqslant a b+a c+b c$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Prove: Using the local Cauchy inequality, we get: $(a+b+1)$ $\left(a+b+c^{2}\right) \geqslant(a+b+c)^{2}$. Therefore, $\frac{1}{a+b+1} \leqslant$ $\frac{a+b+c^{2}}{(a+b+c)^{2}}$. Similarly, we can get the other two inequalities, so: $\frac{1}{a+b+1}$
$$\begin{array}{l}
+\frac{1}{b+c+1}+\frac{1}{c+a+1} \leqslant \\
\fra... | null | Inequalities | proof | Yes | Yes | inequalities | false | 736,860 |
Note: This problem is from the 2007 Balkan Mathematical Olympiad, and it is almost identical to the 20th inequality in reference [3]: For positive real numbers $x, y, z$ satisfying the relation $x+y+z \geqslant x y + x z + y z$, prove that $\frac{1}{1+x+y}+\frac{1}{1+y+z}+\frac{1}{1+z+x} \leqslant 1$.
For the second p... | Proof: By the local Cauchy inequality, we have: \(\left(x^{2}+y+z\right)(1+y+z) \geqslant(x+y+z)^{2}\). Therefore, \(\frac{1}{\sqrt{x^{2}+y+z}} \leqslant \frac{\sqrt{1+y+z}}{x+y+z}\), and similarly, we can obtain the other two inequalities. Also, by the Cauchy inequality, \(3\left(x^{2}+y^{2}+z^{2}\right) \geqslant(x+y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,861 |
$\begin{array}{l}\quad \text { Example } 1 \frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\cdots \\ +\frac{64}{1+x^{64}} .\end{array}$ | Analysis: The characteristic of solving this type of problem is to first observe and determine the conclusion.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
The translation is as follows:
Analysis: The characteristic of s... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,862 |
Example 1. Let positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$. Find
$$\begin{array}{l}
\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}} \\
+\cdots+\frac{a_{n}}{1+a_{1}+\cdots+a_{n-1}}
\end{array}$$
the minimum value. (82 West Germany Competition Problem) | Let the sum of the original expression be $S$, then
$$S=\sum_{k=1}^{n} \frac{a_{k}}{2-a_{k}}=2 \sum_{k=1}^{n} \frac{1}{2-a_{k}}-n \cdot(1)$$
Since $2-a_{k}>0(k=1,2 ; \cdots, n)$, by the Cauchy-Schwarz inequality,
$$\sum_{k=1}^{n} \frac{1}{2-a_{k}} \geqslant \frac{n^{2}}{\sum^{n}}=\frac{n^{2}}{2 n-1}$$
Combining (1) a... | \frac{n}{2 n-1} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,865 |
Example 4. Let $a, b ; c$ be positive numbers. Prove that:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2} .$$ | To prove, it is known that the inequality to be proved is equivalent to
$$\begin{array}{l}
2\left(a^{3}+b^{3}+c^{3}\right) \geqslant a^{2} b+a^{2} c+b^{2} a+b^{2} c \\
+c^{2} a+c^{2} b .
\end{array}$$
By the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b \\
= & a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,866 |
Example 7. Find all real numbers $a$ such that there exist non-negative real numbers $x_{\mathrm{k}}, k=1,2,3,4,5$, satisfying the relations
$$\begin{array}{c}
\sum_{k=1}^{5} k x_{\mathrm{k}}=a, \quad \sum_{k=1}^{5} k^{3} x_{k}=a^{2} \\
\sum_{k=1}^{5} k^{5} x_{k}=a^{3}
\end{array}$$ | If there is such a real number $a$ that makes the above three equations hold for the corresponding $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, then we have
On the other hand, by the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(\sum_{k=1}^{5} k^{3} x_{\mathrm{k}}\right)^{2}=\left(\sum_{\mathrm{k}=1}^{5} k^{\fra... | 0,1,4,9,16,25 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,868 |
Theorem 2 For $a_{i j}>0(i=1,2, \cdots, m, j=1$, $2, \cdots, n)$, we have
$$\left(\sum_{i=1}^{m} \prod_{j=1}^{n} a_{i j}\right)^{n} \leqslant \prod_{j=1}^{n} \sum_{i=1}^{m} a_{i j}^{n} .$$ | Prove: Let $\sum_{i=1}^{m} a_{i j}^{n}=A_{j}^{n}$.
By the Arithmetic-Geometric Mean Inequality, we have
$$\begin{array}{l}
\sum_{i=1}^{m} \prod_{j=1}^{n} \frac{a_{i j}}{A_{j}} \leqslant \sum_{i=1}^{m}\left(\frac{1}{n} \sum_{j=1}^{n} \frac{a_{i j}^{n}}{A_{j}^{n}}\right) \\
=\frac{1}{n} \sum_{j=1}^{n} \sum_{i=1}^{m} \fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,871 |
Example 1 Let the side lengths of $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$ be $a, b, c$ and $a_{1}, b_{1}, c_{1}$, and their areas be $S$ and $S_{1}$. Prove:
$$\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)\left(a_{1}^{2}+b_{1}^{2}+c_{1}^{2}\right)- \\
2\left(a^{2} a_{1}^{2}+b^{2} b_{1}^{2}+c^{2} c_{1}^{2}\... | Prove: From the Cauchy-Schwarz inequality and the formula for the area of a triangle,
$$16 S^{2}=\left(a^{2}+b^{2}+c^{2}\right)^{2}-2\left(a^{4}+b^{4}+c^{4}\right),$$
we get
$$\begin{array}{l}
16 S S_{1}+2\left(a^{2} a_{1}^{2}+b^{2} b_{1}^{2}+c^{2} c_{1}^{2}\right) \\
=4 S \cdot 4 S_{1}+\sqrt{2} a^{2} \cdot \sqrt{2} a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,874 |
Example 2 Prove:
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n}>\frac{2 n}{3 n+1}$$ | $$\begin{array}{l}
\text { Prove: } 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n} \\
=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}+\frac{1}{2 n}\right)- \\
2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right) \\
=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots... | \frac{2 n}{3 n+1} | Inequalities | proof | Yes | Yes | inequalities | false | 736,875 |
Example 4 Given the inequality $(x+y)\left(\frac{1}{x}+\frac{a}{y}\right) \geqslant 9$ holds for any positive real numbers $x, y$. Then the minimum value of the positive real number $a$ is ( ).
(A)2
(B) 4
(C)6
(D) 8 | Solution: By Cauchy-Schwarz inequality, we can obtain
$$\begin{array}{l}
(x+y)\left(\frac{1}{x}+\frac{a}{y}\right) \geqslant\left(\sqrt{x} \cdot \frac{1}{\sqrt{x}}+\sqrt{y} \cdot \frac{\sqrt{a}}{\sqrt{y}}\right)^{2} \\
=(1+\sqrt{a})^{2}
\end{array}$$
When $x=1, y=\sqrt{a}$, $(x+y)\left(\frac{1}{x}+\frac{a}{y}\right)$ ... | B | Inequalities | MCQ | Yes | Yes | inequalities | false | 736,877 |
Example 7 Find the real solutions of the system of equations:
$$\left\{\begin{array}{l}
(x-2)^{2}+\left(y+\frac{3}{2}\right)^{2}+(z-6)^{2}=64 \\
(x+2)^{2}+\left(y-\frac{3}{2}\right)^{2}+(z+6)^{2}=25
\end{array}\right.$$ | Solving: By adding and subtracting the two given equations, we get
$$\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=\left(\frac{3}{2}\right)^{2} \\
-8 x+6 y-24 z=39 .
\end{array}\right.$$
By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
39=-8 x+6 y-24 z \\
\leqslant \sqrt{(-8)^{2}+6^{2}+(-24)^{2}} \cdot \sqrt{x... | \left(-\frac{6}{13}, \frac{9}{26},-\frac{18}{13}\right) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,878 |
Example 8 Given that $a, b$ are positive numbers, $n$ is a positive integer, and $\frac{\sin ^{4} \theta}{a}+\frac{\cos ^{4} \theta}{b}=\frac{1}{a+b}$. | Prove: $\frac{\sin ^{2 n} \theta}{a^{n-1}}+\frac{\cos ^{2 n} \theta}{b^{n-1}}=\frac{1}{(a+b)^{n-1}}$.
Proof: By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
1=(a+b)\left(\frac{\sin ^{4} \theta}{a}+\frac{\cos ^{4} \theta}{b}\right) \\
\geqslant\left(\sqrt{a} \cdot \frac{\sin ^{2} \theta}{\sqrt{a}}+\sqrt{b} ... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,879 |
Example 1 Given that for any integer $x$, the trinomial $a x^{2}+$ $b x+c$ is a perfect square. Prove: there must be
$$a x^{2}+b x+c=(d x+e)^{2}$$ | To prove: Let $f(x)=a x^{2}+b x+c$.
We need to prove: $a$, $b$, $c$ are integers, and $b^{2}=4 a c$.
It is easy to see that $c=f(0)$ is a perfect square, and
$$\begin{array}{l}
2 b=f(1)-f(-1) \\
2 a=f(1)+f(-1)-2 c
\end{array}$$
are integers.
If $b$ is not an integer, then $2 b$ is odd.
Let $2 b=2 n+1$.
Thus, $4 b \equ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,880 |
Example 1 Given $x, y \in \mathbf{R}_{+}$, and $\frac{1}{x}-\frac{1}{y}=1$, find the maximum value of $x-y$. | From the second conclusion above, we get
$$\begin{aligned}
x-y & =(x-y)\left(\frac{9}{x}-\frac{1}{y}\right) \\
& =\left[(\sqrt{x})^{2}-(\sqrt{y})^{2}\right]\left[\left(\frac{3}{\sqrt{x}}\right)^{2}-\left(\frac{1}{\sqrt{y}}\right)^{2}\right] \\
& \leqslant(3-1)^{2}=4
\end{aligned}$$
That is, \( x-y \leqslant 4 \) (when... | 4 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,881 |
Example 2 Find the range of the function $y=\sqrt{2 x-3}-\sqrt{x-2}$. | The domain of the function is $[2,+\infty)$, so $y>0$. From the second conclusion above, we get
$$\begin{array}{l}
y^{2}=(\sqrt{2 x-3}-\sqrt{x-2})^{2}=\left(\sqrt{2} \cdot \sqrt{x-\frac{3}{2}}-1 \cdot \sqrt{x-2}\right)^{2} \\
\geqslant(2-1)\left[\left(x-\frac{3}{2}\right)-(x-2)\right]=\frac{1}{2}
\end{array}$$
That is... | \left[\frac{\sqrt{2}}{2},+\infty\right) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,882 |
Question As shown in Figure 1, an ant crawls up from the bottom of the tree along the trunk. At each fork, the ant randomly chooses a path. What is the probability that it crawls to point $D$? | Because the ant choosing paths $H F, H G$ is equally likely, and after reaching point $G$, choosing paths $G D, G E$ is also equally likely, the probability of it reaching point $D$ is $\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$.
In fact, the probability of the ant reaching points $A, B, C$ is $\frac{1}{2} \times \fr... | \frac{1}{4} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 736,883 |
Question (3): As shown in Figure 3, if $F D$ is connected, what is the probability that it reaches $D$?
untranslated part:
(3) : 如图 3 所示, 如果 $F D$ 是相连的, 它色到 $D$处的概率是多少?
translated part:
(3): As shown in Figure 3, if $F D$ is connected, what is the probability that it reaches $D$? | Solve: The probability that it slinks to $D$ is $\frac{1}{2} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{2}=\frac{3}{8}$. | \frac{3}{8} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 736,885 |
Example 7 (Mathematics Bulletin 1996(1) Problem 991) Let $a, b, c$ be the lengths of the three sides of $\triangle ABC$, $s$ be the semi-perimeter, and $\Delta$ be the area. Prove:
$$(s-a)^{4}+(s-b)^{4}+(s-c)^{4} \geqslant \Delta^{2} .$$ | Proof: Let $x=s-a, y=s-b, z=s-c$, then $s=$ $x+y+z, \Delta^{2}=x y z(x+y+z)$, hence
$$(12) \Leftrightarrow x^{4}+y^{4}+z^{4} \geqslant x y z(x+y+z)$$
By Theorem 2 and the mean inequality, we get
$$\begin{array}{c}
\frac{x^{4}}{x+y+z}+\frac{y^{4}}{x+y+z}+\frac{z^{4}}{x+y+z} \geqslant \frac{(x+y+z)^{4}}{3^{3}(x+y+z)} \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,888 |
Example 2 (1987 CMO Training Team Problem) Let $a, b, c \in$ $\mathbb{R}^{+}$, prove that:
$$a^{5}+b^{5}+c^{5} \geqslant a^{3} b c+b^{3} c a+c^{3} a b .$$ | Proof: Since $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$, by Theorem 2
$$\frac{a^{4}}{b c}+\frac{b^{4}}{c a}+\frac{c^{4}}{a b} \geqslant \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{b c+c a+a b} \geqslant a^{2}+b^{2}+c^{2}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,891 |
Example 3 (Mathematical Olympiad Problems for High School, Intermediate Mathematics, 2003(4) 12) Let $a, b, c \in R^{+}$, prove that:
$$\frac{b^{2}}{a}+\frac{c^{2}}{b}+\frac{a^{2}}{c} \geqslant \sqrt{3}\left(a^{2}+b^{2}+c^{2}\right) .$$ | Proof: By the AM-GM inequality, we have $a^{3}+c^{2} a \geqslant 2 a^{2} c, b^{3}+a^{2} b \geqslant 2 a b^{2}, c^{3}+b^{2} c \geqslant 2 b c^{2}$, hence $a^{3}+b^{3}+c^{3}+a^{2} b+b^{2} c+c^{2} a \geqslant 2\left(a b^{2}+b c^{2}+c a^{2}\right)$, which means $\left(a^{2}+b^{2}+c^{2}\right)(a+b+c) \geqslant 3\left(a b^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,892 |
Example 4 (Mathematical Bulletin) 1994(1) Problem 8>1) Given $\alpha, \beta$ $\in\left(0, \frac{\pi}{2}\right), n \in N^{*}$, prove:
$$\frac{\sin ^{n+2} \alpha}{\cos ^{n} \beta}+\frac{\cos ^{n+2} \alpha}{\sin ^{n} \beta} \geqslant 1$$ | Proof: Clearly $\sin \alpha, \cos \alpha, \sin \beta, \cos \beta, \sin (\alpha+\beta) \in(0$,
1). By Theorem 2,
$$\begin{array}{c}
(9) \text { left }=\frac{\left(\sin ^{2} \alpha\right)^{n}}{(\sin \alpha \cos \beta)^{n}}+\frac{(\cos 2)^{n}}{(\cos \alpha \sin \beta)^{n}} \\
\geqslant \frac{\left(\sin ^{2} \alpha+\cos ^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,893 |
Example 5 (26th $IMO$ Preliminary Problem) Given that $a, b, c$ are positive numbers, $s=(a+b+c) / 2, n \in \boldsymbol{N}^{*}$, prove:
$$\frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b} \geqslant\left(\frac{2}{3}\right)^{n-2} s^{n-1} .$$ | Proof: By Theorem 2, we get
(10) Left $\geqslant \frac{(a+b+c)^{n}}{3^{n-2}(2 a+2 b+2 c)}=$ (10) Right. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,894 |
[Math Problem 294.4] For $n=0,1,2, \cdots$ find the sum:
$$S_{n}=\sum_{k=0}^{n} \operatorname{arccot}\left(k^{2}-k+4+\frac{6}{k^{2}-k-1}\right)$$
and find the value of $\lim S_{n}$. | Solution: It is easy to see that $S_{0}=\operatorname{arccot}(-2)=\pi-\operatorname{arccot} 2, S_{1}=$ $2 \operatorname{arccot}(-2)=2 \pi-2 \operatorname{arccot} 2$. For $n=2,3, \cdots$, it is easy to see that
$$\left.S_{n}=S_{1}+\sum_{k=2}^{n} \arctan \frac{A_{k}-1}{\left(A_{k}+1\right.}\right)\left(A_{k}+2\right),$$
... | \frac{9}{4} \pi - \arctan 3 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,896 |
Example 1-(2007 National Competition Henan Preliminary Question) Given $7 \sin \alpha+24 \cos \alpha=25$, find the value of $\tan \alpha$. | Analysis: From the known equation and the Cauchy-Schwarz inequality, we get
\[1 \cdot \left(7^{2} + 24^{2}\right) = \left(\sin ^{2} \alpha + \cos ^{2} \alpha\right) \left(7^{2} + 24^{2}\right) \geqslant (7 \sin \alpha + 24 \cos \alpha)^{2} = 25^{2}.\]
The equality in the above equation holds if and only if \(\frac{\si... | \frac{7}{24} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,897 |
Example 2 (1992 "Friendship Cup" International Mathematics Invitational for Grade 9) Let $a, b, c, x, y, z \in R, a^{2}+b^{2}+c^{2}=25, x^{2}+$ $y^{2}+z^{2}=36, a x+b y+c z=30$, find the value of $\frac{a+b+c}{x+y+z}$. | Analysis: From the known equation and the Cauchy-Schwarz inequality, we get
\[ 25 \cdot 36 = \left(a^{2} + b^{2} + c^{2}\right)\left(x^{2} + y^{2} + z^{2}\right) \geqslant (a x + b y + c z)^{2} = 30^{2}. \]
The equality in the above inequality holds if and only if \(\frac{a}{x} = \frac{b}{y} = \frac{c}{z}\). From this,... | \frac{5}{6} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,898 |
Example 11 (33rd $I M O$ problem) If $x, y, z>1$, and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$, prove $\sqrt{x+y+z} \geqslant \sqrt{x-1}+$ $\sqrt{y-1}+\sqrt{z-1}$. | Analysis: From the known equation, we get $\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}=1-$ $\frac{1}{x}+1-\frac{1}{y}+1-\frac{1}{z}=3-2=1$. From this and the Cauchy-Schwarz inequality, we have $\left(\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}\right)(x+y+z) \geqslant$ $(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1})^{2}$, i.e., $1 \cdo... | \sqrt{x+y+z} \geqslant \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} | Inequalities | proof | Yes | Yes | inequalities | false | 736,899 |
Example 12 (36th $I M O$ problem) Let $a, b, c \in \mathbb{R}^{+}$, and $abc = 1$. Try to prove: $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}$. | Given $a b c=1$, we know the original inequality is equivalent to
$$
\frac{b^{2} c^{2}}{a(b+c)}+\frac{a^{2} c^{2}}{b(a+c)}+\frac{a^{2} b^{2}}{c(a+b)} \geqslant \frac{3}{2}.
$$
From this and the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
{\left[\frac{b^{2} c^{2}}{a(b+c)}+\frac{a^{2} c^{2}}{b(a+c)}+\frac{a^{2}... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,900 |
Example 4 (1992 "Friendship Cup" International Mathematics Invitational for Grade 8) Find three real numbers $x, y, z$, such that they simultaneously satisfy the following equations: $2 x+3 y+z=13,4 x^{2}+9 y^{2}+z^{2}-2 x+15 y+$ $3 z=82$. | Analysis: By adding and completing the square of the two equations, we get $(2 x)^{2}+$ $(3 y+3)^{2}+(z+2)^{2}=108$. Write the first equation as $2 x$ $+(3 y+3)+(z+2)=18$. From this and the Cauchy-Schwarz inequality, we have $108 \cdot 3=\left[(2 x)^{2}+(3 y+3)^{2}+(z+2)^{2}\right]\left(1^{2}\right.$ $\left.+1^{2}+1^{2... | x=3, y=1, z=4 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,902 |
L5- (3rd "Hope Cup" for Senior High School, 1002nd Edition) If $a \sqrt{1-b^{2}}+b \sqrt{1-a^{2}}=1$, prove that $a^{2}+b^{2}=1$. | Analysis: From the known equation and the Cauchy-Schwarz inequality, we get
\[1 \cdot 1 = \left[a^{2} + \left(\sqrt{1-a^{2}}\right)^{2}\right]\left[\left(\sqrt{1-b^{2}}\right)^{2} + b^{2}\right] \geqslant \left(a \sqrt{1-b^{2}} + b \sqrt{1-a^{2}}\right)^{2} = 1^{2}.\]
The equality in the above equation holds if and on... | a^{2} + b^{2} = 1 | Algebra | proof | Yes | Yes | inequalities | false | 736,903 |
Example 6-(1996 Hong Kong International Mathematical Olympiad Problem) Given $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, and $\frac{\sin ^{4} \alpha}{\cos ^{2} \beta}+\frac{\cos ^{4} \alpha}{\sin ^{2} \beta}=1$, prove that $\alpha+$ $\beta=\frac{\pi}{2}$. | $$\begin{array}{l}
\quad 1 \cdot 1=\left[\left(\frac{\sin ^{2} \alpha}{\cos \beta}\right)^{2}+\left(\frac{\cos ^{2} \alpha}{\sin \beta}\right)^{2}\right]\left(\cos ^{2} \beta+\sin ^{2} \beta\right) \\
\geqslant\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}=1^{2} \text {. The equality holds if and only if } \\
\frac... | \alpha + \beta = \frac{\pi}{2} | Algebra | proof | Yes | Yes | inequalities | false | 736,904 |
Example 7 (1990 First "Hope Cup" National Mathematics Invitational Contest备用题) Given $x, y, z \in R^{+}$, and $\frac{x^{2}}{1+x^{2}}+\frac{y^{2}}{1+y^{2}}+\frac{z^{2}}{1+z^{2}}$ $=2$, find the maximum value of $w=\frac{x}{1+x^{2}}+\frac{y}{1+y^{2}}+\frac{z}{1+z^{2}}$. | Given the equation, we have $\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{1}{1+z^{2}}=1$ $-\frac{x^{2}}{1+x^{2}}+1-\frac{y^{2}}{1+y^{2}}+1-\frac{z^{2}}{1+z^{2}}=3-2=1$. From this and the Cauchy-Schwarz inequality, we get $2 \cdot 1=\left[\left(\frac{x}{\sqrt{1+x^{2}}}\right)^{2}+\right.$ $\left.\left(\frac{y}{\sqrt{1+y^{2... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,905 |
8-(22nd $I M O$ Problem) Let $P$ be any point inside triangle $A B C$, and let the distances from $P$ to the three sides $B C, C A, A B$ be $d_{1}, d_{2}, d_{3}$ respectively, with $B C=a, C A=b, A B=c$. Find the minimum value of $u=\frac{a}{d_{1}}+\frac{b}{d_{2}}+\frac{c}{d_{3}}$. | Given the area of triangle $A B C$ is $S$, then $a d_{1}+b d_{2}+$ $c d_{3}=2 S$. From this and the Cauchy-Schwarz inequality, we have $2 S \cdot u=\left(a d_{1}+b d_{2}\right.$ $\left.+c d_{3}\right)\left(\frac{a}{d_{1}}+\frac{b}{d_{2}}+\frac{c}{d_{3}}\right) \geqslant(a+b+c)^{2}$, which means $u \geqslant$ $\frac{(a+... | \frac{(a+b+c)^{2}}{2 S} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,906 |
Example 9 (1980 Leningrad Mathematical Olympiad) Let $a, b, c, d \in \mathbb{R}^{+}$, and $a+b+c+d=1$, then $\sqrt{4 a+1}+$ $\sqrt{4 b+1}+\sqrt{4 c+1}+\sqrt{4 d+1}<6$. | Analysis: From the known equation, we get $4 a+1+4 b+1+4 c+1+$ $4 d+1=8$. From this and the Cauchy-Schwarz inequality, we have $8 \cdot 4=$ $\left[(\sqrt{4 a+1})^{2}+(\sqrt{4 b+1})^{2}+(\sqrt{4 c+1})^{2}+\right.$ $\left.(\sqrt{4 d+1})^{2}\right]\left(1^{2}+1^{2}+1^{2}+1^{2}\right) \geqslant$ $(\sqrt{4 a+1}+\sqrt{4 b+1}... | \sqrt{32}<6 | Inequalities | proof | Yes | Yes | inequalities | false | 736,907 |
L 10 (24th All-Union Mathematical Olympiad) Given that $a_{1}, a_{2}, \cdots, a_{n}$ are all positive numbers and their sum is 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} .$$ | Analysis: According to the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left[\left(\frac{a_{1}}{\sqrt{a_{1}+a_{2}}}\right)+\right. \\
\left.\left(\frac{a_{2}}{\sqrt{a_{2}+a_{3}}}\right)^{2}+\cdots+\left(\frac{a_{n}}{\sqrt{a_{n}+a_{1}}}\right)^{2}\right] . \\
{\left[\left(\sqrt{a_{1}+a_{2}}\right)^{2}+\left(\... | \frac{1}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,908 |
Theorem The function $y=\frac{a}{\sin x}+\frac{b}{\cos x}, x \in(0$, $\left.\frac{\pi}{2}\right), a, b$ are positive constants, then $y_{\text {min }}=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}$. | Proof: Let $m, n$ be undetermined positive constants. By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(\frac{a}{\sin x}+\frac{b}{\cos x}\right)(m \sin x+n \cos x) \\
\geqslant(\sqrt{a m}+\sqrt{b n})^{2}, \\
\left(m^{2}+n^{2}\right)\left(\sin ^{2} x+\cos ^{2} x\right) \\
\geqslant(m \sin x+n \cos x)^{2... | \left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}} | Algebra | proof | Yes | Yes | inequalities | false | 736,911 |
Example 2 Find the minimum value of the function $y=\frac{2 \sqrt{2}}{\sin x}+\frac{1}{\cos x}, x \in$ $\left(0, \frac{\pi}{2}\right)$. | From the theorem, we get $y_{\min }=\left[(2 \sqrt{2})^{\frac{2}{3}}+1\right]^{\frac{3}{2}}$ $=3 \sqrt{3}$. | 3 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,913 |
Example 3 The line $l$ passes through a fixed point $P(a, b)$ in the first quadrant and intersects the positive halves of the two coordinate axes at points $A, B$ respectively. Find the minimum value of the line segment $|A B|$.
untranslated text remains the same as the source, only the example has been translated. | As shown in Figure 1, let $P M \perp x$-axis at $M$, $P N \perp y$-axis at $N$, and $\angle O A B=\theta$. Then, we have $|A B|=|A P|+|P B|=\frac{b}{\sin \theta}+\frac{a}{\cos \theta}$. According to the theorem, we have
$$|A B|_{\min }=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}$$ | \left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,914 |
Example 1 (2006 China Mathematical Olympiad National Training Team Exam (3) Second Question) Given that $x_{1}, x_{2}, \cdots, x_{n}$ are positive numbers, and satisfy $\sum_{i=1}^{n} x_{i}=1$, prove:
$$\left(\sum_{i=1}^{\infty} \sqrt{x_{i}}\right)\left(\sum_{i=1}^{n} \frac{1}{\sqrt{1+x_{i}}}\right) \leqslant \frac{n^{... | Proof: Let $1+x_{i}=y$, then $x_{i}=y_{i}-1, y_{i}>1(i=1,2, \cdots, n), \sum_{i=1}^{n} y_{i}=n+1$, the original inequality is equivalent to
$$\left(\sum_{i=1}^{n} \sqrt{y_{i}-1}\right)\left(\sum_{i=1}^{n} \frac{1}{\sqrt{y_{i}}}\right) \leqslant \frac{n^{2}}{\sqrt{n+1}}$$
By the Cauchy-Schwarz inequality, we have
$$\be... | \frac{n^{2}}{\sqrt{n+1}} | Inequalities | proof | Yes | Yes | inequalities | false | 736,916 |
Example 2 (2004 China Western Mathematical Olympiad, Problem 8) Prove that for any real numbers $a, b, c$,
$$1<\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}} \leqslant \frac{3 \sqrt{2}}{2}$$ | $$\begin{array}{l}
\left(\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}}\right)^{2} \\
=\left[\sqrt{a^{2}+c^{2}} \cdot \frac{a}{\sqrt{\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)}}+\left[\sqrt{b^{2}+a^{2}}\right.\right. \\
\left.\frac{b}{\sqrt{\left(b^{2}+c^{2}\right)\left(b^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,917 |
Example 3 (2006 China Mathematical Olympiad National Training Team Test (4) Question 2): Let $x, y, z>0, x+y+z=1$, prove that: $\square$
$$\frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+z x}}+\frac{z x}{\sqrt{z x+x y}} \leqslant \frac{\sqrt{2}}{2} .$$ | $$\begin{array}{l}
\left(\frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+z x}}+\frac{z x}{\sqrt{z x+x y}}\right)^{2} \\
=\left(\frac{x \sqrt{y}}{\sqrt{x+z}}+\frac{y \sqrt{z}}{\sqrt{y+x}}+\frac{z \sqrt{x}}{\sqrt{z+y}}\right)^{2} \\
=\left[\sqrt{x+y} \cdot \frac{x \sqrt{y}}{\sqrt{(x+y)(x+z)}}+\sqrt{y+z}\right. \\
\left.\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,918 |
Example 4—(2005 National High School Mathematics League Additional Question 2) Let positive numbers $a, b, c, x, y, z$ satisfy $c y + b z = a, a z + c x = b, b x + a y = c$, find the minimum value of the function $f(x, y, z) = \frac{x^{2}}{1+x} + \frac{y^{2}}{1+y} + \frac{z^{2}}{1+z}$. | Given the conditions, we have $b(a z+c x) + c(b x+a y) - a(c y+b z) = b^{2} + c^{2} - a^{2}$, which simplifies to $2 b c x = b^{2} + c^{2} - a^{2}$. Therefore, $x = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$,
Similarly, $y = \frac{c^{2} + a^{2} - b^{2}}{2 c a}$, and $z = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$.
Thus,
$$\begin{... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,919 |
Example 5 (2001 42nd International Mathematical Olympiad (IMO) Problem 2) For all positive real numbers $a, b, c$, prove:
$$\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1$$ | Prove that by the equivalent form (2) of the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}} \\
=\frac{a^{2}}{a \sqrt{a^{2}+8 b c}}+\frac{b^{2}}{b \sqrt{b^{2}+8 c a}}+\frac{c^{2}}{c \sqrt{c^{2}+8 a b}} \\
\geqslant \frac{(a+b+c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,920 |
Example 1 (1998 Iran Mathematical Olympiad) If $x, y, z \geqslant 1$, and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$, prove: $\sqrt{x+y+z} \geqslant$ $\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$ | Note that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$, by Cauchy-Schwarz inequality we have
$$\begin{array}{l}
\quad \sqrt{x+y+z} \sqrt{\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}} \geqslant \sqrt{x-1}+ \\
\sqrt{y-1}+\sqrt{z-1}
\end{array}$$
and $\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}=3-\left(\frac{1}{x}+\frac{1}{y}+\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,923 |
Example 2 (2006 National Training Team Exam Question) Let $x, y, z$ $\in \mathbf{R}^{+}$, and $x+y+z=1$, prove: $\frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+x z}}$ $+\frac{x z}{\sqrt{x z+x y}} \leqslant \frac{\sqrt{2}}{2}$. | Prove that by the AM-GM inequality, we have $\frac{x y}{x+y} \leqslant \frac{x+y}{4}, \frac{x z}{x+z} \leqslant \frac{x+z}{4}, \frac{y z}{y+z} \leqslant \frac{y+z}{4}$, and $3(x y+y z+x z) \leqslant(x+y+z)^{2}$. Well, you are the boss, you can even think of using the Cauchy-Schwarz inequality to get
$$\begin{aligned}
&... | \frac{\sqrt{2}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,924 |
Example 3 (2004 China Western Mathematical Olympiad) Prove that for any positive real numbers $a, b, c$, we have $1<\frac{a}{\sqrt{a^{2}+b^{2}}}+$ $\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}} \leqslant \frac{3 \sqrt{2}}{2}$. | Proof of the left side of the inequality is straightforward. In fact,
$$\frac{a}{\sqrt{a^{2}+b^{2}}}+
\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+$$
$$\frac{c}{a+b+c}=1.$$
Below, we prove the right side. By the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,925 |
Example 4 (42nd IMO Problem) For all positive real numbers $a$,
$$b, c \text {, prove that } \frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1 \text {. }$$ | Prove that by Cauchy-Schwarz inequality,
$$\begin{aligned}
& \left(a \sqrt{a^{2}+8 b c}+b \sqrt{b^{2}+8 c a}+c \sqrt{c^{2}+8 a b}\right) \\
& \cdot\left(\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}}\right) \\
\geqslant & (a+b+c)^{2}
\end{aligned}$$
Again by Cauchy-Schwarz inequ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,926 |
Example 5 (2002 Asia Pacific Mathematical Olympiad) Let $x, y, z$ be positive numbers, and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$. Prove that: $\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geqslant \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z}$. Is it really that strenuous? | $$\begin{array}{l}
\text { Prove that because } \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1, \\
\text { so } x y+y z+z x=x y z, \\
\sum \sqrt{x+y z}=\sum \sqrt{x \frac{x y z}{x y+y z+z x}+y z} \\
=\sum \sqrt{\frac{y z(x+y)(x+z)}{x y+y z+z x}} \\
=\sqrt{\frac{1}{x y+y z+z x}} \sum \sqrt{y z(x+y)(x+z)} \\
\geqslant \sqrt{\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,927 |
Example 6 (2002 Vietnam Olympiad Problem) Let $x, y, z$ be real numbers, and $x^{2}+y^{2}+z^{2}=9$, prove the inequality: $2(x+y+$ $z)-x y z \leqslant 10 . \quad$ You are copying the standard answer
Translate the text above into English, please keep the original text's line breaks and format, and output the translatio... | Assume without loss of generality that $x^{2} \geqslant y^{2} \geqslant z^{2}$, so $x^{2} \geqslant 3,6 \geqslant y^{2} + z^{2} \geqslant 2 y z$. Using the Cauchy-Schwarz inequality, we get
$$\begin{aligned}
& (2(x+y+z)-x y z)^{2} \\
= & (2(y+z)+x(2-y z))^{2} \\
\leqslant & \left((y+z)^{2}+x^{2}\right)\left(2^{2}+(2-y ... | null | Inequalities | proof | Yes | Yes | inequalities | false | 736,928 |
Example 7 (2006 National Training Team Problem) Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive numbers, and $\sum_{i=1}^{n} x_{i}=1$, prove that:
$$\left(\sum_{i=1}^{n} \sqrt{x_{i}}\right)\left(\sum_{i=1}^{n} \frac{1}{\sqrt{1+x_{i}}}\right) \leqslant \frac{n^{2}}{\sqrt{n+1}}$$ | $$\begin{array}{l}
\sqrt{x_{1}} \frac{1}{\sqrt{n}}+\sqrt{x_{2}} \frac{1}{\sqrt{n}}+\cdots+\sqrt{x_{n}} \frac{1}{\sqrt{n}} \\
\leqslant \sqrt{\frac{1}{n}+x_{2}+x_{3}+\cdots+x_{n}} \cdot \sqrt{x_{1}+\frac{n-1}{n}}, \\
\text { hence } \frac{\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}}}{\sqrt{1+x_{1}}} \\
\leqslant \sqrt{... | \left(\sum_{i=1}^{n} \sqrt{x_{i}}\right)\left(\sum_{i=1}^{n} \frac{1}{\sqrt{1+x_{i}}}\right) \leqslant \frac{n^{2}}{\sqrt{n+1}} | Inequalities | proof | Yes | Yes | inequalities | false | 736,929 |
Example 8 (2005 Serbian Olympiad Problem) Given that $x, y, z$ are positive integers, prove that: $\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}} \geqslant$ $\sqrt{\frac{3}{2}(x+y+z)}$. | Prove that by Cauchy-Schwarz inequality,
$$\left(\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\right.$$
$$\left.\frac{z}{\sqrt{x+y}}\right)(x \sqrt{y+z}+y \sqrt{z+x}+z \sqrt{x+y}) \geqslant(x+$$
$$y+z)^{2}$$
Again by Cauchy-Schwarz inequality,
$$\begin{aligned}
& x \sqrt{y+z}+y \sqrt{z+x}+z \sqrt{x+y} \\
= & \sqrt{x} \sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,930 |
Example $\quad$ Let $\sqrt[10]{2} a+\sqrt[10]{3} b=1, \sqrt[5]{2} a+\sqrt[5]{3} b=1$, prove that: $2 a^{10}+3 b^{10} \geqslant \frac{1}{5 \times 2^{8}}$ | Proof: By the theorem, we have $2 a^{10}+3 b^{10}=\frac{1}{5}\left(2 \cdot 1^{10}+3\right.$.
$$\begin{array}{l}
\left.1^{10}\right) \cdot\left(2 \cdot a^{10}+3 \cdot b^{10}\right) \geqslant \frac{1}{5 \times 2^{8}}\left(2^{\frac{1}{5}} a+3^{\frac{1}{5}} b\right)^{10}= \\
\frac{1}{5 \times 2^{8}}
\end{array}$$ | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,932 |
Example 10 Proof: For any positive real numbers $a, b, c, d$, we have $\frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b}$ $\geqslant 0$, and determine the conditions for equality.
(49th IMO Preliminary Problem) | Proof: Let \( A=\frac{(a-b)(a-c)}{a+b+c}, B=\frac{(b-c)(b-d)}{b+c+d}, C=\frac{(c-d)(c-a)}{c+d+a}, D=\frac{(d-a)(d-b)}{d+a+b} \).
Then \( 2 A=A^{\prime}+A^{\prime \prime} \), where \( A^{\prime}=\frac{(a-c)^{2}}{a+b+c}, A^{\prime \prime}=\frac{(a-c)(a-2 b+c)}{a+b+c} \).
Similarly, we have \( 2 B=B^{\prime}+B^{\prime \pr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,936 |
Example 1: Among the numbers $2008, 2009, 2010$, which can be written in the form $x^{3}+y^{3}+z^{3}-3 x y z$? (where $x, y, z$ are all positive integers) | Solve: Because $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z) \left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)=(x+y+z) - \frac{1}{2}\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$,
If $2008=x^{3}+y^{3}+z^{3}-3 x y z$, then $(x+y+z) \cdot\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]=4016 = 2^{4} \cdot 251$. Since $x+y+z$ and $\left[(x-y)^{2}+(... | 2008 \text{ and } 2009 \text{ can be expressed, but } 2010 \text{ cannot be expressed.} | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 736,937 |
Example 1 Let $a, b, c, d$ be real numbers, prove the inequality: $(a+b+c+d)^{2} \leqslant 3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+6 a b$. | Analysis We cannot directly use the Cauchy-Schwarz inequality, so we perform a completion of the square on the right side.
Proof By the Cauchy-Schwarz inequality, we have $[(a+b)+c+d)]^{2} \leqslant 3\left[(a+b)^{2}+c^{2}+d^{2}\right]$, and rearranging yields the desired result.
Note The greatest skill in applying the ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,938 |
Example 2 Let $x, y, z$ be positive numbers, prove: $\frac{1+y z+z x}{(1+x+y)^{2}}+\frac{1+z x+x y}{(1+y+z)^{2}}+\frac{1+x y+y z}{(1+z+x)^{2}} \geqslant 1$. | Analysis: There are too many denominators, and they are also complex. Use the Cauchy-Schwarz inequality locally to unify the denominators.
Proof: By the Cauchy-Schwarz inequality, we have $[z(x+y)+1]\left(\frac{x+y}{z}+1\right) \geqslant(x+y+1)^{2}$, so $\frac{1+y z+z x}{(1+x+y)^{2}} \geqslant \frac{z}{x+y+z}$. Similar... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,939 |
Example 4 Let $a, b, c, d$ be positive numbers, prove: $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b} \geqslant 0$. | Analyzing the selection of elements in the Cauchy inequality, they are often positive numbers because the right side involves taking a square root. Therefore, add 4 to both sides of the inequality and then proceed with the proof.
Prove that
\[
\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b} \geqslant 0... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,940 |
Example 6 Let $n \geqslant 2, a_{1}, a_{2}, \cdots, a_{n}$ be $n$ positive real numbers, satisfying: $\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right) \leqslant\left(n+\frac{1}{2}\right)^{2}$. Prove: $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\} \leqslant 4 \mi... | Analyzing from $n=2$ people's hands, for $n \geqslant 3$, sort the elements appropriately, and then simplify using the Cauchy-Schwarz inequality.
Proof: By symmetry, without loss of generality, assume $m=a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}=M$, we need to prove $M \leqslant 4 m$.
When $n=2$, the condi... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,942 |
Example 7 Let $a, b, c$ be real numbers. Prove that: $\sqrt{2\left(a^{2}+b^{2}\right)}+\sqrt{2\left(b^{2}+c^{2}\right)}+\sqrt{2\left(c^{2}+a^{2}\right)} \geqslant \sqrt{3(a+b)^{2}+3(b+c)^{2}+3(c+a)^{2}}$.
(2004 Polish Mathematical Olympiad Problem) | Analysis: The conclusion of this problem is strong, so we first use the method of analysis, squaring both sides and then considering, and later locally adopting the Cauchy-Schwarz inequality to achieve the purpose of simplification.
Proof: Squaring both sides, we only need to prove
\[2 \sqrt{\left(a^{2}+b^{2}\right)\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,943 |
Example 9 Let $a, b, c$ be positive numbers, and $a+b+c=3$, prove: $\frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+\frac{1}{2+c^{2}+a^{2}} \leqslant \frac{3}{4}$. | Analyze transforming the inequality into an equivalent inequality for proof.
Proof: By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
{\left[\left(2+a^{2}+b^{2}\right)+\left(2+b^{2}+c^{2}\right)+\left(2+c^{2}+a^{2}\right)\right]\left(\frac{a^{2}+b^{2}}{2+a^{2}+b^{2}}+\frac{b^{2}+c^{2}}{2+b^{2}+c^{2}}+\frac{c... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 736,945 |
Example 1 Given $a, \beta \in\left(0, \frac{\pi}{2}\right), n \in \mathrm{N}$, prove: $\frac{\sin ^{n+2} \alpha}{\cos ^{n} \beta}+\frac{\cos ^{\sin 2} \alpha}{\sin ^{n} \beta} \geqslant 1$. (Mathematics Bulletin $1994(1)$ Question 871) | Prove that under the conditions: $\sin \alpha, \cos \alpha, \sin \beta, \cos \beta, \sin (\alpha + \beta) \in (0,1)$
From Proposition 2, we get: $\frac{\sin ^{n+2} \alpha}{\cos ^{2} \beta}+\frac{\cos ^{n+2} \alpha}{\sin ^{n} \beta}$
$$\begin{array}{l}
=\frac{\left(\sin ^{2} \alpha\right)^{n+1}}{(\sin \alpha \cos \beta... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,947 |
Example 2 Let $a, \in \mathbf{R}^{+}, i=1,2, \cdots, n$, and $a_{1}+a_{2}$ $+\cdots+a_{n}=1$,
Prove: $\frac{a_{1}}{2-a_{1}}+\frac{a_{3}}{2-a_{3}}+\cdots+\frac{a_{n}}{2-a_{n}} \geqslant \frac{n}{2 n-1}$. | Proof from Proposition 2:
$$\begin{array}{l}
\frac{a_{1}}{2-a_{1}}+\frac{a_{2}}{2-a_{2}}+\cdots+\frac{a_{n}}{2-a_{n}} \\
\geqslant n^{1+1-1} \frac{a_{1}+a_{2}+\cdots+a_{n}}{2 n-\left(a_{1}+a_{3}+\cdots+a_{n}\right)} \\
=\frac{n}{2 n-1} .
\end{array}$$ | \frac{n}{2 n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 736,948 |
Example 5 Given $a_{1}, a_{2}, \cdots, a_{\pi}, x_{1}, x_{2}, \cdots x_{n} \in \mathrm{R}^{+}$, and $x_{1}+x_{2}+\cdots+x_{\pi}=1$, then we have
$$\begin{array}{l}
\frac{a_{1}}{x_{1}{ }^{m}}+\frac{a_{2}}{x_{2}^{m}}+\cdots+\frac{a_{n}}{x_{n}^{m}} \\
\geqslant\left(\sqrt[m+1]{a_{1}}+\sqrt[m+1]{a_{2}}+\cdots+\sqrt[m+1]{a_... | $$\begin{array}{l}
\frac{a_{1}}{x_{1}^{m}}+\frac{a_{2}}{x_{2}^{m}}+\cdots+\frac{a_{n}}{x_{n}^{m}} \\
=\underbrace{\left(x_{1}+x_{2}+\cdots+x_{2}\right) \cdots\left(x_{1}+\cdots+x_{n}\right)}_{m \uparrow\left(x_{1}+x_{2}+\cdots+x_{n}\right)} \\
\cdot\left(\frac{a_{1}}{x_{1}^{m}}+\frac{a_{2}}{x_{2}^{m}}+\cdots+\frac{a_{n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,951 |
Question 1. Given $a_{1}, a_{2}, x_{1}, x_{2} \in \mathrm{R}^{+}$, and $x_{1}+x_{2}=1$, prove:
$$\frac{a_{1}}{x_{1}}+\frac{a_{2}}{x_{2}} \geqslant\left(\sqrt{a_{1}}+\sqrt{a_{2}}\right)^{2}$$ | Solution 1, by the AM-GM inequality:
$$\begin{array}{l}
\frac{a_{1}}{x_{1}}+\frac{a_{2}}{x_{2}}=\frac{a_{1}\left(x_{1}+x_{2}\right)}{x_{1}}+\frac{a_{2}\left(x_{1}+x_{2}\right)}{x_{2}}= \\
a_{1}+a_{2}+\frac{a_{1} x_{2}}{x_{1}}+\frac{a_{2} x_{1}}{x_{2}} \geqslant a_{1}+a_{2}+2 \sqrt{a_{1} a_{2}} \\
=\left(\sqrt{a_{1}}+\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,952 |
Theorem 3 Let $a_{1} \in \mathbf{R}, i=1,2, \cdots, n, n \in \mathbf{N} ; k_{0}, k$ be non-negative integers not greater than $n$, $s, r \in \mathbf{R}, r \neq 0$, and $\left|\frac{s}{r}\right| \geqslant 2$, then
$$\sum_{1=1}^{n} \frac{a_{i}^{s}}{\sum_{j=k_{0}}^{k_{0}+k} a_{i+s}^{r}} \geqslant \frac{n^{2-\frac{s}{r}}}{... | To prove (8). In (7), let $b_{1}=\sum_{i=k_{0}}^{t_{0}^{+t}} a_{i+s}^{r}$. According to (7), we get
$$\sum_{i=1}^{n} \frac{a_{1}^{s}}{\sum_{j=k_{0}}^{k_{0}^{+k}} a_{i+j}^{r}}=\sum_{i=1}^{n} \frac{\left(a_{r}^{r}\right)^{\frac{s}{r}}}{\sum_{j=k_{0}}^{t_{0}^{+k}} a_{i+1}^{r}} \geqslant n^{2-\frac{1}{r}} \frac{\left(\sum_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,957 |
Proposition 1: Given $a_{j i} \in \mathbf{R}^{+}, d_{j} \in \mathbf{R}^{+}(i=1,2$, $\cdot n, j=1,2, \cdots, m)$ and $\sum_{j=1}^{m} \alpha_{j}=1$, then we have
$$\begin{array}{l}
\sum_{i=1}^{n} a_{1 i}^{a_{1}} a_{2 i}^{a_{2}} \cdots a_{m}^{a_{m}} \leqslant \\
\left(\sum_{i=1}^{n} a_{1 i}\right)^{a_{1}}\left(\sum_{i=1}^... | Proof: Use mathematical induction on $m$.
1) When $m=2$, the proposition holds by Lemma 1.
2) Assume that when $m=k$, the proposition holds, then when $m=k+1$, since $\sum_{j=1}^{k+1} \alpha_{j}=1$, let $\sum_{j=2}^{k+1} \alpha_{j}=s$, then $a_{1}+s=1$, note that $\frac{1}{s}\left(\alpha_{2}+\cdots+\alpha_{k+1}\right)=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,958 |
Theorem 9 (Hölder's Inequality) For any real numbers \(a_{i}, b_{i}(i=1,2, \cdots, k)\), and rational number \(r>1\), the following holds:
$$\sum_{i=1}^{k}\left|a_{i} b_{i}\right| \leqslant\left(\sum_{i=1}^{k}\left|a_{i}\right|^{r}\right)^{\frac{1}{r}}\left(\sum_{i=1}^{k}\left|b_{i}\right|^{\frac{r}{r-1}}\right)^{\frac... | $\begin{array}{c}\text { Prove, without loss of generality, }\left(\sum_{i=1}^{k}\left|a_{i}\right|^{\prime}\right)^{\frac{1}{r}},\left(\sum_{i=1}^{k}\left|b_{i}\right|^{\frac{r}{r-1}}\right)^{\frac{r-1}{r}}>0 \text {, in Young's inequality } x y \leqslant \frac{1}{r} x^{r}+\left(1-\frac{1}{r}\right) y^{\frac{r}{r-1}} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,961 |
Theorem $\mathbf{1 3}^{[22,4]}$ (Newman's Inequality) Let $n \geqslant 2$ be a positive integer, and let $a_{i}>0(i=1,2, \cdots, n)$ be real numbers such that $\sum_{i=1}^{n} a_{i}=1$. Then we have $\prod_{i=1}^{n}\left(\frac{1}{a_{i}}-1\right) \geqslant(n-1)^{n}$ | Prove that
$$\frac{1}{a_{i}}-1=\frac{\sum_{k=1}^{n} a_{k}}{a_{i}}-1=\frac{\sum_{k \neq i} a_{k}}{a_{i}} \geqslant \frac{(n-1)\left(\prod_{k \neq i} a_{k}\right)^{\frac{1}{(n-1)}}}{a_{i}}$$
Therefore,
$$\prod_{i=1}^{n}\left(\frac{1}{a_{i}}-1\right) \geqslant(n-1)^{n}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,963 |
Theorem 14 ${ }^{[2,4]}$ (Klamkin Inequality) Let $n \geqslant 2$ be a positive integer, and let $a_{i}>0(i=1,2, \cdots, n)$ be real numbers such that $\sum_{i=1}^{n} a_{i}=1$. Then we have (1) $\prod_{i=1}^{n}\left(1+\frac{1}{a}\right) \geqslant(n+1)^{n} ;(2) \prod_{i=1}^{n}\left(\frac{1+a_{i}}{1-a}\right) \geqslant\l... | Proof (1) Using Chrystal's inequality from Theorem 12, we get
Since
$$\begin{array}{c}
\prod_{i=1}^{n}\left(1+\frac{1}{a_{i}}\right) \geqslant\left(1+\left(\prod_{i=1}^{n} \frac{1}{a_{i}}\right)^{\frac{1}{n}}\right)^{n}, \\
\left(\prod_{i=1}^{n} \frac{1}{a_{i}}\right)^{\frac{1}{n}} \geqslant \frac{n}{\sum_{i=1}^{n} a_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,964 |
Theorem $15^{[3-4]}$ If $a_{1}, a_{2}, \cdots, a_{n}$ are all positive numbers, and $\alpha<0<\beta$, then the following holds:
$$\left(\frac{a_{1}^{\alpha}+a_{2}^{\alpha}+\cdots+a_{n}^{\alpha}}{n}\right)^{\frac{1}{\alpha}} \leqslant\left(a_{1} a_{2} \cdots a_{n}\right)^{\frac{1}{n}} \leqslant\left(\frac{a_{1}^{\beta}+... | Prove that using the geometric-arithmetic mean inequality, we get
$$\left(a_{1} a_{2} \cdots a_{n}\right)^{\frac{1}{n}} \leqslant\left(\frac{a_{1}^{\beta}+a_{2}^{\beta}+\cdots+a_{n}^{\beta}}{n}\right)^{\frac{1}{\beta}},\left(\frac{a_{1}^{\alpha}+a_{2}^{\alpha}+\cdots+a_{n}^{\alpha}}{n}\right)^{\frac{1}{\alpha}} \leqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,965 |
Example $5^{[2,4]}$ Let $a_{1}, a_{2}>0$, and $a_{1}+a_{2}=1$, then it holds that $\frac{1+a_{1}}{1-a_{1}} \cdot \frac{1+a_{2}}{1-a_{2}} \geqslant 3^{2}$. | Prove that by using the condition and the three-variable geometric-arithmetic mean inequality, we get
$$\frac{1+a_{1}}{1-a_{1}} \cdot \frac{1+a_{2}}{1-a_{2}}=\frac{a_{1}+a_{2}+a_{1}}{a_{2}} \cdot \frac{a_{1}+a_{2}+a_{2}}{a_{1}} \geqslant \frac{3 \sqrt[3]{a_{1}^{2} a_{2}} \cdot 3 \sqrt[3]{a_{1} a_{2}^{2}}}{a_{1} a_{2}}=... | 3^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,967 |
Example $\mathbf{6}^{[1.4]}$ Let $a_{1}, a_{2}, a_{3}>0$, and $a_{1}+a_{2}+a_{3}=1$, then it holds that $\frac{1+a_{1}}{1-a_{1}} \cdot \frac{1+a_{2}}{1-a_{2}} \cdot \frac{1+a_{3}}{1-a_{3}} \geqslant 2^{3}$. | Prove $\begin{array}{l} \frac{1+a_{1}}{1-a_{1}} \cdot \frac{1+a_{2}}{1-a_{2}} \cdot \frac{1+a_{3}}{1-a_{3}}=\frac{\left(a_{1}+a_{2}\right)+\left(a_{1}+a_{3}\right)}{a_{2}+a_{3}} \cdot \frac{\left(a_{1}+a_{2}\right)+\left(a_{2}+a_{3}\right)}{a_{1}+a_{3}} \cdot \frac{\left(a_{1}+a_{3}\right)+\left(a_{2}+a_{3}\right)}{a_{... | 2^3 | Inequalities | proof | Yes | Yes | inequalities | false | 736,968 |
Example 1 Positive real numbers $x, y, z$ satisfy $x y z \geqslant 1$ Prove:
$$\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$$ | $$\begin{array}{l}
\text { Prove: The original inequality is equivalent to: } \\
\quad\left(\frac{x^{2}-x^{5}}{x^{5}+y^{2}+z^{2}}+1\right)+\left(\frac{y^{2}-y^{5}}{y^{5}+z^{2}+x^{2}}+1\right)+ \\
\left(\frac{z^{2}-z^{5}}{z^{5}+x^{2}+y^{2}}+1\right) \leqslant 3 \\
\quad \text { i.e., } \frac{x^{2}+y^{2}+z^{2}}{x^{5}+y^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,969 |
Example 2 Given $a, b, c > 0$ and $\frac{1}{a+b+1}+\frac{1}{b+c+1}+$ $\frac{1}{c+a+1} \geqslant 1$, prove: $a+b+c \geqslant ab+ac+bc$ | Prove: Using the local Cauchy inequality, we get: \((a+b+1)(a + b + c^{2}) \geqslant (a+b+c)^{2}\). Therefore, \(\frac{1}{a+b+1} \leqslant \frac{a+b+c^{2}}{(a+b+c)^{2}}\), and similarly for the other two expressions, so: \(\frac{1}{a+b+1} + \frac{1}{b+c+1} + \frac{1}{c+a+1} \leqslant \frac{a+b+c^{2}+a^{2}+b+c+a+b^{2}+c... | a+b+c \geqslant ab+ac+bc | Inequalities | proof | Yes | Yes | inequalities | false | 736,970 |
Example 3 Let $x, y, z$ be non-negative numbers, and $x^{2}+y^{2}+z^{2}=3$ Prove: $\frac{x}{\sqrt{x^{2}+y+z}}+\frac{y}{\sqrt{y^{2}+z+x}}+\frac{z}{\sqrt{z^{2}+x+y}}$ $\leqslant \sqrt{3}$ | Proof: By the local Cauchy inequality, we have: $\left(x^{2}+y+z\right)$ $(1+y+z) \geqslant(x+y+z)^{2}$. Therefore, $\frac{1}{\sqrt{x^{2}+y+z}} \leqslant$ $\frac{\sqrt{1+y+z}}{x+y+z}$, and similarly, we can obtain the other two inequalities. Also, by the Cauchy inequality $3\left(x^{2}+y^{2}+z^{2}\right) \geqslant(x+y+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,971 |
Example 1 Through an internal point $O$ of $\triangle ABC$, draw three lines parallel to the sides: $DE \parallel BC, FG \parallel CA, HI \parallel AB$. Points $D, G, E, F, H, I$ are all on the sides of $\triangle ABC$. Let $S_{1}$ represent the area of hexagon $DGHFEI$, and $S_{2}$ represent the area of $\triangle ABC... | Proof: Let the sides of $\triangle ABC$ be $a, b, c$, $IF = x$, $EH = y$, $GD = z$. According to the problem, $\triangle OEH \sim \triangle BCA$.
$$\therefore \frac{y}{b} = \frac{OE}{a} = \frac{CF}{a}, (\because OE = CF)$$, similarly, $\frac{z}{c} = \frac{BI}{a}$,
$$\therefore \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = ... | proof | Geometry | proof | Yes | Yes | inequalities | false | 736,974 |
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