problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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Example 7 Let $x_{k}>0(k=1,2, \cdots, n)$, and $x_{1}+x_{2}+$ $\cdots+x_{n}=1$.
Prove: $\frac{1}{\sqrt{1-x_{1}}}+\frac{1}{\sqrt{1-x_{2}}}+\cdots+$
$$\frac{1}{\sqrt{1-x_{n}}} \geqslant n \sqrt{\frac{n}{n-1}} .$$ | Proof: Let $f(x)=\frac{1}{\sqrt{1-x}}(0<x<1)$, then $f''(x)=\frac{3}{4}(1-x)^{-\frac{5}{2}}>0$, so $f(x)$ is a convex function on $(0,1)$.
By Jensen's inequality
$$\begin{array}{l}
f\left(\frac{1}{n} \sum_{i=1}^{n} x_{i}\right) \leqslant \frac{1}{n} \sum_{i=1}^{n} f\left(x_{i}\right), \\
\text { i.e., } f\left(\frac{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,123 |
Example 8 Let $a_{i}>0(i=1,2, \cdots, n)$ and $\sum_{i=1}^{n} a_{i}=1$.
Prove: $\prod_{i=1}^{n}\left(1+\frac{1}{a_{i}}\right) \geqslant(n+1)^{n}$. | Prove that the original inequality is equivalent to
$$\sum_{i=1}^{n} \ln \left(1+\frac{1}{a_{i}}\right) \geqslant n \ln (1+n)$$
Let \( f(x)=\ln \left(1+\frac{1}{x}\right)(00 \), so \( f(x) \) is a convex function on \( (0,1) \).
By Jensen's inequality,
$$\begin{array}{l}
f\left(\frac{1}{n} \sum_{i=1}^{n} x_{i}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,124 |
Example 9 (Klamkin Inequality) Let $a>0(i=$ $1,2, \cdots, n)$ and $\sum_{i=1}^{n} a_{i}=1$.
Prove: $\prod_{i=1}^{n}\left(a_{i}+\frac{1}{a_{i}}\right) \geqslant\left(\frac{1}{n}+n\right)^{n}$. | Prove that the original inequality is equivalent to
$$\sum_{i=1}^{n} \ln \left(a_{i}+\frac{1}{a_{i}}\right) \geqslant n \ln \left(\frac{1}{n}+n\right) .$$
Let $f(x)=\ln \left(x+\frac{1}{x}\right)(0<x<1)$. Since $f''(x) > 0$, $f(x)$ is a convex function on $(0,1)$.
By Jensen's inequality,
$$\begin{array}{l}
f\left(\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,125 |
Example 1 There exist positive numbers $x_{1}, x_{2}, \cdots, x_{n}$ less than 1 and $x_{1}+x_{2}+\cdots+x_{n}=1$, prove:
$$\frac{1}{x_{1}-x_{1}^{3}}+\frac{1}{x_{2}-x_{2}^{3}}+\cdots+\frac{1}{x_{n}-x_{n}^{3}}>4 .$$
(2010 Zhejiang University Independent Admission Examination Question)
Generalization There exist positive... | $$\frac{1}{x_{1}-x_{1}^{3}}+\frac{1}{x_{2}-x_{2}^{3}}+\cdots+\frac{1}{x_{n}-x_{n}^{3}} \geqslant \frac{n^{4}}{n^{2}-1}$$
Proof: Let $g(x)=\frac{1}{x-x^{3}}(00)$,
thus $g(x)$ is a strictly convex function on $\mathbf{R}$. According to Jensen's inequality, we have
$$\begin{array}{l}
g\left(x_{1}\right)+g\left(x_{2}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,126 |
Example 2 Let $x, y$ be real numbers, and $x+y=1$. Prove that for any positive integer $n$, we have $x^{2 n}+y^{2 n} \geqslant \frac{1}{2^{2 n-1}}$.
(2009 Tsinghua University Independent Admission Examination Question)
Generalization Let $x_{1}, x_{2}, \cdots, x_{m}$ be real numbers, and $x_{1}+x_{2}+\cdots+$ $x_{m}=1$... | Prove that for $g(x)=x^{2 n}$, we have
$$g^{\prime}(x)=2 n \cdot x^{2 n-1},$$
Therefore, $g^{\prime \prime}(x)=(2 n)(2 n-1) \cdot x^{2 n-2} \geqslant 0$, so $g(x)$ is a strictly convex function on $\mathbf{R}$. According to Jensen's inequality, we get
$$\begin{array}{l}
g\left(x_{1}\right)+g\left(x_{2}\right)+\cdots+g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,127 |
Example 3 Let $a, b, c \in \mathbf{R}^{+}$, and $a+b+c=1$, prove:
$$\left(a+\frac{1}{a}\right) \cdot\left(b+\frac{1}{b}\right) \cdot\left(c+\frac{1}{c}\right) \geqslant \frac{1000}{27} .$$
(2008 Nanjing University Independent Admission Examination Question)
Generalization Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}... | Prove that for $g(x)=\ln \left(x+\frac{1}{x}\right)(x>0)$, we have $g^{\prime}(x)=x-\frac{1}{x^{3}}$, and therefore
$$g^{\prime \prime}(x)=1+\frac{4}{x^{4}} \geqslant 0$$
Thus, $g(x)$ is a strictly convex function on the interval $(0,+\infty)$. According to Jensen's inequality, we get
$$\begin{array}{l}
g\left(x_{1}\r... | \left(x_{1}+\frac{1}{x_{1}}\right) \cdot\left(x_{2}+\frac{1}{x_{2}}\right) \cdot\left(x_{3}+\frac{1}{x_{3}}\right) \geqslant \frac{1000}{27} | Inequalities | proof | Yes | Yes | inequalities | false | 737,128 |
Theorem: As shown in Figure 1, for the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>$ $0)$, the directrix is $L: x=\frac{a^{2}}{c}, F$ is the corresponding focus to $L$, $B O$ $=b$, and the line $B F$ intersects $L$ at $Q$. Then $\sqrt{\frac{|B F|}{|B Q|}}=e$ (a constant). | Prove: Connect BF, intersecting line $L$ at point $Q$, as shown in Figure 1, let the intersection point of line $L$ and the $x$-axis be $N$.
Obviously, $\triangle F N Q \sim \triangle F O B$, so | proof | Geometry | proof | Yes | Yes | inequalities | false | 737,130 |
If $a+b=1, a>0, b>0$, prove: $a^{b-1} \cdot b^{a-1} \leq 2$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Prove: $a^{b-1} \cdot b^{a-1}=a^{b-(a+b)} \cdot b^{a-(a+b)}=$ $\frac{1}{a^{a} b^{b}}$.
From the proof process above:
$$\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^{a_{1}+a_{2}+\cdots+a_{n}} \leq a_{1}^{a_{1}} a_{2}^{a_{2}} \cdots a_{n}^{a_{n}}$$
we get: $\left(\frac{a+b}{2}\right)^{a+b} \leq a^{a} b^{b}, \frac{1}... | null | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,131 |
Prove the mean inequality: If $a_{1}, a_{2}, \cdots, a_{n}>0$ then
$$\frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}} \leq \sqrt[n]{\prod_{k=1}^{n} a_{k}} \leq \frac{1}{n} \sum_{k=1}^{n} a_{k}$$ | Proof: Let $f(x)=\ln x, x>0$, we find the first and second derivatives of $f(x)$ to be:
$$f^{\prime}(x)=\frac{1}{x}, f^{\prime \prime}(x)=-\frac{1}{x^{2}}$$
It can be seen that the function $f(x)=1 \mathrm{n} x$ has $f^{\prime \prime}(x)<0$ when $x>0$, indicating that $f(x)$ is a strictly concave function. Since $f^{\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,132 |
Example 1 If $a, b, c$ are positive numbers, and $a b c=1$, then
$$\frac{a^{2}}{2+a}+\frac{b^{2}}{2+b}+\frac{c^{2}}{2+c} \geqslant 1 .$$ | Prove that from $a b c=1$ we get $\ln a+\ln b+\ln c=0$, let $x_{1}=\ln a, x_{2}=\ln b, x_{3}=\ln c$,
then $x_{1}+x_{2}+x_{3}=0, \frac{a^{2}}{2+a}+\frac{b^{2}}{2+b}+\frac{c^{2}}{2+c}=$ $\frac{\mathrm{e}^{2 x_{1}}}{2+\mathrm{e}^{x_{1}}}+\frac{\mathrm{e}^{2 x_{2}}}{2+\mathrm{e}^{x_{2}}}+\frac{\mathrm{e}^{2 x_{3}}}{2+\mat... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,133 |
Example 3 If $a, b, c$ are positive numbers, and $a b c=1$, then
$$\begin{array}{l}
\frac{a^{2}}{1+\sqrt{2} a}+\frac{b^{2}}{1+\sqrt{2} b}+\frac{c^{2}}{1+\sqrt{2} c} \leqslant \frac{2 \sqrt{2}}{3}(a+b+c) \\
+\sqrt{2}-3
\end{array}$$ | $$\begin{array}{l}
\text { Prove that from } a b c=1 \text { we get } \ln a+\ln b+\ln c=0 \text {, let } x_{1} \\
=\ln a, x_{2}=\ln b, x_{3}=\ln c, \\
\text { then } x_{1}+x_{2}+x_{3}=0 \text {, therefore } \\
\left(\frac{a^{2}}{1+\sqrt{2} a}+\frac{b^{2}}{1+\sqrt{2} b}+\frac{c^{2}}{1+\sqrt{2} c}\right)-\frac{2 \sqrt{2}... | \frac{a^{2}}{1+\sqrt{2} a}+\frac{b^{2}}{1+\sqrt{2} b}+\frac{c^{2}}{1+\sqrt{2} c} \leqslant \frac{2 \sqrt{2}}{3}(a+b+c)+\sqrt{2}-3 | Inequalities | proof | Yes | Yes | inequalities | false | 737,135 |
Example 1 (31st IMO Problem) Let $a, b, c, d \geqslant 0$, and $a b+b c+c d+d a=1$. Prove that:
$$\begin{array}{l}
\frac{a^{3}}{b+c+d}+\frac{b^{3}}{c+d+a}+\frac{c^{3}}{d+a+b}+ \\
\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3}
\end{array}$$ | First, prove the case where $a, b, c, d$ are all non-zero.
From the given theorem: $2(a b+a c+a d+b c+b d+c d) \leqslant 3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$.
$$\begin{aligned}
& \text { The left side of (1) }=\frac{a^{4}}{a b+a c+a d}+\frac{b^{4}}{b c+b d+a b} \\
+ & \frac{c^{4}}{c d+a c+b c}+\frac{d^{4}}{a d+b d+c ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,137 |
Example 2 (28th IMO Preliminary Question) If $a, b, c$ are the side lengths of $\triangle A B C$, $2 p=a+b+c, n \in \mathbf{N}_{+}$, prove:
$$\frac{a^{n}}{b+c}+\frac{b^{n}}{a+c}+\frac{c^{n}}{a+b} \geqslant\left(\frac{2}{3}\right)^{n-2} p^{n-1}$$ | Prove that from the given theorem,
$$4 p^{2}=(a+b+c)^{2} \geqslant 3(a b+b c+a c).$$
(2) The left side of the equation $=\frac{a^{n+1}}{a b+a c}+\frac{b^{n+1}}{a b+b c}+\frac{c^{n+1}}{a c+b c}$
$\geqslant \frac{\left(a^{\frac{n+1}{2}}+b^{\frac{n+1}{2}}+c^{\frac{n+1}{2}}\right)^{2}}{2(a b+b c+a c)}$ (Cauchy-Schwarz ineq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,138 |
Example 3 Let $a_{i}>0, i=1,2,3, \cdots, n, k \in$ $\mathbf{N}_{+}$, prove: | $$\begin{array}{l}
\left(\frac{a_{1}}{a_{2}+a_{3}+\cdots+a_{n}}\right)^{k}+\left(\frac{a_{2}}{a_{1}+a_{3}+\cdots+a_{n}}\right)^{k} \\
+\cdots+\left(\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n-1}}\right)^{k} \geqslant \frac{n}{(n-1)^{k}}
\end{array}$$
Proof: Let $S=a_{1}+a_{2}+\cdots+a_{n}$. By Jensen's inequality, we have
$$... | \frac{n}{(n-1)^{k}} | Inequalities | proof | Yes | Yes | inequalities | false | 737,139 |
Example 4 Let $a_{i}>0, S=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}$, $\prod_{i=1}^{n} a_{i}=1(i=1,2, \cdots n)$, prove that when $k \geqslant 1$ we have $\sum_{i=1}^{n} \frac{1}{a_{i}^{k}\left(S-\frac{1}{a_{i}}\right)} \geqslant \frac{n}{n-1}$ (This problem is a generalization of the 36th IMO problem: Giv... | $$\begin{array}{l}
\text { Prove } \sum_{i=1}^{n} \frac{1}{a_{i}^{k}\left(S-\frac{1}{a_{i}}\right)}=\sum_{i=1}^{n} \frac{\frac{1}{a_{i}^{k+1}}}{\frac{1}{a_{i}}\left(S-\frac{1}{a_{i}}\right)}{ }^{\prime} \\
=\sum_{i=1}^{n} \frac{\left[\left(\frac{1}{a_{i}}\right)^{\frac{k+1}{2}}\right]^{2}}{\frac{1}{a_{i}}\left(S-\frac{... | \frac{n}{n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 737,140 |
Example 1 Let $a_{i}>0(i=1,2, \cdots, n), \sum_{i=1}^{n} a_{i}=1$, prove: $\left(1+\frac{1}{a_{1}}\right)\left(1+\frac{1}{a_{2}}\right) \cdots\left(1+\frac{1}{a_{n}}\right) \geqslant(n+1)^{n}$. | $$\text { Prove } \begin{aligned}
\because 1+\frac{1}{a_{1}} & =1+\underbrace{\frac{1}{n a_{1}}+\frac{1}{n a_{1}}+\cdots+\frac{1}{n a_{1}}}_{n \uparrow} \\
& \geqslant(n+1)\left(\frac{1}{n a_{1}}\right)^{\frac{n}{n+1}} .
\end{aligned}$$
Similarly, $1+\frac{1}{a_{2}} \geqslant(n+1)\left(\frac{1}{n a_{2}}\right)^{\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,141 |
Example 2 Let $a_{i}>0(i=1,2, \cdots, n), \sum_{i=1}^{n} a_{i}=1$, prove: $\left(\frac{1}{a_{1}}-1\right)\left(\frac{1}{a_{2}}-1\right) \cdots\left(\frac{1}{a_{n}}-1\right) \geqslant(n-1)^{n}$. | Prove $\because 1-a_{1}=a_{2}+a_{3}+\cdots+a_{n}$
$$\geqslant(n-1)\left(a_{2} a_{3} \cdots a_{n}\right)^{\frac{1}{n-1}}$$
Similarly, $1-a_{2} \geqslant(n-1)\left(a_{1} a_{3} \cdots a_{n}\right)^{\frac{1}{n-1}}, \cdots$,
$$1-a_{n} \geqslant(n-1)\left(a_{1} a_{2} \cdots a_{n-1}\right)^{\frac{1}{n-1}}$$
Multiplying thes... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,142 |
Example 3 Let $a_{i}>0(i=1,2, \cdots, n), \sum_{i=1}^{n} a_{i}=1, k \in$ N , prove:
$$\begin{aligned}
& \left(a_{1}{ }^{k}+\frac{1}{a_{1}{ }^{k}}\right)\left(a_{2}{ }^{k}+\frac{1}{a_{2}{ }^{k}}\right) \cdots\left(a_{n}{ }^{k}+\frac{1}{a_{n}{ }^{k}}\right) \\
\geqslant & \left(n^{k}+\frac{1}{n^{k}}\right)^{n} .
\end{ali... | $$\begin{array}{l}
\text { Prove } \quad \because a_{1}^{k}+\frac{1}{a_{1}^{k}}=a_{1}^{k}+ \\
\underbrace{\frac{1}{n^{2 k} a_{1}^{k}}+\frac{1}{n^{2 k} a_{1}^{k}}+\cdots+\frac{1}{n^{2 k} a_{1}^{k}}} \geqslant\left(n^{2 k}+1\right) \\
{ }^{2^{2 k} \uparrow} \\
{\left[\frac{a_{1}^{k}}{\left(n^{2 k} a_{1}^{k}\right)^{2 k}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,143 |
Example 4 Let $a_{i}>0(i=1,2, \cdots, n), \sum_{i=1}^{n} a_{i}=1, k \in$ N , prove:
$$\begin{array}{l}
\quad\left(1+a_{1}^{k}+\frac{1}{a_{1}^{k}}\right)\left(1+a_{2}^{k}+\frac{1}{a_{2}^{k}}\right) \cdots\left(1+a_{n}^{k}+\right. \\
\left.\frac{1}{a_{n}{ }^{k}}\right) \geqslant\left(1+n^{k}+\frac{1}{n^{k}}\right)^{n} .
... | $$\begin{array}{l}
\text { Prove } \because 1+a_{1}^{k}+\frac{1}{a_{1}^{k}}=\underbrace{\frac{1}{n^{k}}+\frac{1}{n^{k}}+\cdots+\frac{1}{n^{k}}}_{n^{k} \uparrow}+a_{1}^{k} \\
+\underbrace{\frac{1}{n^{2 k} a_{1}^{k}}+\cdots+\frac{1}{n^{2 k} a_{1}^{k}}}_{n^{2 k} \uparrow} \\
\geqslant\left(n^{2 k}+n^{k}+1\right)\left[\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,144 |
Example 5 Let $a_{i}>0(i=1,2, \cdots, n), \sum_{i=1}^{n} a_{i}=1$, prove that $\left(\frac{1}{a_{1}}-a_{1}\right)\left(\frac{1}{a_{2}}-a_{2}\right) \cdots\left(\frac{1}{a_{n}}-a_{n}\right) \geqslant\left(\frac{1}{n}-n\right)^{n}$. | $$\begin{array}{l}
\text { Note that } \\
\left(\frac{1}{a_{1}}-a_{1}\right)\left(\frac{1}{a_{2}}-a_{2}\right) \cdots\left(\frac{1}{a_{n}}-a_{n}\right)= \\
\frac{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right)}{a_{1} a_{2} \cdots a_{... | \left(n-\frac{1}{n}\right)^{n} | Inequalities | proof | Yes | Yes | inequalities | false | 737,145 |
Example 1 Given $x, y, z > 0$, and $x+y+z=1$. Prove: $\left(\frac{1}{x^{2}}+x\right)\left(\frac{1}{y^{2}}+y\right)\left(\frac{1}{z^{2}}+z\right) \geqslant\left(\frac{28}{3}\right)^{3}$. | Prove: Let $g=\left(\frac{1}{x^{2}}+x\right)\left(\frac{1}{y^{2}}+y\right)\left(\frac{1}{z^{2}}+z\right)$, then $\ln g=\ln \left(\frac{1}{x^{2}}+x\right)+\ln \left(\frac{1}{y^{2}}+y\right)+\ln \left(\frac{1}{z^{2}}+z\right)$.
Let $f(x)=\ln \left(\frac{1}{x^{2}}+x\right)(0<x<1)$. Then, $f'(x) = \frac{1 - \frac{2}{x^3}}... | \left(\frac{28}{3}\right)^{3} | Inequalities | proof | Yes | Yes | inequalities | false | 737,148 |
Example 1 If $a^{2}+b^{2}+a b+b c+c a<0$, then $a^{2}+b^{2}<c^{2}$. | $$\begin{aligned}
0 & >2 a^{2}+2 b^{2}+2(a b+b c+c a) \\
& =a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a+a^{2}+b^{2}-c^{2} \\
& =(a+b+c)^{2}+a^{2}+b^{2}-c^{2} \\
& \geq a^{2}+b^{2}-c^{2}, \\
\therefore & a^{2}+b^{2}<c^{2} .
\end{aligned}$$
Prove: For the known inequality, multiplying both sides by 2, we transform it as follows... | a^{2}+b^{2}<c^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,150 |
Example 3 Given $2 x^{2}+y^{2}-4 x \leq 0$, prove that $y^{2} \leq 4 x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Example 3 Given $2 x^{2}+y^{2}-4 x \leq 0$, prove that $y^{2} \leq 4 x$. | $\begin{array}{l}\text { Prove: } \because y^{2}-4 x \leq 2 x^{2}+y^{2}-4 x \leq 0, \\ \therefore \quad y^{2} \leq 4 x .\end{array}$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,154 |
Example 7 Let real numbers $a_{1}, a_{2}, b_{1}, b_{2}, c_{1}, c_{2}$ satisfy the inequalities $a_{1} a_{2} >0, a_{1} c_{1} \geq b_{1}^{2}, a_{2} c_{2} \geq b_{2}^{2}$.
Prove: $\quad\left(a_{1}+a_{2}\right)\left(c_{1}+c_{2}\right) \geq\left(b_{1}+b_{2}\right)^{2}$. | Proof: Clearly $a_{1} c_{2} \geq 0, a_{2} c_{1} \geq 0$.
$$\begin{array}{l}
\because a_{1 c_{2}+a_{2} c_{1}} \geq 2 \quad \overline{a_{1} c_{2} a_{2} c_{1}} \geq 2 \quad \overline{b_{1}{ }^{2} b_{2}{ }^{2}} \geq 2 b_{1} b_{2}, \\
\therefore \quad\left(a_{1}+a_{2}\right)\left(c_{1}+c_{2}\right) \\
=a_{1 c_{1}}+a_{2 c_{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,156 |
Example 9 Given $x^{2}+y^{2}-4 x+3<0$, prove that $y^{2}<3 x$.
---
The translation maintains the original text's line breaks and format as requested. | Prove: By completing the square, we get $(x-2)^{2}+y^{2}<1$.
$$\text { Let }\left\{\begin{array}{l}
x=2+r \cos \theta, \\
y=r \sin \theta,
\end{array}\right.$$
$\because f(t)=-t^{2}-3_{t}+1$ is a decreasing function on $t \in[-1,1]$,
$$\begin{aligned}
\therefore \quad f(\cos \theta) & \leq f(-1)=3 . \\
\therefore \quad... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,158 |
Example 10 Let $x, y, z$ be real numbers, and satisfy $x+y+z=a$, $x^{2}+y^{2}+z^{2} \leq \frac{a^{2}}{2}(a>0)$, prove that $0 \leq x \leq \frac{2}{3} a, 0 \leq y \leq \frac{2}{3} a$, $0 \leq z \leq \frac{2}{3} a$. | Proof: Let $P(x, y)$ be a point on the plane. According to the problem, the line $l: \quad x+y+(z-a)=0$ and the circle $C: x^{2}+y^{2}=\frac{a^{2}}{2}-z^{2}$ have at least one common point $P$. Therefore, the distance from the center of the circle $C$ at $(0,0)$ to the line $l$ is no greater than the radius of the circ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,159 |
2. Set $A=\{x|| x-2 |<3, x \in \mathrm{R}\}, B=\left\{x \mid x^{2}\right.$ $+(1-a) x-a<0, x \in \mathrm{R}\}$, if $B \subseteq A$, then the range of real number $a$ is ( ).
A. $[-1,5]$
B. $(-1,5)$
C. $[-1,5)$
D. $(-1,5]$ | 2. $\mathrm{A} ;$ | A | Inequalities | MCQ | Yes | Yes | inequalities | false | 737,162 |
16. Given functions $f(x), g(x)(x \in \mathrm{R})$, let the solution set of the inequality $|f(x)|+|g(x)|<a (a>0)$ be $M$, and the solution set of the inequality $|f(x)+g(x)|<a (a>0)$ be $N$, then the relationship between the solution sets $M$ and $N$ is ( ).
A. $M \subseteq N$
B. $M=N$
C. $N \varsubsetneqq M$
D. $M \v... | 1). A;
The above text has been translated into English, maintaining the original text's line breaks and format. | A | Inequalities | MCQ | Yes | Yes | inequalities | false | 737,168 |
3. The graph of the function $y=f(x)$ consists of two segments of a circle with the origin as the center and a radius of 1, as shown in Figure 6-1. Then the solution set of the inequality $f(x) > f(-x) + x$ is $(\quad)$.
A. $\left[-1,-\frac{2 \sqrt{5}}{5}\right) \cup(0,1]$
B. $[-1,0) \cup\left(0, \frac{2 \sqrt{5}}{5}\r... | 3. Ci
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | MCQ | Yes | Yes | inequalities | false | 737,173 |
21. The following size relationship is correct ( ).
A. $0.4^{3}<3^{0.4}<\log _{4} 0.3$
B. $0.4^{3}<\log _{4} 0.3<3^{0.4}$
C. $\log _{1} 0.3<0.4^{3}<3^{0.4}$
D. $\log _{4} 0.3<3^{0.4}<0.4^{3}$ | 21. C.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | MCQ | Yes | Yes | inequalities | false | 737,174 |
22. The solution set of the inequality $\frac{2 x-1}{|x+1|} \leqslant 0$ is $\qquad$ | 22. $\left\{x \mid x<-1\right.$ or $\left.-1<x \leqslant \frac{1}{2}\right\}$; | \left\{x \mid x<-1\right. \text{ or } \left.-1<x \leqslant \frac{1}{2}\right\} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,175 |
23. The solution set of the inequality $\left|\frac{x}{x-2}\right|>\frac{x}{2-x}$ is | $\begin{array}{l}\text { 23. }(-\infty, 0) \\ \cup(2,+\infty) \text {; }\end{array}$ | (-\infty, 0) \cup (2,+\infty) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,176 |
24. If there exists a positive real number $x$, such that the inequality $\frac{\ln x}{1+x} \geqslant \ln \left(\frac{k x}{1+x}\right)$ holds, then the range of real number $k$ is | 24. $0<k \leqslant 2$; | 0<k \leqslant 2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,177 |
26. Given that $y=f(x)$ is a decreasing function on the domain $(-1,1)$, and $f(1-a)<f(2a-1)$, then the range of values for $a$ is $\qquad$. | $26.0<a<\frac{2}{3}$ | 0<a<\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,179 |
27. If the inequality $a x^{2}+2 x-5 \geqslant 0$ has only one solution, then the real number $a=$ $\qquad$ . | 27. $-\frac{1}{5} ;$ | -\frac{1}{5} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,180 |
29. Given two positive numbers $x, y$ satisfying $x+y=4$, the range of real number $m$ that makes the inequality $\frac{1}{x}+\frac{4}{y} \geqslant m$ always true is $\qquad$. | 29. $\left(-\infty, \frac{9}{4}\right]$; | \left(-\infty, \frac{9}{4}\right] | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,182 |
4. The solution set of the inequality $\sqrt{5-x} \geqslant x+1$ is ( ).
A. $\{x \mid-4 \leqslant x \leqslant 1\}$
B. $\{x \mid x \leqslant-1\}$
C. $\{x \mid x \leqslant 1\}$
D. $\{x \mid-1 \leqslant x \leqslant 1\}$ | 4. C;
;
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | MCQ | Yes | Yes | inequalities | false | 737,184 |
31. Given the set $A=\left\{x \left\lvert\, \frac{x-2 a}{x-\left(a^{2}+1\right)}<0\right.\right\}, B=\{x \mid x$ $<5 a+7\}$, if $A \cup B=B$, then the range of real number $a$ is $\qquad$ . | 31. $[-1,6]$; | [-1,6] | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,185 |
34. $y=\frac{3+x+x^{2}}{1+x}(x>0)$ The minimum value is $\qquad$ | 34. $-1+2 \sqrt{3}$ | -1+2 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,188 |
36. Let the domain of the function $f(x)$ be R. If there exists a constant $m>0$ such that $|f(x)| \leqslant m|x|$ holds for all real numbers $x$, then $f(x)$ is called an $F$ function. Given the following functions:
(1) $f(x)=0$; (2) $f(x)=x^{2}$; (3) $f(x)=\sqrt{2}(\sin x + \cos x)$; (4) $f(x)=\frac{x}{x^{2}+x+1}$; (... | 36. (1)(4)(5). | (1)(4)(5) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,190 |
38. Given $\boldsymbol{a}=(1, x), \boldsymbol{b}=\left(x^{2}+x,-x\right), m$ is a constant and $m \leqslant-2$, find the range of $x$ that satisfies $\boldsymbol{a} \cdot \boldsymbol{b}+2>m\left(\frac{2}{\boldsymbol{a} \cdot \boldsymbol{b}}+1\right)$. | $$\begin{array}{l}
\text { 38. } \because \boldsymbol{a}=(1, x), \boldsymbol{b}=\left(x^{2}+x,-x\right), \\
\quad \therefore \boldsymbol{a} \cdot \boldsymbol{b}=x^{2}+x-x^{2}=x . \\
\text { Therefore, } \boldsymbol{a} \cdot \boldsymbol{b}+2>m\left(\frac{2}{\boldsymbol{a} \cdot \boldsymbol{b}}+1\right) \\
\quad \Leftrig... | (m,-2) \cup(0,+\infty) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,192 |
39. Solve the inequality with respect to $x$: $a\left|x^{2}-1\right|>a+2$ $(a<0)$ | 39. (1) When $-20, \therefore$ the solution set of the inequality is $\varnothing$.
$$\begin{array}{l}
\text { (3) When } a0, \therefore-1-\frac{2}{a}0,2+\frac{2}{a}>0, \\
\therefore \sqrt{-\frac{2}{a}}<x<\sqrt{2+\frac{2}{a}} \text { or }-\sqrt{2+\frac{2}{a}}<x
\end{array}$$
$$<-\sqrt{-\frac{2}{a}}$$
In summary, when ... | null | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,193 |
40. Given $a>b>c, a+b+c=1, a^{2}+b^{2}+c^{2}=3$, prove: $b+c<\frac{1}{2}$. | 40. From $(a+b+c)^{2}=1, a^{2}+b^{2}+c^{2}=3$, we get $ab+bc+ca=-1$. Given $a>b>c$, we know that at least $c<1$. Also, since $a>b$, we have $a+a>a+b>1 \Rightarrow a>\frac{1}{2}, \therefore b+c<\frac{1}{2}$. | b+c<\frac{1}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,194 |
5. The solution set of the inequality $f(x)=a x^{2}-x-c>0$ is $\{x \mid-2<x<1\}$, then the graph of the function $y=f(-x)$ (Figure 6-2) is $(\quad)$. | 5. B;
;
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,195 |
41. Given the function $f(x)=\log _{2}(x+m)$, and $f(0)$, $f(2)$, $f(6)$ form an arithmetic sequence. If $a$, $b$, $c$ are pairwise distinct positive numbers, and $a$, $b$, $c$ form a geometric sequence, determine the relationship between $f(a)+f(c)$ and $2 f(b)$, and prove your conclusion. | 41. Given the function $f(x)=\log _{2}(x+m)$, and $f(0)$, $f(2)$, $f(6)$ form an arithmetic sequence, so $2 f(2)=f(0)+f(6)$,
i.e., $\log _{2}(2+m)=\log _{2} m+\log _{2}(6+m)$,
solving this gives $m=2$.
Also, $a$, $b$, $c$ are distinct positive numbers, and $a$, $b$, $c$ form a geometric sequence, $\therefore b^{2}=a c... | f(a)+f(c)>2 f(b) | Algebra | proof | Yes | Yes | inequalities | false | 737,196 |
42. Does there exist a constant $c$, such that the inequality $\frac{x}{2 x+y}+\frac{y}{x+2 y}$ $\leqslant c \leqslant \frac{x}{x+2 y}+\frac{y}{2 x+y}$ holds for any positive real numbers $x, y$? Prove your conclusion. | 42. When $x=y$, from the known inequality we get $c=\frac{2}{3}$.
Below, the proof is divided into two parts:
(1) First, prove $\frac{x}{2 x+y}+\frac{y}{x+2 y} \leqslant \frac{2}{3}$,
This inequality
$\Leftrightarrow 3 x(x+2 y)+3 y(2 x+y) \leqslant 2(2 x+y)(x+2 y)$
$\Leftrightarrow 2 x y \leqslant x^{2}+y^{2}$, this i... | \frac{2}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 737,197 |
43. Given $a>0, b>0$, and $\frac{1}{a}+\frac{2}{b}=1$.
(1) Find the minimum value of $a+b$;
(2) If line $l$ intersects the $x$-axis and $y$-axis at points $A(a, 0)$, $B(0, b)$, respectively, find the minimum value of the area of $\triangle O A B$. | 43. (1) $3+2 \sqrt{2} ;(2) S_{\min }=4$ | 4 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,198 |
44. For the inequality about $x$, $\left|x-\frac{1}{2}(a+1)^{2}\right| \leqslant \frac{1}{2}(a-1)^{2}$, and $x^{2}-3(a+1) x+2(3 a+1) \leqslant 0$ where $(a \in \mathrm{R})$, the solution sets are $A$ and $B$ respectively, and $A \cap B=A$. Find the range of values for $a$. | 44. From the inequality $\left|x-\frac{1}{2}(a+1)^{2}\right| \leqslant \frac{1}{2}(a-1)^{2}$,
we get $A=\left\{x \mid 2 a \leqslant x \leqslant a^{2}+1\right\}$.
From $x^{2}-3(a+1) x+2(3 a+1) \leqslant 0(a \in R)$,
we get $(x-2)[x-(3 a+1)] \leqslant 0$.
When $a \geqslant \frac{1}{3}$, $B=\{x \mid 2 \leqslant x \leqsla... | 1 \leqslant a \leqslant 3 \text{ or } a=-1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,199 |
45. Given the proposition $p$ : the equation $a^{2} x^{2}+a x-2=0$ has a solution in $[-1$, 1]; proposition $q$ : there is only one real number $x$ that satisfies the inequality $x^{2} + 2 a x + 2 a \leqslant 0$, if the proposition " $p$ or $q$ " is false, find the range of values for $a$. | 45. From $a^{2} x^{2}+a x-2=0$, we get $(a x+2)(a x-1)=0$, obviously $a \neq 0, \therefore x=-\frac{2}{a}$ or $x=\frac{1}{a}$.
$\because x \in[-1,1]$, hence $\left|\frac{2}{a}\right| \leqslant 1$ or $\left|\frac{1}{a}\right| \leqslant 1, \therefore|a|$ $\geqslant 1$. "There is only one real number satisfying $x^{2}+2 a... | -1<a<0 \text{ or } 0<a<1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,200 |
46. Find the range of real numbers $\lambda$ such that the inequality $\left|\frac{1-a b \lambda}{a \lambda-b}\right|>1$ holds for all real numbers $a, b$ satisfying $|a|<1,|b|<1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directl... | 46. To make the inequality $\left|\frac{1-a b \lambda}{a \lambda-b}\right|>1$ hold for all real numbers $a, b$ satisfying $|a||a \lambda-b|$ for $|a|(a \lambda-b)^{2}$ for $|a|0$ for $|a|<1,|b|<1$, it is sufficient that $\lambda^{2}<\frac{1}{a^{2}}$, it is sufficient that $\lambda^{2} \leqslant 1$, it is sufficient tha... | null | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,201 |
47. In congested traffic areas, to ensure traffic safety, the distance $d$ (meters) between motor vehicles and the speed $v$ (km/h) must follow the relationship $d \geqslant \frac{1}{2500} a v^{2}$ (where $a$ (meters) is the vehicle length, and $a$ is a constant), and it is also stipulated that $d \geqslant \frac{a}{2}... | 47. (1) From $\frac{a}{2} \geqslant \frac{1}{2500} a v^{2}$, we get $025 \sqrt{2}$, $Q=-\frac{1000}{a\left(\frac{1}{v}+\frac{v}{2500}\right)} \leqslant \frac{25000}{a}$,
$\therefore$ When $v=50$, $Q$ is maximized at $\frac{25000}{a}$. | v=50, Q=\frac{25000}{a} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,202 |
48. Let the function $y=f(x)$ be a decreasing function defined on $\mathrm{R}_{+}$, and it satisfies $f(x y)=f(x)+f(y), f\left(\frac{1}{3}\right)=1$,
(1) Find the value of $f(1)$;
(2) If $f(x)+f(2-x)<2$, find the range of $x$. | \(\begin{array}{l}\text { 48. }(1) \because f(1)=f(1)+f(1), \\ \therefore f(1)=0 . \\ (2) \because f(x)+f(2-x)=f(x(2-x)), \\ \therefore f(x(2-x))=0, \\ 2-x>0, \\ x(2-x)>\frac{1}{9},\end{array} \text { solving this yields } \quad 0<x<1+\frac{\sqrt{10}}{3} .\right.\end{array}\) | 0<x<1+\frac{\sqrt{10}}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,203 |
49. A waterworks' reservoir contains 400 tons of water. The waterworks can inject 60 tons of water into the reservoir per hour, while the reservoir continuously supplies water to residential areas, with the total water supply in $t$ hours being $120 \sqrt{6 t}$ tons $(0 \leqslant t \leqslant 24)$.
(1) From the start of... | 49. (1) Let the amount of water in the reservoir after $t$ hours be $y$ tons, then $y=400+60 t-120 \sqrt{6 t}$;
Let $\sqrt{6 t}=x$; then $x^{2}=6 t$, i.e., $y=400+10 x^{2}-120 x=10(x-6)^{2}+40$, so when $x=6$, i.e., $t=6$, $y_{\text {min }}=40$, which means from the start of water supply to the 6th hour, the water lev... | 8 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,204 |
50. To perform large-scale computations using computers, people currently generally adopt the following two methods:
The first traditional method is to build a supercomputer. This method was widely used in the past, but people gradually found that building a single supercomputer is not cost-effective, as its computing... | 50. Let the invested capital be $x$ million yuan, and the computing power achieved by the two methods be $y_{1}, y_{2}$ MIPS,
then $y_{1}=k_{1} \sqrt{x}$, substituting $x=100, y_{1}=6000$ into the above equation gives $k_{1}=600$,
and $y_{2}=k_{2}(x-600)$, when $x-600=5$, $y_{2}=300$ substituting into the above equat... | 900 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,205 |
51. Given the function $f(x)=\frac{x^{2}+c}{a x+b}$ is an odd function, $f(1) < f(3)$, and the solution set of the inequality $0 \leqslant f(x) \leqslant \frac{3}{2}$ is $[-2,-1] \cup [2,4]$.
(1) Find $a, b, c$;
(2) Does there exist a real number $m$ such that the inequality $f(-2+\sin \theta) < -m^{2}+\frac{3}{2}$ hol... | 51. (1) $f(x)$ is an odd function $\Leftrightarrow f(-x)=f(x)$ for all $x$ in the domain $\Leftrightarrow b=0$, thus $f(x)=\frac{1}{a}\left(x+\frac{c}{x}\right)$. Also, $\left\{\begin{array}{l}f(2) \geqslant 0, \\ f(-2) \geqslant 0\end{array} \Leftrightarrow\left\{\begin{array}{l}f(2) \geqslant 0, \\ -f(2) \geqslant 0\... | no real number m | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,207 |
7. When $0<a<1$, which of the following is true?
A. $a^{a}>a^{a^{a}}$
B. $a^{a}>a^{a^{a}}>a$
C. $a^{a^{a}}>a>a^{a}$
D. $a^{a}>a>a^{a^{a}}$ | 7. B;
;
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Inequalities | MCQ | Yes | Yes | inequalities | false | 737,208 |
9. In $\triangle A B C$, the lengths of the three sides are $a, b, c$. If $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ form an arithmetic sequence, then the angle opposite to $b$ is ( ).
A. Acute angle
B. Right angle
C. Obtuse angle
D. Cannot be determined | 9. A ;
The above text has been translated into English, maintaining the original text's line breaks and format. | A | Geometry | MCQ | Yes | Yes | inequalities | false | 737,210 |
Example 1 If $x, y, z \in \mathrm{R}^{+}, x y z \geqslant 1$, prove:
$$\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z} \geqslant 2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) .$$ | Analysis: Transform the left expression and then apply the 3-variable mean inequality. In fact,
$$\begin{aligned}
\sum \frac{y+z}{x} & =\sum x \cdot \sum \frac{1}{x}-3 \\
& =\frac{1}{3} \sum x \cdot \sum \frac{1}{x}-3+\frac{2}{3} \sum \frac{1}{x} \cdot \sum x \\
& \geqslant 3-3+\frac{2}{3} \sum \frac{1}{x} \cdot 3 \sqr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,213 |
Example 10 Solve the equation $\cos \frac{x^{2}+x}{2}-2^{x-1}-2^{-x-1}=0$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the request is to translate the given text, here is the translation:
Example 10 Solve t... | Analysis: Transform the original equation into
$$\cos \frac{x^{2}+x}{2}=2^{x-1}+2^{-x-1} .$$
It seems difficult to proceed with further transformations, so we might consider "inequality" instead of "equality". Thus, we have
$$2^{x-1}+2^{-x-1} \geqslant 2 \sqrt{2^{x-1} \cdot 2^{-x-1}}=1$$
and $\cos \frac{x^{2}+x}{2} \... | x=0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,214 |
1. For a triangle with side lengths $a$, $b$, and $c$, its area equals $\frac{1}{4}$, and the radius of the circumscribed circle is 1. If $s=\sqrt{a}+\sqrt{b}+\sqrt{c}$ and $t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, then the relationship between $s$ and $t$ is ( ).
A. $s>t$
B. $s<t$
C. $s=t$
D. Uncertain | 1. B;
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | inequalities | false | 737,215 |
3. The maximum area of an inscribed trapezoid with the major axis of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ as its base is ( ).
A. $\frac{\sqrt{3}}{2} a b$
B. $\frac{3 \sqrt{3}}{4} a b$
C. $\frac{\sqrt{3}}{6} a^{2}$
D. $\frac{\sqrt{3}}{8} a^{2}$ | 3. B;
;
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | inequalities | false | 737,217 |
4. The number of elements in the set $\left((x, y) \left\lvert\, \lg \left(x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}\right)=\lg x+\lg y\right.\right\}$ is ( ).
A. 0
B. 1
C. 2
D. More than 2 | 4. B;
;
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | inequalities | false | 737,218 |
6. Let $0<\beta<\alpha<\frac{\pi}{2}$, and $\operatorname{tg} \alpha=2 \operatorname{tg} \beta$, then the range of $\alpha-\beta$ is $\qquad$ . | 6. $\left(0, \frac{\pi}{6}\right]$ | \left(0, \frac{\pi}{6}\right] | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,220 |
Example 2 Given positive real numbers $a, b, c$ satisfying $a+b+c=1$, prove that: $a^{2}+b^{2}+c^{2}+2 \sqrt{3 a b c} \leqslant 1$. | Analysis: By the 2-variable mean inequality, we have
$$\begin{array}{l}
a^{2} b^{2}+b^{2} c^{2} \geqslant 2 a b^{2} c, \\
b^{2} c^{2}+c^{2} a^{2} \geqslant 2 a b c^{2}, \\
c^{2} a^{2}+a^{2} b^{2} \geqslant 2 a^{2} b c \text {. }
\end{array}$$
Adding them up, we get $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \geqslant a b^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,222 |
Example 3 Let the three sides of $\triangle A B C$ be $a, b, c$. Prove that
$$\begin{array}{l}
\frac{a}{\sqrt{2 b^{2}+2 c^{2}-a^{2}}}+\frac{b}{\sqrt{2 c^{2}+2 a^{2}-b^{2}}} \\
+\frac{c}{\sqrt{2 a^{2}+2 b^{2}-c^{2}}} \geqslant \sqrt{3} \text {. }
\end{array}$$ | Analysis: By using local substitution and then applying the 2-variable mean inequality to prove it.
$$\begin{array}{l}
\text { Let } x=\frac{a}{\sqrt{2 b^{2}+2 c^{2}-a^{2}}}, y=\frac{b}{\sqrt{2 c^{2}+2 a^{2}-b^{2}}} \\
z= \frac{c}{\sqrt{2 a^{2}+2 b^{2}-c^{2}}},
\end{array}$$
Then $x^{2}=\frac{a^{2}}{2 b^{2}+2 c^{2}-a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,223 |
Example 4 Given $x>-1$, and $x \neq 0, 2 \leqslant n \in \mathrm{N}$, prove:
$$(1+x)^{n}>1+n x .$$ | $$\begin{array}{l}
\because \quad x \neq 0 \Rightarrow 1+x \neq 1, \\
\therefore \quad(1+x)^{n}+(n-1) \\
=(1+x)^{n}+1+1+\cdots+1 \\
>n \quad \sqrt[n]{(1+x)^{n} \cdot 1^{n-1}} \\
=n(1+x), \\
\therefore \quad(1+x)^{n}>1+n x .
\end{array}$$
Analysis: This is the Bernoulli's Inequality. The proof below uses the n-variable... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,224 |
Example 5 Let $0<\theta<\pi$, find the maximum value of $y=\sin \frac{\theta}{2}(1+\cos \theta)$. | Analysis: Eliminate the trigonometric function through substitution.
Let $\sin \frac{\theta}{2}=x, x \in(0,1)$, then
$$y=2 \sin \frac{\theta}{2}\left(1-\sin ^{2} \frac{\theta}{2}\right)=2 x\left(1-x^{2}\right)$$
Transforming, we get
$$\begin{aligned}
1 & =x^{2}+\frac{y}{2 x} \\
& =x^{2}+\frac{y}{4 x}+\frac{y}{4 x} \\
... | \frac{4 \sqrt{3}}{9} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,225 |
Example 6 Given $x, y, z \in \mathrm{R}^{+}, x+y+z=1$, find the minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
This line is not translated as it is a note or instruction and should remain in the original language as per the request. | $$\begin{array}{l}
\text { Analysis: } \because x, y, z \in \mathbb{R}^{+}, x+y+z=1, \\
\therefore \quad \frac{1}{x}+\frac{4}{y}+\frac{9}{z} \\
=(x+y+z)\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right) \\
=\frac{y+z}{x}+\frac{4 z+4 x}{y}+\frac{9 x+9 y}{z}+14 \\
=\left(\frac{y}{x}+\frac{4 x}{y}\right)+\left(\frac{4 z}{y}... | 36 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,226 |
Example 7 Let real numbers $a$, $b$, $c$ satisfy
$$\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.$$
Find the range of real number $a$. | $$\begin{array}{l}
\left\{\begin{array}{l}
b^{2}+c^{2}=-a^{2}+14 a-13, \\
2 b c=2 a^{2}-16 a+14
\end{array}\right. \\
\because \quad b^{2}+c^{2} \geqslant 2 b c, \\
\therefore \quad-a^{2}+14 a-13 \geqslant 2 a^{2}-16 a+14,
\end{array}$$
that is, $a^{2}-10 a+9 \leqslant 0$,
thus $1 \leqslant a \leqslant 9$.
To find the... | 1 \leqslant a \leqslant 9 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,227 |
Example 8 If $a, b, c$ are the length, width, and height of a rectangular prism, and $a+b-$ $c=1$, it is known that the length of the diagonal of the rectangular prism is 1, and $a>b$, try to find the range of values for the height $c$.
The above text is translated into English, please retain the original text's line ... | Analysis: From the formula for the length of the diagonal of a cuboid, we get $a^{2}+b^{2}+c^{2}=1$
Thus $\left\{\begin{array}{l}a^{2}+b^{2}=1-c^{2}, \\ a+b=1+c .\end{array}\right.$
$\because\left(\frac{a+b}{2}\right)^{2}b)$,
$\therefore\left(\frac{1+c}{2}\right)^{2}0)$,
Solving, we get $0<c<\frac{1}{3}$.
The transfor... | 0<c<\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,228 |
Example 9 Find all integer pairs $(x, y)$ that satisfy $\left(x^{2}+y^{2}\right)(x+y-3)=2 x y$.
The equation to solve is $\left(x^{2}+y^{2}\right)(x+y-3)=2 x y$. | Analysis: $\because x^{2}+y^{2} \geqslant 2|x y|$, the equality holds when $x= \pm y$, and at the same time, $x+y-3$ is an integer,
$\therefore$ for the original equation to hold, there should be the following four cases:
$$\left\{\begin{array} { l }
{ x = y , } \\
{ x + y - 3 = 1 ; } \\
{ x y = 0 , } \\
{ x + y - 3 =... | (2,2),(0,3),(3,0),(0,0) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,229 |
Proposition If $h_{a} 、 h_{b} 、 h_{i} 、 r$ are the lengths of the altitudes and the inradius of a triangle, $0<\lambda \leqslant 2$, then
$$\frac{1}{h_{a}-\lambda r}+\frac{1}{h_{b}-\lambda r}+\frac{1}{h_{i}-\lambda r} \geqslant \frac{3}{(3-\lambda) r},$$
with equality if and only if the triangle is equilateral. | Proof: $\because \frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}=\frac{1}{r}$,
$$\therefore \frac{h_{a}-\lambda r}{h_{a}}+\frac{h_{b}-\lambda r}{h_{b}}+\frac{h_{c}-\lambda r}{h_{c}}=3-\lambda$$
Thus, $\frac{h_{a}}{h_{a}-\lambda r}+\frac{h_{b}}{h_{b}-\lambda r}+\frac{h_{c}}{h_{c}-\lambda r} \geqslant \frac{9}{3-\lambda... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,230 |
Question 1 Let $x, y, z$ be non-negative real numbers, prove:
$$\begin{array}{l}
3 x y z+x^{3}+y^{3}+z^{3} \geqslant 2\left((x y)^{\frac{3}{2}}+(y z)^{\frac{3}{2}}+\right. \\
\left.(z x)^{\frac{3}{2}}\right) \text {. }
\end{array}$$ | Prove that using the variant (3) of Schur's inequality and the AM-GM inequality, we have
$$3 x y z+\sum_{\text {cyclic }} x^{3} \geqslant \sum_{\text {cyclic }}\left(x^{2} y+x y^{2}\right) \geqslant \sum_{\text {cyclic }} 2(x y)^{\frac{3}{2}}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,232 |
Question 2 Let $a, b, c$ be positive real numbers, and $a b+b c+c a=$ 3, prove:
$$a^{3}+b^{3}+c^{3}+6 a b c \geqslant 9 .$$ | Prove that from the variant (3) we can get
$$a^{3}+b^{3}+c^{3}+6 a b c \geqslant(a+b+c)(a b+b c+c a)$$
We know
$$\begin{array}{c}
\quad(a+b+c)^{2} \geqslant 3(a b+b c+c a)=9 \\
\Rightarrow a+b+c \geqslant 3, \text { so } a^{3}+b^{3}+c^{3}+6 a b c \geqslant 9 .
\end{array}$$
The equality in the original inequality hol... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,233 |
Question 11 Let $x, y, z$ be positive real numbers, prove:
$$\frac{x^{3}+y^{3}+z^{3}}{3 x y z}+\frac{3 \sqrt[3]{x y z}}{x+y+z} \geqslant 2$$ | To prove in (7), we replace $x, y, z$ with $\sqrt{x}, \sqrt{y}, \sqrt{z}$, obtaining
$$x+y+z+3 \sqrt[3]{x y z} \geqslant 2(\sqrt{x y}+\sqrt{y z}+\sqrt{z x})$$
or $3 \sqrt[3]{x y z} \geqslant 2(\sqrt{x y}+\sqrt{y z}+\sqrt{z x})-(x+y+z)$. We have $\frac{3 \sqrt[3]{x y z}}{x+y+z} \geqslant$
$$\frac{2(\sqrt{x y}+\sqrt{y z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,234 |
Question 12 Prove: For any positive real numbers $a, b, c$, the following inequality holds
$$\begin{array}{l}
\quad \frac{a+b+c}{3}-\sqrt[3]{a b c} \leqslant \max \left\{(\sqrt{a}-\sqrt{b})^{2},(\sqrt{b}-\right. \\
\left.\sqrt{c})^{2},(\sqrt{c}-\sqrt{a})^{2}\right\}
\end{array}$$ | Proof 1 is obvious:
$$\begin{array}{l}
\frac{(\sqrt{a}-\sqrt{b})^{2}+(\sqrt{b}-\sqrt{c})^{2}+(\sqrt{c}-\sqrt{a})^{2}}{3} \leqslant \\
\max \left\{(\sqrt{a}-\sqrt{b})^{2},(\sqrt{b}-\sqrt{c})^{2},(\sqrt{c}-\sqrt{a})^{2}\right\}
\end{array}$$
We prove a stronger inequality:
$$\begin{aligned}
& a+b+c-3 \sqrt[3]{a b c} \le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,235 |
Question 3 (IMO 1984) Given that $x, y, z$ are non-negative real numbers satisfying $x + y + z = 1$, prove:
$$0 \leqslant x y + y z + z x - 2 x y z \leqslant \frac{7}{27}$$ | Prove the first part first:
$$\begin{array}{c}
x y+y z+z x-2 x y z \\
=(x+y+z)(x y+y z+z x)-2 x y z \\
=x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2}+x y z
\end{array}$$
Next, prove the second part:
$$\begin{array}{l}
\quad x y+y z+z x-2 x y z \leqslant \frac{7}{27} \\
\quad \Leftrightarrow(x+y+z)(x y+y z+z x)-2 x y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,236 |
Question 5 Find all positive integers $k$, such that for any positive numbers $a, b, c$ satisfying $abc=1$, the following inequality holds:
$$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+3 k \geqslant(k+1)(a+b+c)$$ | Let $a=b=\frac{1}{n+1}, c=(n+1)^{2}\left(n \in \mathbf{N}^{*}\right)$,
from (11) we get
$$k \leqslant \frac{n^{2}+2 n+1+\frac{1}{(n+1)^{4}}-\frac{2}{n+1}}{n^{2}+2 n+\frac{2}{n+1}-1}$$
and $\lim _{n+\infty} \frac{n^{2}+2 n+1+\frac{1}{(n+1)^{4}}-\frac{2}{n+1}}{n^{2}+2 n+\frac{2}{n+1}-1}=1$,
so $k \leqslant 1$, since $k$... | k=1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,238 |
Problem 6 (2008 IMO China National Training Team Test) Let $x, y, z \in \mathbf{R}^{+}$, prove that:
$$\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}>2 \sqrt[3]{x^{3}+y^{3}+z^{3}} .$$ | Prove that if $\frac{x y}{z}=a^{2}, \frac{y z}{x}=b^{2}, \frac{z x}{y}=c^{2}$, then $y=a b$, $z=b c, x=c a$. The original inequality is equivalent to
$$\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)^{3}>8\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}\right) . \\
\text { Left side }=\sum a^{6}+3 \sum\left(a^{4} b^{2}+a^{2} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,239 |
Question 7 (2004 Asia Pacific Mathematical Olympiad) Let $a, b, c$ be positive real numbers, prove that:
$$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geqslant 9(a b+b c+c a)$$ | $$\begin{array}{l}
\text { Prove that (14) is equivalent to } \\
a^{2} b^{2} c^{2}+2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+4\left(a^{2}+b^{2}+\right. \\
\left.c^{2}\right)+8 \geqslant 9(a b+b c+c a) .
\end{array}$$
We know that $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$;
$$\begin{array}{l}
\quad\left(a^{2} b^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,240 |
Question 8 For any positive numbers $a, b, c$, prove that the following inequality holds:
$$a^{2}+b^{2}+c^{2}+2 a b c+1 \geqslant 2(a b+b c+c a)$$ | Prove that the solution uses a variant of Schur's inequality (6) $2(a b+b c+c a)-\left(a^{2}+b^{2}+c^{2}\right) \leqslant \frac{9 a b c}{a+b+c}$ and the AM-GM inequality
$$2 a b c+1=a b c+a b c+1 \geqslant 3 \sqrt[3]{a^{2} b^{2} c^{2}}$$
From (16) and (17), inequality (15) is transformed into
$$3 \sqrt[3]{a^{2} b^{2} c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,241 |
Question 9 (2001 IMO Romanian National Team Selection Test) Let $a, b, c$ be positive real numbers, prove:
$$\sum_{c \text { cyclic }}(b+c-a)(c+a-b) \leqslant \sqrt{a b c}(\sqrt{a}+\sqrt{b}+$$
$$\sqrt{c})$$ | To prove that through simple calculation we can verify
$$\begin{array}{l}
\quad \sum_{\text {cyclic }}(b+c-a)(c+a-b)=2(a b+b c+c a) \\
-\left(a^{2}+b^{2}+c^{2}\right)
\end{array}$$
From (7), we only need to prove
$$3 \sqrt[3]{a^{2} b^{2} c^{2}} \leqslant \sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c})$$
This can be obtained... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,242 |
Question 10 (2004 Romanian National Mathematical Olympiad Preliminary) Let $a, b, c$ be positive real numbers, prove:
$$\begin{array}{r}
\sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c})+(a+b+c)^{2} \geqslant \\
4 \sqrt{3 a b c(a+b+c)}
\end{array}$$ | Proof From problem 9, we have
$$\begin{aligned}
& \sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c}) \geqslant 2(a b+b c+c a)-\left(a^{2}\right. \\
+ & \left.b^{2}+c^{2}\right)
\end{aligned}$$
Now, we only need to prove
$$a b+b c+c a \geqslant \sqrt{3 a b c(a+b+c)}$$
This is obviously true. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,243 |
Theorem 1 Let $\mathrm{M} \in M_{n}(C), \lambda_{I}, \lambda_{2}, \cdots, \lambda_{n}$ be the eigenvalues of $M$, $M \neq 0$, and partition the matrix $M$ as in (2). If $\|B\|_{F} \neq 0$, $\|C\|_{F} \neq 0$, then
$$\sum_{i=1}^{n}\left|\lambda_{i}\right|^{2} \leq\|M\|_{F}^{2}-\max _{1 \leq k \leq n-1}\left\{\left(\|B\|... | Prove that if $\|B\|_{F} \neq 0, \|C\|_{F} \neq 0$, let
$$P=\left(\begin{array}{cc}
p I_{k \times k} & 0 \\
0 & I_{(n-k) \times (n-k)}
\end{array}\right)$$
then $M_{1}=P M P^{-1}$, and
$$\begin{aligned}
\left\|M_{1}\right\|_{F}^{2} & =\|A\|_{F}^{2}+p^{2}\|B\|_{F}^{2}+\frac{1}{p^{2}}\|C\|_{F}^{2}+\|D\|_{F}^{2} \\
& =\|... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,244 |
Schur's Inequality ${ }^{[1]}$ Let $x, y, z$ be non-negative real numbers, then
$$x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y) \geqslant 0,$$
with equality holding if and only if $x=y=z$.
For the sake of simplicity in writing, we use $\sum$ to denote cyclic sums, then Schur's Inequality can be briefly written as:
$$\sum x^{3}-... | To prove that since the left side of the equation to be proven is a symmetric polynomial in $x, y, z$, we can assume without loss of generality that $x \geqslant y \geqslant z \geqslant 0$. Thus,
$$\begin{aligned}
\sum x^{3}-\sum x^{2}(y+z)+3 x y z & =\sum x(x-y)(x-z) \geqslant y(x-y)(x-z)+y(y-x)(y-z) \\
& =y(x-y)^{2} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,247 |
Example 1 Let $a, b, c \in R^{+}$, prove: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}$.
(1963 Moscow Competition Problem) | To prove that the inequality can be transformed into $\sum a(c+a)(b+a) \geqslant \frac{3}{2} \Pi(a+b)$ (Note: $\Pi$ represents the cyclic product).
$$\begin{aligned}
\text { and } \sum a(c+a)(b+a)-\frac{3}{2} \prod(a+b) & =\left[\sum a^{3}+\sum a^{2}(b+c)+3 a b c\right]-\frac{3}{2}\left[\sum a^{2}(b+c)+2 a b c\right] ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,250 |
Example 2 Given that $x, y, z$ are non-negative real numbers, and $x+y+z=1$. Prove that: $0 \leqslant y z+z x+x y-2 x y z \leqslant \frac{7}{27}$.
(25th IMO Problem) | Prove on one hand: $y z+z x+x y-2 x y z=(y z+z x+x y)(x+y+z)-2 x y z$
$$=\sum x^{2}(y+z)+x y z \geqslant 0$$
On the other hand: 7-27 $(y z+z x+x y-2 x y z)=7(x+y+z)^{3}-27[(y z+z x+x y)(x+y+z)-2 x y z]$
$$\begin{array}{l}
=7 \sum x^{3}-6 \sum x^{2}(y+z)+15 x y z \\
\left.\geqslant 6 \sum x^{3}-6 \sum x^{2}(y+z)+18 x y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,251 |
Example 3 Let $a, b, c$ be non-negative real numbers, and $abc=1$, prove that: $\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1$.
(41st IMO Problem) | Without loss of generality, let $a=\frac{x}{y}, b=\frac{y}{z}$, then $c=\frac{z}{x}$ (where $x, y, z \in R^{+}$).
The inequality to be proven is: $(x-y+z)(y+z-x)(z-x+y) \leqslant x y z$.
By $x y z-(x-y+z)(y+z-x)(z-x+y)=\sum x^{3}-\sum x^{2}(y+z)+3 x y z$
and (1), it follows that $(x-y+z)(y+z-x)(z-x+y) \leqslant x y z$.... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,252 |
Example 1 Let $a, b, c \in R^{+}$, prove: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}$. | To prove that the inequality can be transformed into $\sum a(c+a)(b+a) \geqslant \frac{3}{2} \prod(a+b)$ (Note: $\prod$ represents the cyclic product).
$$\begin{aligned}
\text { and } \sum a(c+a)(b+a)-\frac{3}{2} \prod(a+b) & =\left[\sum a^{3}+\sum a^{2}(b+c)+3 a b c\right]-\frac{3}{2}\left[\sum a^{2}(b+c)+2 a b c\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,256 |
Example 2 Given that $x, y, z$ are non-negative real numbers, and $x+y+z=1$. Prove: $0 \leqslant y z+z x+x y-2 x y z \leqslant \frac{7}{27}$. | Prove on one hand: $y z+z x+x y-2 x y z=(y z+z x+x y)(x+y+z)-2 x y z$
$$=\sum x^{2}(y+z)+x y z \geqslant 0$$
$$\begin{array}{l}
=7 \sum x^{3}-6 \sum x^{2}(y+z)+15 x y z \\
\left.\geqslant 6 \sum x^{3}-6 \sum x^{2}(y+z)+18 x y z \text { (since } \sum x^{3} \geqslant 3 x y z\right) \\
\geqslant 0 \text { (by (1) ). }
\en... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,257 |
Example 3 Let $a, b, c$ be non-negative real numbers, and $abc=1$, prove that: $\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1$.
| Without loss of generality, let $a=\frac{x}{y}, b=\frac{y}{z}$, then $c=\frac{z}{x}$ (where $x, y, z \in R^{+}$).
The inequality to be proven is: $(x-y+z)(y+z-x)(z-x+y) \leqslant x y z$.
By $x y z-(x-y+z)(y+z-x)(z-x+y)=\sum x^{3}-\sum x^{2}(y+z)+3 x y z$
and (1), it follows that $(x-y+z)(y+z-x)(z-x+y) \leqslant x y z$.... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,258 |
Example 1 (2000 - IMO) Let $a, b, c$ be positive real numbers, and satisfy $abc=1$. Prove:
$$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1$$ | Prove that given $abc=1$, we can let $a=\frac{y}{z}, b=\frac{z}{x}, c=\frac{x}{y}, x, y, z>0$, then $\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1 \Leftrightarrow$ $(y-z+x)(z-x+y)(x-y+z) \leqslant xyz$, which is the Schur inequality for $r=1$.
Given $abc=1$, we can se... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,263 |
Example 2 (2008 - Adapted from China National Training Team Test) If $\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}>M \sqrt[3]{x^{3}+y^{3}+z^{3}}$ holds for any positive real numbers $x, y, z$, find the maximum value of the real number $M$.
保持源文本的换行和格式如下：
Example 2 (2008 - Adapted from China National Training Team Test) ... | Let $x=y=t>0, z=1$, then $t^{2}+2 > M \sqrt[3]{2 t^{3}+1}$, i.e., $M < \frac{t^{2}+2}{\sqrt[3]{2 t^{3}+1}}$. Let $f(t) = \frac{t^{2}+2}{\sqrt[3]{2 t^{3}+1}}$, it is easy to know that $f(t)$ is increasing on $\left(0, \frac{-1+\sqrt{5}}{2}\right)$, decreasing on $\left(\frac{-1+\sqrt{5}}{2}, 1\right)$, and increasing on... | 2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,264 |
Example 4 Given the sides of $\triangle A B C$ are $a, b, c$, and the semi-perimeter is $p$, prove: $\sum a^{3}(p-a) \leqslant a b c p$, equality holds if and only if $\triangle A B C$ is an equilateral triangle. | Prove $\quad \sum a^{3}(p-a) \leqslant a b c p \Leftrightarrow \sum a^{2}(a-b) (a-c) \geqslant 0$ (Schur's inequality for $r=2$ case) is proved. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,265 |
Example 5 (Problem 1830 from "Mathematics Bulletin") $a, b, c \in \mathbf{R}_{+}$, and $a+b+c=2$, prove:
$$\sum\left(\frac{1-a}{a}\right)\left(\frac{1-b}{b}\right) \geqslant \frac{3}{4}$$ | $$\begin{array}{l}
\text { Prove } \quad \sum\left(\frac{1-a}{a}\right)\left(\frac{1-b}{b}\right) \geqslant \frac{3}{4} \\
\Leftrightarrow \sum\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right) \geqslant \frac{3}{4} \\
\Leftrightarrow \sum\left(\frac{1}{a b}-\frac{2}{c}\right)+\frac{9}{4} \geqslant 0 \\
\Leftrightarr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,266 |
Example 6 (adapted from a problem in *Crux Mathematicorum*) Let $\lambda (\lambda > -1)$ be a real constant, and $x, y, z$ be non-negative real numbers, satisfying $\sum x = 1, \sum xy > 0$. Find the maximum value of $M$ such that the inequality $(1+\lambda x)(1+\lambda y)(1+\lambda z) \geqslant M(1-x)(1-y)(1-z)$ alway... | Given $x, y, z \in [0,1)$, let $F(x, y, z)=$
$$\frac{(1+\lambda x)(1+\lambda y)(1+\lambda z)}{(1-x)(1-y)(1-z)}=\lambda^{2}+(\lambda+1)$$
$\frac{\lambda^{2} x y z+1}{x y+y z+z x-x y z}$, the problem is to find the minimum value of $F(x, y, z)$.
We conjecture that $F(x, y, z) \geqslant F\left(\frac{1}{3}, \frac{1}{3}, \... | M_{\text {max }}=\begin{cases} \frac{(\lambda+3)^{3}}{8}, & \text{if } \lambda^{2} \leqslant 5 \\ (\lambda+2)^{2}, & \text{if } \lambda^{2} > 5 \end{cases} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,267 |
Example 7 (Problem 1421 from "Mathematics Bulletin") In an acute $\triangle ABC$, prove that:
$$\cos (A-B) \cos (B-C) \cos (C-A) \geqslant 8 \cos A \cos B \cos C$$ | Prove that in $\triangle ABC$, it always holds that $\tan A + \tan B + \tan C = \tan A \tan B \tan C$,
then
\[
\frac{\cos C}{\cos (A-B)} = -\frac{\cos A \cos B - \sin A \sin B}{\cos A \cos B + \sin A \sin B} = \frac{\tan A \tan B - 1}{\tan A \tan B + 1} = \frac{\tan A \tan B \tan C - \tan C}{\tan A \tan B \tan C + \t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,268 |
Professor An Zhenping proposed "Twenty-six Beautiful Inequalities" in the January-February 2010 issue of the *Teaching Reference for Middle School Mathematics*. Now, using the Schur inequality, we will prove the 5th and 18th beautiful inequalities.
The 5th beautiful inequality: Let $x, y, z$ be positive real numbers, ... | Prove that the original inequality is equivalent to $4 x(y+z x)(z+x y)+4 y(x+y z)(z+x y)+$ $4 z(x+y z)(y+z x) \leqslant 9(x+y z)(y+z x)(z+x y)$, simplifying and organizing yields $x^{2} y^{2}+$ $y^{2} z^{2}+z^{2} x^{2}+5 x y z\left(x^{2}+y^{2}+z^{2}\right)+9 x^{2} y^{2} z^{2} \geqslant 3 x y z$.
Since $x+y+z=1$, the i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,271 |
In this article, the symbol $\sum f(x, y, z)$ denotes cyclic summation.
The 18th beautiful inequality: Let $x, y, z$ be positive real numbers, and satisfy $x+y+z=1$. Prove that $\frac{x}{2 x^{2}+y^{2}+z^{2}}+\frac{y}{x^{2}+2 y^{2}+z^{2}}+\frac{z}{x^{2}+y^{2}+2 z^{2}} \leqslant \frac{9}{4}$. | $$\begin{array}{l}
\text { Prove that the original inequality is equivalent to } 4 x\left(x^{2}+2 y^{2}+z^{2}\right)\left(x^{2}+y^{2}+2 z^{2}\right)+4 y\left(2 x^{2}+y^{2}+\right. \\
\left.z^{2}\right)\left(x^{2}+y^{2}+2 z^{2}\right)+4 z\left(2 x^{2}+y^{2}+z^{2}\right)\left(x^{2}+2 y^{2}+z^{2}\right) \leqslant 9\left(2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,272 |
Example 1 (IMO US National Team Training Question) Prove: In any acute triangle $ABC$, we have $\cot ^{3} A+\cot ^{3} B+\cot ^{3} C$ $+6 \cot A \cot B \cot C \geqslant \cot A+\cot B+\cot C$. | Prove: Let $\cot A=x, \cot B=y, \cot C=z$, since $x y$ $+y z+z x=1$, it is sufficient to prove the following homogeneous inequality: $x^{3}+y^{3}+z^{3}+6 x y z \geqslant(x+y+z)(x y+y z+z x)$. This is equivalent to a variant of Schur's inequality (2). | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,273 |
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