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Example 7 Given $x, y, z \in \mathbf{R}^{+}$, prove:
$$\frac{x z}{x^{2}+x z+y z}+\frac{x y}{y^{2}+x y+x z}+\frac{y z}{z^{2}+y z+x y} \leqslant 1$$ | Proof: Assume $x y z=1$.
First, prove $\quad \frac{x z}{x^{2}+x z+y z} \leqslant \frac{1}{1+y+\frac{1}{z}}$.
(1) is equivalent to
$$x^{2}+x z+y z \geqslant x z+1+x$$
which is $x^{2}+\frac{1}{x} \geqslant 1+x$, or $(x-1)^{2}(x+1) \geqslant 0$. Hence (1) holds.
Also, since
$$\frac{1}{1+y+\frac{1}{z}}=\frac{z}{y z+z+1}$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,291 |
Example 8 Let real numbers $a, b, c$ satisfy: $a+b+c=3$. Prove:
$$\frac{1}{5 a^{2}-4 a+11}+\frac{1}{5 b^{2}-4 b+11}+\frac{1}{5 c^{2}-4 c+11} \leqslant \frac{1}{4}$$ | Prove that if $a, b, c$ are all less than $\frac{9}{5}$, then it can be proven that
$$\frac{1}{5 a^{2}-4 a+11} \leqslant \frac{1}{24}(3-a)$$
In fact,
$$\text { (1) } \begin{aligned}
& \Leftrightarrow(3-a)\left(5 a^{2}-4 a+11\right) \geqslant 24 \\
& \Leftrightarrow 5 a^{3}-19 a^{2}+23 a-9 \leqslant 0 \\
& \Leftrightar... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,292 |
Example 9 Let $n(\geqslant 3)$ be an integer, prove that for positive real numbers $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$, the inequality
$$\frac{x_{n} x_{1}}{x_{2}}+\frac{x_{1} x_{2}}{x_{3}}+\cdots+\frac{x_{n-1} x_{n}}{x_{1}} \geqslant x_{1}+x_{2}+\cdots+x_{n}$$
holds. | First, we prove a lemma: If \(0 < x \leqslant y, 0 < a \leqslant 1\), then
\[ x + y \leqslant a x + \frac{y}{a} \]
In fact, from \(a x \leqslant x \leqslant y\) we get \((1 - a)(y - a x) \geqslant 0\), i.e.,
\[ a^2 x + y \geqslant a x + a y \]
Therefore,
\[ x + y \leqslant a x + \frac{y}{a} \]
Now, let \((x, y, a) =... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,293 |
Example 10 Let $n$ be a positive integer, and $n \geqslant 3$. Also, let $a_{1}, a_{2}, \cdots, a_{n}$ be real numbers, where $2 \leqslant a_{i} \leqslant 3, i=1,2, \cdots, n$. If we take $S=a_{1}+a_{2}+\cdots+a_{n}$, prove that:
$$\frac{a_{1}^{2}+a_{2}^{2}-a_{3}^{2}}{a_{1}+a_{2}-a_{3}}+\frac{a_{2}^{2}+a_{3}^{2}-a_{4}^... | Prove that
\[
\frac{a_{i}^{2}+a_{i+1}^{2}-a_{i+2}^{2}}{a_{i}+a_{i+1}-a_{i+2}}=a_{i}+a_{i+1}+a_{i+2}-\frac{2 a_{i} a_{i+1}}{a_{i}+a_{i+1}-a_{i+2}}.
\]
Notice that \(1=2+2-3 \leqslant a_{i}+a_{i+1}-a_{i+2} \leqslant 3+3-2=4\), and from \((a_{i}-2)(a_{i+1}-2) \geqslant 0\) we get \(-2 a_{i} a_{i+1} \leqslant-4(a_{i}+a_{i+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,294 |
Let $x_{1}, x_{2}, \cdots, x_{n}$ all be positive real numbers, prove that:
$$\begin{aligned}
& \frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}} \\
\geqslant & 2\left(\frac{1}{x_{1}+x_{2}}+\frac{1}{x_{2}+x_{3}}+\cdots+\frac{1}{x_{n-1}+x_{n}}+\frac{1}{x_{n}+x_{1}}\right) .
\end{aligned}$$ | 1. It is not difficult to prove: $\frac{1}{x_{i}}+\frac{1}{x_{i+1}} \geqslant \frac{4}{x_{i}+x_{i+1}}\left(i=1,2, \cdots, n, x_{n+1}=x_{1}\right)$, adding up $n$ inequalities yields the original inequality holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,295 |
2 Let $a, b, c \in \mathbf{R}^{+}$, prove:
$$\frac{1}{a^{3}+b^{3}+a b c}+\frac{1}{b^{3}+c^{3}+a b c}+\frac{1}{c^{3}+a^{3}+a b c} \leqslant \frac{1}{a b c} .$$ | 2. Since $a^{3}+b^{3} \geqslant a^{2} b+b^{2} a$, it follows that $\frac{1}{a^{3}+b^{3}+a b c} \leqslant \frac{1}{a b(a+b+c)}$. Similarly, $\frac{1}{b^{3}+c^{3}+a b c} \leqslant \frac{1}{b c(a+b+c)}$; $\frac{1}{c^{3}+a^{3}+a b c} \leqslant \frac{1}{c a(a+b+c)}$. Adding the three inequalities yields the original inequal... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,296 |
Given $0 \leqslant x, y, z \leqslant 1$, solve the equation:
$$\frac{x}{1+y+z x}+\frac{y}{1+z+x y}+\frac{z}{1+x+y z}=\frac{3}{x+y+z} .$$ | 3. It is not difficult to prove: $\frac{x}{1+y+z x} \leqslant \frac{1}{x+y+z}$, $\frac{y}{1+z+x y} \leqslant \frac{1}{x+y+z}$, $\frac{z}{1+x+y z} \leqslant \frac{1}{x+y+z}$. Therefore, if the equality holds, it is easy to get $x=y=z=1$.
Note: This problem can also be solved as follows: First prove $\frac{x}{1+y+z x} \... | x=y=z=1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,297 |
4 Let $a, b, c \in \mathbf{R}^{+}$, and $a b c=1$. Prove:
$$\sum_{\mathrm{oc}} \frac{a b}{a^{5}+b^{5}+a b} \leqslant 1$$
and ask when the equality holds? | 4. Since $a^{5}+b^{5}-a^{2} b^{2}(a+b)=\left(a^{2}-b^{2}\right)\left(a^{3}-b^{3}\right) \geqslant 0$, it follows that $a^{5}+b^{5} \geqslant a^{2} b^{2}(a+b)$, thus
$$\begin{aligned}
\frac{a b}{a^{5}+b^{5}+a b} & =\frac{a b \cdot a b c}{a^{5}+b^{5}+a b \cdot a b c}=\frac{a^{2} b^{2} c}{a^{5}+b^{5}+a^{2} b^{2} c} \\
& \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,298 |
51 Given $\alpha, \beta>0, x, y, z \in \mathbf{R}^{+}, x y z=2004$. Find the maximum value of $u$, where, $u=$
$$\sum_{\mathrm{oc}} \frac{1}{2004^{\alpha+\beta}+x^{\alpha}\left(y^{2 \alpha+3 \beta}+z^{2 \alpha+3 \beta}\right)} .$$ | $$\begin{array}{c}
\text { 5. First, it is not difficult to prove that } y^{2 \alpha+3 \beta}+z^{2 \alpha+3 \beta} \geqslant y^{\alpha+2 \beta} z^{\alpha+\beta}+y^{\alpha+\beta} z^{2 \alpha+\beta} . \\
\text { Also, since } 2004^{\alpha+\beta}+x^{\alpha}\left(y^{2 \alpha+3 \beta}+z^{2 \alpha+3 \beta}\right)=x^{\alpha}\... | 2004^{-(\alpha+\beta)} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,299 |
6 Let $x_{1}, x_{2}, \cdots, x_{n}$ all be positive numbers, and $x_{1}+x_{2}+\cdots+x_{n}=a$. For $m, n \in \mathbf{Z}^{+}, m$, $n>1$, prove: $\frac{x_{1}^{m}}{a-x_{1}}+\frac{x_{2}^{m}}{a-x_{2}}+\cdots+\frac{x_{n}^{m}}{a-x_{n}} \geqslant \frac{a^{m-1}}{(n-1) n^{m-2}}$. | 6. By the AM-GM inequality, $\frac{x_{i}^{m}}{a-x_{i}}+\frac{\left(a-x_{i}\right) a^{m-2}}{(n-1)^{2} n^{n-2}}+\underbrace{\frac{a^{m-1}}{(n-1) n^{n-1}}+\cdots+\frac{a^{n-1}}{(n-1) n^{n-1}}}_{m-2 \uparrow} \geqslant$
$$\begin{array}{l}
m \cdot \frac{x_{i} \cdot a^{m-2}}{(n-1) n^{n-2}} . \\
\quad \text { Hence } \sum_{i=... | \frac{a^{m-1}}{(n-1) n^{m-2}} | Inequalities | proof | Yes | Yes | inequalities | false | 738,300 |
7 Given $a, b, c \in \mathbf{R}^{+}$, prove:
(1) $\sqrt[3]{\frac{a}{b+c}}+\sqrt[3]{\frac{b}{c+a}}+\sqrt[3]{\frac{c}{a+b}}>\frac{3}{2}$;
(2) $\sqrt[3]{\frac{a^{2}}{(b+c)^{2}}}+\sqrt[3]{\frac{b^{2}}{(c+a)^{2}}}+\sqrt[3]{\frac{c^{2}}{(a+b)^{2}}} \geqslant \frac{3}{\sqrt[3]{4}}$. | 7. (1) $\sqrt[3]{\frac{a}{b+c}}=\frac{a}{\sqrt[3]{a \cdot a \cdot(b+c)}} \geqslant \frac{3 a}{2 a+b+c}>\frac{3 a}{2 a+2 b+2 c}$, similarly, we have $\sqrt[3]{\frac{b}{a+c}}>\frac{3 b}{2 a+2 b+2 c} ; \sqrt[3]{\frac{c}{a+b}}>\frac{3 c}{2 a+2 b+2 c}$. Adding the three inequalities, we get the original inequality holds.
(2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,301 |
8 Let $n \geqslant 2$ be a positive integer, and let $n$ positive numbers $v_{1}, v_{2}, \cdots, v_{n}$ satisfy the following two conditions:
(1) $v_{1}+v_{2}+\cdots+v_{n}=1$;
(2) $v_{1} \leqslant v_{2} \leqslant \cdots \leqslant v_{n} \leqslant 2 v_{1}$.
Find the maximum value of $v_{1}^{2}+v_{2}^{2}+\cdots+v_{n}^{2}... | 8. The condition can be generalized to: $v_{1}+v_{2}+\cdots+v_{n}=1, v_{1} \leqslant v_{2} \leqslant \cdots \leqslant v_{n} \leqslant r v_{1}$.
Below is the proof: $v_{1}^{2}+v_{2}^{2}+\cdots+v_{n}^{2} \leqslant \frac{(r+1)^{2}}{4 m}$, where $r>1, r \in \mathbf{R}^{+}$.
For any $j \in \mathbf{Z}^{+}, 1 \leqslant j \le... | \frac{(r+1)^{2}}{4 m} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,303 |
9 Given $a>1, b>1, c>1$, prove:
(1) $\frac{a^{5}}{b^{2}-1}+\frac{b^{5}}{c^{2}-1}+\frac{c^{5}}{a^{2}-1} \geqslant \frac{26}{6} \sqrt{15}$;
(2) $\frac{a^{5}}{b^{3}-1}+\frac{b^{5}}{c^{3}-1}+\frac{c^{5}}{a^{3}-1} \geqslant \frac{5}{2} \sqrt[3]{50}$. | 9. (1) Using the mean inequality, we have
$$\begin{array}{l}
\quad \frac{a^{5}}{b^{2}-1}+\frac{25(5+\sqrt{15})}{12}(b-1)+\frac{25(5-\sqrt{15})}{12}(b+1)+\frac{25}{18} \sqrt{15}+ \\
\frac{25}{18} \sqrt{15} \geqslant \frac{125}{6} a \text {, hence } \frac{a^{5}}{b^{2}-1} \geqslant \frac{125}{6}(a-b)+\frac{25}{18} \sqrt{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,304 |
10. (Reverse Cauchy Inequality - Polya-Szego Inequality)
Let \(0 < m_{1} \leqslant a_{i} \leqslant M_{1}, 0 < m_{2} \leqslant b_{i} \leqslant M_{2}, i=1,2, \cdots, n\), then:
$$\frac{\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{i=1}^{n} b_{i}^{2}\right)}{\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2}} \leqslant \frac... | 10. From the known conditions, $\frac{m_{2}}{M_{1}} \leqslant \frac{b_{i}}{a_{i}} \leqslant \frac{M_{2}}{m_{1}}, \frac{m_{2}}{M_{1}} a_{i} \leqslant b_{i} \leqslant \frac{M_{2}}{m_{1}} a_{i}$, then
$$\left(b_{i}-\frac{M_{2}}{m_{1}} a_{i}\right)\left(b_{i}-\frac{m_{2}}{M_{1}} a_{i}\right) \leqslant 0$$
Therefore, $b_{i... | \frac{1}{4} \cdot\left(\sqrt{\frac{M_{1} M_{2}}{m_{1} m_{2}}}+\sqrt{\frac{m_{1} m_{2}}{M_{1} M_{2}}}\right)^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 738,305 |
12 Let $n$ sets $S_{1}, S_{2}, \cdots, S_{n}$ consist of non-negative integers, and let $x_{i}$ be the sum of all elements in $S_{i}$. Prove: If for some natural number $k, 1<k<n$, we have
$$\sum_{i=1}^{n} x_{i} \leqslant \frac{1}{k+1} \cdot\left[k \frac{n(n+1)(2 n+1)}{6}-(k+1)^{2} \frac{n(n+1)}{2}\right] \text {, }$$
... | 12. By contradiction, if for any $i, j, t, l \in\{1,2, \cdots, n\}$ (at least three are distinct) we have $x_{i}-x_{t} \neq x_{l}-x_{j}$. We need to prove: for any $k, 1< k < n$, the following holds:
$$
x_{m}>\frac{k}{k+1} \cdot\left[k \cdot \frac{n(n+1)(2 n+1)}{6}-(k+1)^{2} \frac{n(n+1)}{2}\right]
$$
Assume without l... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 738,307 |
Example 1 Let $a_{1}=2, a_{n+1}=\frac{a_{n}}{2}+\frac{1}{a_{n}}(n=1,2, \cdots, 2004)$. Prove:
$$\sqrt{2}<a_{2005}<\sqrt{2}+\frac{1}{2005}$$ | Prove that by the AM-GM inequality, we have
$$a_{2005}=\frac{a_{2004}}{2}+\frac{1}{a_{2004}}>2 \sqrt{\frac{a_{2004}}{2} \cdot \frac{1}{a_{2004}}}=\sqrt{2} .$$
Below, we prove: for all positive integers $n$, we have
$$\sqrt{2}<a_{n}<\sqrt{2}+\frac{1}{n}$$
When $n=1$, (1) is clearly true.
Assume that when $n=k$, we hav... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,308 |
Example 2 Let $a>0$, prove that for any positive integer $n$, the inequality holds:
$$\frac{1+a^{2}+\cdots+a^{2 n}}{a+a^{3}+\cdots+a^{2 n-1}} \geqslant \frac{n+1}{n}$$ | Prove for $n$ using mathematical induction.
When $n=1$, $\frac{1+a^{2}}{a} \geqslant 2$, so the inequality holds.
Assume that when $n=k$ the inequality holds, i.e.,
$$A=\frac{1+a^{2}+\cdots+a^{2 k}}{a+a^{3}+\cdots+a^{2 k-1}}>\frac{k+1}{k}$$
Our goal is to prove:
$$B=\frac{1+a^{2}+\cdots+a^{2 k+2}}{a+a^{3}+\cdots+a^{2 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,309 |
Example 5 Prove: For all positive integers $n$, we have
$$\sqrt{1^{2}+\sqrt{2^{2}+\sqrt{3^{3}+\cdots+\sqrt{n^{2}}}}}<2 .$$ | Prove the following:
$$\sqrt{k^{2}+\sqrt{(k+1)^{2}+\cdots+\sqrt{n^{2}}}}<k+1$$
When $k=n$, (1) is obviously true.
Assume that (1) holds for $k=n, n-1, \cdots, m$, then for $k=m-1$, we need to prove
$$\sqrt{(m-1)^{2}+\sqrt{m^{2}+\cdots+\sqrt{n^{2}}}}<m-1+1=m.$$
Using the induction hypothesis,
the left side of the abov... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,312 |
Example 3 Let $a, b, c \in \mathbf{R}^{+}$, prove that for any real numbers $x, y, z$, we have:
$$\begin{aligned}
& x^{2}+y^{2}+z^{2} \\
\geqslant & 2 \sqrt{\frac{a b c}{(a+b)(b+c)(c+a)}}\left(\sqrt{\frac{a+b}{c}} x y+\sqrt{\frac{b+c}{a}} y z+\sqrt{\frac{c+a}{b}} z x\right) .
\end{aligned}$$
And determine the necessar... | Prove that the left side equals the right side:
$$\begin{aligned}
= & {\left[\frac{b}{b+c} x^{2}+\frac{a}{c+a} y^{2}-2 \sqrt{\frac{a b}{(b+c)(c+a)}} x y\right] } \\
& +\left[\frac{c}{c+a} y^{2}+\frac{b}{a+b} z^{2}-2 \sqrt{\frac{b c}{(c+a)(a+b)}} y z\right] \\
& +\left[\frac{c}{b+c} x^{2}+\frac{a}{a+b} z^{2}-2 \sqrt{\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,313 |
Example 19 Find the maximum value $m_{n}$ of the function $f_{n}\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\frac{x_{1}}{\left(1+x_{1}+\cdots+x_{n}\right)^{2}}+$ $\frac{x_{2}}{\left(1+x_{2}+\cdots+x_{n}\right)^{2}}+\cdots+\frac{x_{n}}{\left(1+x_{n}\right)^{2}}$ (where $x_{i} \geqslant 0$). Express $m_{n}$ in terms of $m_{... | Let $a_{i}=\frac{1}{1+x_{i}+\cdots+x_{n}}, 1 \leqslant i \leqslant n$, and define $a_{n+1}=1$. Then
$$1+x_{i}+x_{i+1}+\cdots+x_{n}=\frac{1}{a_{i}}$$
Also,
$$1+x_{i+1}+x_{i+2}+\cdots+x_{n}=\frac{1}{a_{i+1}}$$
Thus,
$$x_{i}=\frac{1}{a_{i}}-\frac{1}{a_{i+1}}$$
Therefore,
$$\begin{aligned}
f_{n} & =\sum_{i=1}^{n} a_{i}^... | 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,314 |
Example 7 Assume $a_{1}<a_{2}<\cdots<a_{n}$ are real numbers, prove that:
$$a_{1} a_{2}^{4}+a_{2} a_{3}^{4}+\cdots+a_{n} a_{1}^{4} \geqslant a_{2} a_{1}^{4}+a_{3} a_{2}^{4}+\cdots+a_{1} a_{n}^{4} .$$ | Prove by mathematical induction on $n$.
When $n=2$, the inequality holds as an equality.
Assume that for $n-1$, the conclusion holds, i.e.,
$$a_{1} a_{2}^{4}+a_{2} a_{3}^{4}+\cdots+a_{n-1} a_{1}^{4} \geqslant a_{2} a_{1}^{4}+a_{3} a_{2}^{4}+\cdots+a_{1} a_{n-1}^{4}.$$
Consider the case for $n$.
We need to prove: $a_{n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,316 |
Example 8 Given a positive integer $n$, and real numbers $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}, y_{1} \geqslant y_{2} \geqslant \cdots \geqslant y_{n}$, satisfying:
$$\sum_{i=1}^{n} i x_{i}=\sum_{i=1}^{n} i y_{i}$$
Prove: For any real number $\alpha$, we have
$$\sum_{i=1}^{n} x_{i}[i \alpha] \geqslan... | Proof: First, we prove a lemma: For any real number $x$ and positive integer $n$, we have
$$\sum_{i=1}^{n-1}[i \alpha] \leqslant \frac{n-1}{2}[n \alpha]$$
Proof of the lemma: It is sufficient to sum $[i \alpha]+[(n-i) \alpha] \leqslant[n \alpha]$ for $i=1,2, \cdots, n-1$. $\square$
Returning to the original problem, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,317 |
1 Let $a_{1}=3, a_{n}=a_{n-1}^{2}-n(n=2,3, \cdots)$, prove: $a_{n}>0$. | 1. Use mathematical induction to prove $a_{n}>n$.
The text above has been translated into English, maintaining the original text's line breaks and format. | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,320 |
3 Let $\left\{a_{n}\right\}$ be a sequence of positive terms. If $a_{n+1} \leqslant a_{n}-a_{n}^{2}$, prove that for all $n \geqslant 2$,
$$a_{n} \leqslant \frac{1}{n+2}$$ | 3. Use the monotonicity of $y=-x^{2}+x$ on $\left(0, \frac{1}{4}\right)$ to complete the transition from $k$ to $k+1$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,321 |
4 Given the sequence $\left\{a_{n}\right\}, a_{1}=a_{2}=1, a_{n+2}=a_{n+1}+a_{n}$. Prove that for any $n \in \mathbf{N}_{+}$, $\operatorname{arccot} a_{n} \leqslant \operatorname{arccot} a_{n+1}+\operatorname{arccot} a_{n+2}$,
and determine the condition under which equality holds. | 4. Given $a_{n}>0, \operatorname{arccot} a_{n+1}+\operatorname{arccot} a_{n+2} \in(0, \pi), \cot x$ is a decreasing function on $(0, \pi)$, it is sufficient to prove: $\cot \left(\operatorname{arccot} a_{n}\right) \geqslant \cot \left(\operatorname{arccot} a_{n+1}+\operatorname{arccot} a_{n+2}\right)$, which means to p... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,322 |
5 Let $a_{1}, a_{2}, \cdots$ be a sequence of real numbers, and for all $i, j=1,2, \cdots$ satisfy: $a_{i+j} \leqslant a_{i}+a_{j}$, prove that for positive integer $n$, we have
$$a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n} \geqslant a_{n}$$ | 5. Use mathematical induction on $n$. When $n=1$, $a_{1} \geqslant a_{1}$, the inequality obviously holds. Assume that when $n=1,2, \cdots, k-1$, the inequality holds, i.e.,
$$\left\{\begin{array}{l}
a_{1} \geqslant a_{1} \\
a_{1}+\frac{a_{2}}{2} \geqslant a_{2} \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
a_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,323 |
6 Let non-negative integers $a_{1}, a_{2}, \cdots, a_{2004}$ satisfy $a_{i}+a_{j} \leqslant a_{i+j} \leqslant a_{i}+a_{j}+1(1 \leqslant i, j$, $i+j \leqslant 2004)$. Prove: There exists $x \in \mathbf{R}$, for all $n(1 \leqslant n \leqslant 2004)$, such that $a_{n}=$ $[n x]$. | 6. If there exists $x$, such that $a_{n}=[n x]$, then $\frac{a_{n}}{n} \leqslant x \geqslant \frac{a_{m}}{m}$.
The following proof: If $m, n$ are positive integers, and $m, n \leqslant 2004$, then we have
$$m a_{n}+m>n a_{m}$$
When $m=n$, (1) holds; also, when $m=1, n=2$ and $m=2, n=1$, we have $a_{2}+1>2 a_{1}$ and $... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 738,324 |
Example 20 Let $a, b, c$ be positive real numbers, prove that:
$$\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(2 b+c+a)^{2}}{2 b^{2}+(c+a)^{2}}+\frac{(2 c+a+b)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8$$ | To prove that since the left-hand side of the equation is homogeneous, we can assume \(a + b + c = 3\), and thus we only need to prove:
Let
$$\frac{(a+3)^{2}}{2 a^{2}+(3-a)^{2}}+\frac{(b+3)^{2}}{2 b^{2}+(3-b)^{2}}+\frac{(c+3)^{2}}{2 c^{2}+(3-c)^{2}} \leqslant 8 .$$
Then
$$\begin{aligned}
& f(x)=\frac{(x+3)^{2}}{2 x^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,325 |
7 Let $1<x_{1}<2$, for $n=1,2,3, \cdots$, define $x_{n+1}=1+x_{n}-\frac{1}{2} x_{n}^{2}$. Prove: for $n \geqslant 3$, we have $\left|x_{n}-\sqrt{2}\right|<\left(\frac{1}{2}\right)^{n}$. | 7. Let $y_{n}=x_{n}-\sqrt{2}$, below we use mathematical induction to prove $\left|y_{n}\right|<\left(\frac{1}{2}\right)^{n}, n \geqslant 3$. Assume $\left|y_{n}\right|<\left(\frac{1}{2}\right)^{n}$, by the problem, $y_{n+1}+\sqrt{2}=1+y_{n}+\sqrt{2}-\frac{1}{2}\left(y_{n}+\sqrt{2}\right)^{2}$, hence $\left|y_{n+1}\rig... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,326 |
8 Let $a>0$, prove:
$$\sqrt{a+\sqrt{2 a+\sqrt{3 a+\cdots+\sqrt{n a}}}}<\sqrt{a}+1$$ | 8. We prove by induction: When $1 \leqslant k \leqslant n$,
$$\sqrt{k a+\sqrt{(k+1) a+\cdots+\sqrt{n a}}}<1+\sqrt{k a}$$
When $k=n$, the above inequality clearly holds. Suppose $\sqrt{(k+1) a+\sqrt{(k+2) a+\cdots+\sqrt{n a}}}<1+$
$$\begin{array}{l}
\sqrt{(k+1) a}, \text{ then } \sqrt{k a+\sqrt{(k+1) a+\cdots+\sqrt{n a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,327 |
2 Let the sequence $a_{1}, a_{2}, \cdots, a_{2 n+1}$ satisfy: $a_{i}-2 a_{i+1}+a_{i+2} \geqslant 0(i=1,2, \cdots, 2 n-1)$, prove that:
$$\frac{a_{1}+a_{3}+\cdots+a_{2 n+1}}{n+1} \geqslant \frac{a_{2}+a_{4}+\cdots+a_{2 n}}{n}$$ | 2. Let $(k-1)\left(a_{1}+a_{3}+\cdots+a_{2 k-1}\right) \geqslant k\left(a_{2}+a_{4}+\cdots+a_{2 k-2}\right)$, to prove the proposition for $n$, it is only necessary to prove $a_{1}+a_{2}+\cdots+a_{2 k-1}+k a_{2 k+2} \geqslant a_{2}+a_{4}+\cdots+a_{2 k-2}+(k+$ 1) $a_{2 k}$, i.e., $k\left(a_{2 k+1}-a_{2 k}\right) \geqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,328 |
9 If $a_{i}>0(i=1,2,3, \cdots, n)$, and $a_{1} \cdot a_{2} \cdot \cdots \cdot a_{n}=1$, prove:
$$\sum_{i=1}^{n} a_{i} \geqslant n$$ | 9. When $n=1$, $a_{1}=1$, the conclusion holds. Suppose when $n=k$ the conclusion holds. When $n=k+1$, $a_{1} a_{2} \cdots a_{k+1}=1$, by the induction hypothesis, we have
$$\begin{array}{l}
a_{k+1} \cdot a_{1}+a_{2}+\cdots+a_{n} \geqslant k ; a_{k+1} \cdot a_{2}+a_{1}+a_{3}+\cdots+a_{k} \geqslant k ; \cdots, a_{k+1} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,329 |
10 Let the sequence $\left\{a_{n}\right\}$ satisfy: $a_{1}=a_{2}=1, a_{n+2}=a_{n+1}+a_{n}, S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Prove: $S_{n}=\sum_{k=1}^{n} \frac{a_{k}}{2^{k}}<2$. | 10. Let $b_{k}=\frac{a_{k}}{2^{k}}$, then $b_{k+2}=\frac{1}{2} b_{k+1}+\frac{1}{4} b_{k}, b_{1}=\frac{1}{2}, b_{2}=\frac{1}{4}$, hence
$$b_{k+1}=\frac{1}{2} b_{k}+\frac{1}{4} b_{k-1}, b_{k}=\frac{1}{2} b_{k-1}+\frac{1}{4} b_{k-2}, \cdots, b_{3}=\frac{1}{2} b_{2}+\frac{1}{4} b_{1}$$
Adding them up, we get $S_{k+2}-b_{1... | proof | Algebra | proof | Yes | Yes | inequalities | false | 738,330 |
II Let $r_{1}, r_{2}, \cdots, r_{n}$ be real numbers $\geqslant 1$, prove:
$$\frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}+\cdots+\frac{1}{r_{n}+1} \geqslant \frac{n}{\sqrt[n]{r_{1} r_{2} \cdots r_{n}}+1}$$ | 11. When $n=1$, the inequality obviously holds. Below, we prove by induction that the inequality holds when $n=2^{k}$.
When $k=1$, we have
$$\frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}-\frac{2}{\sqrt{r_{1} r_{2}}+1}=\frac{\left(\sqrt{r_{1} r_{2}}-1\right)\left(\sqrt{r_{1}}-\sqrt{r_{2}}\right)^{2}}{\left(r_{1}+1\right)\left(r_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,331 |
12 Let $f(n)$ be defined on the set of positive integers, and satisfy $f(1)=2, f(n+1)=f^{2}(n)-f(n)+1, n=1,2, \cdots$. Prove that for all integers $n>1$, we have
$$1-\frac{1}{2^{2^{n^{-1}}}}<\frac{1}{f(1)}+\frac{1}{f(2)}+\cdots+\frac{1}{f(n)}<1-\frac{1}{2^{2^{n}}}$$ | 12. From the given conditions, we have $\quad \frac{1}{f(n)}=\frac{1}{f(n)-1}-\frac{1}{f(n+1)-1}$, hence $\quad \sum_{k=1}^{n} \frac{1}{f(k)}=\frac{1}{f(1)-1}-\frac{1}{f(n+1)-1}=1-\frac{1}{f(n+1)-1}$.
Now we prove: $2^{2^{n-1}}<f(n+1)-1<2^{2^{n}}$, and thus the conclusion holds.
When $n=1$, $f(2)=3$, then $2<f(2)<4$.
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,332 |
14 Let $a_{1}, a_{2}, \cdots, a_{n}$ be non-negative real numbers, satisfying: $\sum_{i=1}^{n} a_{i}=4, n \geqslant 3$. Prove that: $a_{1}^{3} a_{2}+a_{2}^{3} a_{3}+\cdots+a_{n-1}^{3} a_{n}+a_{n}^{3} a_{1} \leqslant 27$. | 14. Use mathematical induction on $n$. When $n=3$, since the original inequality is cyclically symmetric with respect to $a_{1}, a_{2}, a_{3}$, without loss of generality, assume $a_{1}$ is the largest. If $a_{2}<a_{3}$, then
$$a_{1}^{3} a_{2}+a_{2}^{3} a_{3}+a_{3}^{3} a_{1}-\left(a_{1} a_{2}^{3}+a_{2} a_{3}^{3}+a_{3}... | 27 | Inequalities | proof | Yes | Yes | inequalities | false | 738,334 |
15 Let $n$ be a positive integer, and $x$ be a positive real number. Prove that: $\sum_{k=1}^{n} \frac{x^{k^{2}}}{k} \geqslant x^{\frac{1}{2} n(n+1)}$. | 15. Use mathematical induction on $n$. When $n=1$, the conclusion is obviously true.
Assume that when $n=s$, we have $\sum_{k=1}^{s} \frac{x^{k^{2}}}{k} \geqslant x^{\frac{1}{2} s(s+1)}$. Consider the case when $n=s+1$.
$$\sum_{k=1}^{s+1} \frac{x^{k^{2}}}{k}=\sum_{k=1}^{s} \frac{x^{k^{2}}}{k}+\frac{x^{(s+1)^{2}}}{s+1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,335 |
Example 21 Given $a+b+c>0, a x^{2}+b x+c=0$ has real roots, prove:
$$4 \min \{a, b, c\} \leqslant a+b+c \leqslant \frac{9}{4} \max \{a, b, c\} .$$ | Prove: Without loss of generality, assume $a+b+c=1$, otherwise we can use $\frac{a}{a+b+c} 、 \frac{b}{a+b+c} 、 \frac{c}{a+b+c}$ to replace $a 、 b 、 c$.
First, prove: $\max \{a, b, c\} \geqslant \frac{4}{9}$.
(1) If $b \geqslant \frac{4}{9}$, then the conclusion holds.
(2) If $b < \frac{4}{9}$, then $a + c > \frac{5}{9... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,336 |
16. Let $z_{i}(1 \leqslant i \leqslant n)$ be $n$ complex numbers, $s_{i}=z_{1}+z_{2}+\cdots+z_{i}, 1 \leqslant i \leqslant n$. Prove:
$$\sum_{1 \leqslant i<j \leqslant n}\left|s_{j}-z_{i}\right| \leqslant \sum_{k=1}^{n}\left[(n+1-k) \cdot\left|z_{k}\right|+(k-2)\left|s_{k}\right|\right] .$$ | 16. Use mathematical induction on $n$. When $n=1$, the conclusion is obviously true. Suppose the original inequality holds for $n-1$, i.e., $\sum_{1 \leqslant i<j \leqslant n-1}\left|s_{j}-z_{i}\right| \leqslant \sum_{k=1}^{n-1}\left((n-k)\left|z_{k}\right|+(k-2)\left|s_{k}\right|\right)$, and $\sum_{1 \leqslant i<j \l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,337 |
Example 1 Find the minimum value of the ratio of a three-digit number (in decimal) to the sum of its digits.
Translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | Let the three-digit number be $100x + 10y + z$, where $x, y, z$ are integers, and $1 \leqslant x \leqslant 9$, $0 \leqslant y, z \leqslant 9$. Thus, the ratio is
$$f(x, y, z) = \frac{100x + 10y + z}{x + y + z} = 1 + \frac{99x + 9y}{x + y + z}.$$
In the expression on the right-hand side of (1), only the denominator con... | \frac{199}{19} | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 738,338 |
Example 2 Let $A, B, C$ be non-negative real numbers, $A+B+C=\frac{\pi}{2}$, and $M=\sin A+\sin B+$ $\sin C, N=\sin ^{2} A+\sin ^{2} B+\sin ^{2} C$, then $M^{2}+N \leqslant 3$. | Assume $C \leqslant A, B$, then $C \in\left[0, \frac{\pi}{6}\right]$,
$$\begin{array}{c}
M=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}+\sin C \\
=2 \sin \left(\frac{\pi}{4}-\frac{C}{2}\right) \cos \frac{A-B}{2}+\sin C \\
N=\frac{1}{2}(1-\cos 2 A)+\frac{1}{2}(1-\cos 2 B)+\sin ^{2} C
\end{array}$$
Fix $C$, we have
$$\begin{... | 3 | Inequalities | proof | Yes | Yes | inequalities | false | 738,339 |
Example 3 Let $A, B, C, D$ be four points in space, and connect $AB, AC, AD, BC, BD, CD$ where at most one of these has a length greater than 1. Try to find the maximum value of the sum of the lengths of the six segments.
Connect $AB, AC, AD, BC, BD, CD$ where at most one of these has a length greater than 1, try to f... | Let $AD$ be the longest of the six line segments.
(1) Fixing the lengths of the other five line segments, it is evident that $AD$ attains its maximum value when $A$ and $D$ are opposite vertices of the parallelogram $ABDC$.
(2) Fixing the positions of $B$ and $C$, then $A$ and $D$ must lie within the intersection of tw... | 5 + \sqrt{3} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 738,340 |
Example 4 If $a, b, c$ are non-negative real numbers and $a+b+c=1$, try to find the maximum value of $S=ab+bc+ca-3abc$.
保留源文本的换行和格式，直接输出翻译结果。 | Let's assume $a \geqslant b \geqslant c$, then $c \leqslant \frac{1}{3}$. Fix $c$,
$$S=a b+b c+c a-3 a b c=a b(1-3 c)+c(a+b)$$
Since the value of $a+b$ is fixed, $1-3 c \geqslant 0$, and $a b$ reaches its maximum value when $a=b$, hence $S$ reaches its maximum value when $a=b$.
Adjust $a, b, c$ to $a=b \geqslant c$, ... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,341 |
Example 5 Let non-negative numbers $\alpha, \beta, \gamma$ satisfy $\alpha+\beta+\gamma=\frac{\pi}{2}$, find the minimum value of the function
$$f(\alpha, \beta, \gamma)=\frac{\cos \alpha \cos \beta}{\cos \gamma}+\frac{\cos \beta \cos \gamma}{\cos \alpha}+\frac{\cos \gamma \cos \alpha}{\cos \beta}$$ | Let's assume $\gamma \leqslant \alpha, \beta$, then $\gamma \in\left[0, \frac{\pi}{6}\right]$. Fix $\gamma$, since
$$\begin{aligned}
\frac{\cos \beta \cos \gamma}{\cos \alpha}+\frac{\cos \gamma \cos \alpha}{\cos \beta} & =2 \cos \gamma\left(\frac{\cos ^{2} \gamma}{\sin \gamma+\cos (\alpha-\beta)}+\sin \gamma\right) \\
... | \frac{5}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,342 |
Example 6 Let $x, y, z$ be non-negative real numbers, and $x+y+z=1$. Prove that:
$$y z+z x+x y-2 x y z \leqslant \frac{7}{27}$$ | Prove that it is obvious when $x=y=z=\frac{1}{3}$, the equality in the inequality holds. Without loss of generality, assume $x \geqslant y \geqslant z$, then $x \geqslant \frac{1}{3} \geqslant z$.
Let $\quad x^{\prime}=\frac{1}{3}, y^{\prime}=y, z^{\prime}=x+z-\frac{1}{3}$,
then
$$x^{\prime}+z^{\prime}=x+z, x^{\prime... | \frac{7}{27} | Inequalities | proof | Yes | Yes | inequalities | false | 738,343 |
Example 7 Given non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 3)$ satisfy the inequality: $x_{1}+x_{2}+\cdots+$ $x_{n} \leqslant \frac{1}{2}$, find the minimum value of $\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right)$. | When $x_{1}, x_{2}, \cdots, x_{n-2}, x_{n-1}+x_{n}$ are all constants, since
$$\left(1-x_{n-1}\right)\left(1-x_{n}\right)=1-\left(x_{n-1}+x_{n}\right)+x_{n-1} x_{n}$$
it is evident that the larger $\left|x_{n-1}-x_{n}\right|$, the smaller the value of the above expression. Therefore, for $n \geqslant 3$, let
$$x_{i}^{... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,344 |
Example 8 Let $a, b, c, d \geqslant 0$, and $a+b+c+d=1$, prove that: $b c d+c d a+d a b+a b c \leqslant \frac{1}{27}+\frac{176}{27} a b c d$. | Prove that if $d=0$, then $a b c \leqslant \frac{1}{27}$, the inequality is obviously true. If $a, b, c, d>0$, we only need to prove:
Let
$$f(a, b, c, d)=\sum_{c y c} \frac{1}{a}-\frac{1}{27} \frac{1}{a b c d} \leqslant \frac{176}{27}$$
Then
$$a \leqslant \frac{1}{4} \leqslant b, a^{\prime}=\frac{1}{4}, b^{\prime}=a+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,345 |
Example 22 Non-negative real numbers $a, b, c, d$ satisfy: $a^{2}+b^{2}+c^{2}+d^{2}=4$, prove that: $a^{3}+b^{3}+c^{3}+d^{3}+a b c+b c d+c d a+d a b \leqslant 8$. | Prove that the original inequality is equivalent to
$$\left(a^{3}+b^{3}+c^{3}+d^{3}+a b c+b c d+c d a+d a b\right)^{2} \leqslant\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{3} .$$
Since
$$\begin{array}{l}
a^{3}+b^{3}+c^{3}+d^{3}+a b c+b c d+c d a+d a b=a\left(a^{2}+b c\right)+b\left(b^{2}+c d\right)+ \\
c\left(c^{2}+d a\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,347 |
Example 10: Prove that among all $n$-sided polygons with a fixed perimeter $l$, the regular $n$-sided polygon has the largest area. | Proof (1) First, a concave polygon cannot have the maximum area.
As shown in Figure 8-1, let \( A_{1} A_{2} \cdots A_{n} \) be a concave polygon. When \(\triangle A_{i-1} A_{i} A_{i+1}\) is centrally symmetric to \(\triangle A_{i-1} A_{i}^{\prime} A_{i+1}\) with the midpoint of line segment \(A_{i+1} A_{i-1}\) as the c... | proof | Geometry | proof | Yes | Yes | inequalities | false | 738,348 |
Example 11 If the sum of several positive integers is 2011, find the maximum value of their product. | Since there are only a finite number of different sets of positive integers whose sum is 2011, the maximum value must exist.
Let $x_{1}, x_{2}, \cdots, x_{n}$ all be positive integers, $x_{1}+x_{2}+\cdots+x_{n}=2011$ and the product $u=$ $x_{1} x_{2} \cdots x_{n}$ reaches its maximum value, then
(1) $x_{i} \leqslant 4(... | 2^{2} \times 3^{669} | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 738,349 |
Example 12 Let $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ be a non-decreasing sequence of positive integers. For $m \geqslant 1$, define $b_{m}=\min \left\{n, a_{n} \geqslant m\right\}$, i.e., $b_{m}$ is the smallest value of $n$ such that $a_{n} \geqslant m$. Given that $a_{19}=85$, find
$$a_{1}+a_{2}+\cdots+a_{19}+b_{1}+b... | If there exists an $i$ such that $a_{i}<a_{i+1}(1 \leqslant i \leqslant 18)$, then perform the following adjustment: $a_{i}^{\prime}=a_{i}+1$, $a_{j}^{\prime}=a_{j}(j \neq i)$, and denote the adjusted $b_{j}$ as $b_{j}^{\prime}(j=1,2, \cdots, 85)$.
By definition, $b_{a_{i}+1}=i+1, b_{a_{i}+1}^{\prime}=i=b_{a_{i}+1}-1,... | 1700 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 738,350 |
1 Let $x \geqslant 0, y \geqslant 0, z \geqslant 0, x+y+z=1$, find the maximum and minimum values of $S=2 x^{2}+y+3 z^{2}$. | 1. Fix the value of $z$, first find the maximum and minimum values of $2 x^{2}+y$, then let $z$ vary, find the overall maximum value $S_{\max }=3$, minimum value $S_{\min }=\frac{57}{72}$. | S_{\max }=3, S_{\min }=\frac{57}{72} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,351 |
2 Prove: In $\triangle A B C$, for $0 \leqslant \lambda_{i} \leqslant 2(i=1,2,3)$, we have
$$\sin \lambda_{1} A+\sin \lambda_{2} B+\sin \lambda_{3} C \leqslant 3 \sin \frac{\lambda_{1} A+\lambda_{2} B+\lambda_{3} C}{3}$$ | 2. Suppose $C$ is an acute angle. From $0 \leqslant \lambda_{i} \leqslant 2$, we get $00, \sin \frac{1}{6}\left(\lambda_{1} A+\lambda_{2} B+4 \lambda_{3} C\right)>0, \sin \frac{1}{3}\left(\lambda_{1} A+\lambda_{2} B+\lambda_{3} C\right)>0$.
Thus, $\quad \sin \lambda_{1} A+\sin \lambda_{2} B+\sin \lambda_{3} C+\sin \fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,352 |
3 Given $\theta_{1}, \theta_{2}, \cdots, \theta_{n}$ are all non-negative, and $\theta_{1}+\theta_{2}+\cdots+\theta_{n}=\pi$. Find
$$\sin ^{2} \theta_{1}+\sin ^{2} \theta_{2}+\cdots+\sin ^{2} \theta_{n}$$
the maximum value. | 3. First, consider when $\theta_{1}+\theta_{2}$ is constant, we have $\sin ^{2} \theta_{1}+\sin ^{2} \theta_{2}=\left(\sin \theta_{1}+\sin \theta_{2}\right)^{2}-$ $2 \sin \theta_{1} \sin \theta_{2}=4 \sin ^{2} \frac{\theta_{1}+\theta_{2}}{2} \cos ^{2} \frac{\theta_{1}-\theta_{2}}{2}-\cos \left(\theta_{1}-\theta_{2}\rig... | \frac{9}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,353 |
4. Let $a, b, c, d \geqslant 0$, and $a+b+c+d=4$. Prove:
$$b c d+c d a+d a b+a b c-a b c d \leqslant \frac{1}{2}(a b+a c+a d+b c+b d+c d) .$$ | 4. When $a, b, c, d$ have 1 zero, let's assume $d=0$, then we only need to prove: $a b c \leqslant \frac{1}{2}(a b + b c + c a)$. Since $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + a + b + c \geqslant 6$, we have $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geqslant 2$, hence the above inequality holds.
If $a, b, c, d$ are... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,354 |
5】 Given $x_{i}$ are non-negative real numbers, $i=1,2,3,4 . x_{1}+x_{2}+x_{3}+x_{4}=1$. Let $S=1-$ $\sum_{i=1}^{4} x_{i}^{3}-6 \sum_{1 \leqslant i<j<k \leqslant 4} x_{i} x_{j} x_{k}$, find the range of $S$. | 5. $S=\left(x_{1}+x_{2}+x_{3}+x_{4}\right)^{2}-\sum_{i=1}^{4} x_{i}^{3}-6 \sum_{1 \leqslant i<j<k<4} x_{i} x_{j} x_{k}=3 x_{1}^{2} \left(1-x_{1}\right)+3 x_{2}^{2}\left(1-x_{2}\right)+3 x_{3}^{2}\left(1-x_{3}\right)+3 x_{4}^{2}\left(1-x_{4}\right)$.
Therefore, $S \geqslant 0$, with equality when one of $x_{i} (1 \leqs... | S \in \left[0, \frac{3}{4}\right] | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,355 |
6 Let $x_{1}, x_{2}, \cdots, x_{n}$ be $n$ non-negative real numbers $\left(n>2, n \in \mathbf{N}^{*}\right)$, and
$$\sum_{i=1}^{n} x_{i}=n, \sum_{i=1}^{n} i x_{i}=2 n-2 .$$
Find the maximum value of $x_{1}+4 x_{2}+\cdots+n^{2} x_{n}$. | 6. Let $y_{i}=\sum_{j=i}^{n} x_{j}$, then $y_{1}=n, \sum_{i=1}^{n} y_{i}=2 n-2$, hence $S=\sum_{k=1}^{n} k^{2} x_{k}=\sum_{k=1}^{n} k^{2}$ $\left(y_{k}-y_{k+1}\right)+n^{2} y_{n}=\sum_{k=1}^{n}(2 k-1) y_{k}$.
Since $y_{1}=n, y_{2} \geqslant y_{3} \geqslant \cdots \geqslant y_{n}$, if there exists $i \in\{2,3, \cdots, ... | n^2 - 2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,356 |
7 For non-negative real numbers $x_{i}(i=1,2, \cdots, n)$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$, find the maximum value of $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$. | 7. When $n=1$, $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)=0$.
When $n=2$, $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)=\left(x_{1}^{4}+x_{2}^{4}\right)-\left(x_{1}+x_{2}\right)\left(x_{1}^{4}-x_{1}^{3} x_{2}+x_{1}^{2} x_{2}^{2}-\right.$ $\left.x_{1} x_{2}^{3}+x_{2}^{4}\right)=x_{1}^{3} x_{2}-x_{1}^{2} x_{2}^{2... | \frac{1}{12} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,357 |
Example 23 Given an integer $n \geqslant 4$, for any non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$ satisfying
$$a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n}>0$$
find the maximum value of $\frac{\sum_{i=1}^{n} a_{i}\left(a_{i}+b_{i}\right)}{\sum_{i=1}^{n} b_{i}\left(a_{i}+b_{... | By homogeneity, we may assume $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}=1$. First, when $a_{1}=1, a_{2}=$ $a_{3}=\cdots=a_{n}=0, b_{1}=0, b_{2}=b_{3}=\cdots=b_{n}=\frac{1}{n-1}$, we have $\sum_{i=1}^{n} a_{i}\left(a_{i}+b_{i}\right)=$ $1, \sum_{i=1}^{n} b_{i}\left(a_{i}+b_{i}\right)=\frac{1}{n-1}$, hence
$$\frac{\sum_... | n-1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,358 |
8 Let $a_{1}, a_{2}, a_{3} \geqslant 0$, prove:
$$a_{1}+a_{2}+a_{3}+3 \sqrt[3]{a_{1} a_{2} a_{3}} \geqslant 2\left(\sqrt{a_{1} a_{2}}+\sqrt{a_{2} a_{3}}+\sqrt{a_{3} a_{1}}\right) .$$ | 8. Let $f\left(a_{1}, a_{2}, a_{3}\right)=\left(a_{1}+a_{2}+a_{3}+3 \sqrt[3]{a_{1} a_{2} a_{3}}\right)-2\left(\sqrt{a_{1} a_{2}}+\sqrt{a_{2} a_{3}}+\right.$ $\left.\sqrt{a_{3} a_{1}}\right)$, without loss of generality, we can assume $a_{1} \leqslant a_{2} \leqslant a_{3}$. Now, we make the following adjustments: let
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,359 |
9 Let $a_{1}, a_{2}, \cdots, a_{10}$ be 10 distinct positive integers, with a sum of 2002. Find the minimum value of $a_{1} a_{2}+$ $a_{2} a_{3}+\cdots+a_{10} a_{1}$. | 9. The solution to this problem is carried out in two steps:
(1) First, consider how to arrange $a_{1}, a_{2}, \cdots, a_{10}$ to minimize the corresponding sum when $a_{1}, a_{2}, \cdots, a_{10}$ are determined. First, observe a simple case, i.e., $a_{1}+a_{2}+\cdots+a_{10}=55$, $\left\{a_{1}, a_{2}, \cdots, a_{10}\ri... | 6065 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,360 |
10 non-negative real numbers $a, b, c$ satisfy: $a b + b c + a = 1$, find the minimum value of $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}$. | 10. Let $f(a, b, c)=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$, and assume without loss of generality that $a \leqslant b \leqslant c$. We first prove: $f\left(0, a+b, c^{\prime}\right) \leqslant f(a, b, c)$. Here $c^{\prime}=\frac{1}{a+b}, a b+b c+c a=1$. In fact, $f\left(0, a+b, c^{\prime}\right) \leqslant f(a, b, c)... | \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,361 |
11 Let $a_{1}, a_{2}, \cdots, a_{2001}$ be non-negative real numbers, satisfying:
(1) $a_{1}+a_{2}+\cdots+a_{2001}=2$;
(2) $a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2000} a_{2001}+a_{2001} a_{1}=1$.
Find the extremum of $S=a_{1}^{2}+a_{2}^{2}+\cdots+a_{2001}^{2}$. | 11. First, calculate the maximum value of $f\left(a_{1}, a_{2}, \cdots, a_{2001}\right)=a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2000} a_{2001}+a_{2001} a_{1}$.
Lemma: There exists an $i \in\{1,2, \cdots, 2001\}$ such that $a_{i}>a_{i+4}$ (with $a_{2001+i}=a_{i}$).
Proof: If not, then $a_{1} \leqslant a_{5} \leqslant a_{9} \... | S_{\max }=2, S_{\min }=\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,362 |
12 Let $x_{1}, x_{2}, \cdots, x_{n}$ all be no less than 0, and $\sum_{i=1}^{n} x_{i}=1$, find the maximum value of the sum
$$\sum_{1 \leq i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)$$ | $$\begin{array}{l}
\text { 12. } \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)=\frac{1}{2} \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)+\frac{1}{2} \sum_{1 \leqslant i<j \leqslant n} x_{j} x_{i}\left(x_{j}+\right. \\
\left.x_{i}\right)=\frac{1}{2} \sum_{i \neq j} x_{i}... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,363 |
14. Let $S=\left\{\left.\frac{l}{1997} \right\rvert\, l=0,1,2, \cdots, 1996\right\} . S$ contains three numbers $x, y, z$ that satisfy: $x^{2}+ y^{2}-z^{2}=1$. Find the minimum and maximum values of $x+y+z$. | 14. Let $l=1997 x, m=1997 y, n=1997 z$, then $l, m, n \in\{0,1,2, \cdots, 1996\}$, and $l^{2}+m^{2}-n^{2}=1997^{2}$.
First, let's find the minimum value of $l+m+n$.
Fixing $l$, we have $(m-n)(m+n)=(1997-l)(1997+l)$. To make $m+n$ as small as possible, $m+n$ and $m-n$ should be as close as possible, but $m+n \geqslant ... | 2117 \text{ and } 5741 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 738,365 |
Example 1: Prove that $\left|\sum_{k=1}^{n}(-1)^{k}\left\{\frac{n}{k}\right\}\right| \leqslant 3 \sqrt{n}\left(n \in \mathbf{N}_{+}\right)$. | Analysis: For a monotonically decreasing, alternating sign sequence, we can adopt the following processing method:
Let $a_{1}>a_{2}>\cdots>a_{k}>0$, then we have
$$-a_{1}<a_{1}-a_{2}+a_{3}-a_{4}+\cdots<a_{1} .$$
(1) is not difficult to prove.
Returning to the original problem, two possible methods for estimating the l... | 3 \cdot \sqrt{n} | Number Theory | proof | Yes | Yes | inequalities | false | 738,366 |
Example 2 Real numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfy: $a_{1}+a_{2}+\cdots+a_{n}=0$, and
$$2 a_{k} \leqslant a_{k-1}+a_{k+1}, k=2,3, \cdots, n-1$$
Find the smallest $\lambda(n)$, such that for all $k \in\{1,2, \cdots, n\}$, we have
$$\left|a_{k}\right| \leqslant \lambda(n) \cdot \max \left\{\left... | Solution: First, take $a_{1}=1, a_{2}=-\frac{n+1}{n-1}, a_{k}=\frac{n+1}{n-1}+\frac{2 n(k-2)}{(n-1)(n-2)}, k=3$, $4, \cdots, n$, then it satisfies $a_{1}+a_{2}+\cdots+a_{n}=0$ and $2 a_{k} \leqslant a_{k-1}+a_{k+1}, k=2,3, \cdots$, $n-1$. At this point,
$$\lambda(n) \geqslant \frac{n+1}{n-1}$$
Next, we prove that when... | \frac{n+1}{n-1} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,367 |
Schur's Inequality: Let $x, y, z \in \mathbf{R}^{+}$, then
$$x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y) \geqslant 0$$
(i.e., $\sum_{\infty}[x(x-y)(x-z)] \geqslant 0$.).
Generally, Schur's Inequality is: Let $x, y, z \geqslant 0, r>0$, then
$$\sum_{\propto x} x^{r}(x-y)(x-z) \geqslant 0 .$$ | Proof: Without loss of generality, let $x \geqslant y \geqslant z$, then
$$\begin{aligned}
\text { LHS } & \geqslant x^{r}(x-y)(x-z)-y^{r}(x-y)(y-z) \\
& \geqslant y^{r}(x-y)^{2} \geqslant 0 .
\end{aligned}$$
The following two transformed forms of Schur's inequality are very useful in problem solving:
Transform I: $\q... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,369 |
Example 4 Let $n \geqslant 2$ be a positive integer, and $x_{1}, x_{2}, \cdots, x_{n} \in[0,1]$. Prove that there exists some $i, 1 \leqslant i \leqslant n-1$, such that the inequality
$$x_{i}\left(1-x_{i+1}\right) \geqslant \frac{1}{4} x_{1}\left(1-x_{n}\right)$$
holds. | Let $m=\min \left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, and let $x_{r}=m, 0 \leqslant m \leqslant 1$. We will discuss two cases:
(1) If $x_{2} \leqslant \frac{1}{2}(m+1)$, take $i=1$, then we have
$$\begin{aligned}
x_{1}\left(1-x_{2}\right) & \geqslant x_{1}\left(1-\frac{1+m}{2}\right)=\frac{1}{2} x_{1}(1-m) \\
& \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,370 |
Example 5 Given $x_{1}, x_{2}, x_{3}, x_{4}, y_{1}, y_{2}$ satisfy:
$$\begin{array}{c}
y_{2} \geqslant y_{1} \geqslant x_{4} \geqslant x_{3} \geqslant x_{2} \geqslant x_{1} \geqslant 2, \\
x_{1}+x_{2}+x_{3}+x_{4} \geqslant y_{1}+y_{2},
\end{array}$$
Prove: $x_{1} x_{2} x_{3} x_{4} \geqslant y_{1} y_{2}$. | Prove that keeping $y_{1}+y_{2}$ unchanged, increasing $y_{1}$ to make $y_{1} = y_{2}$, and equal to $\frac{y_{1}+y_{2}}{2}$, then $y_{1} y_{2}$ increases during this process, so we only need to prove the case where $y_{1}=y_{2}=y$.
Given:
$$\begin{array}{c}
y \geqslant x_{4} \geqslant x_{3} \geqslant x_{2} \geqslant ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,371 |
Example 6 Let $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ be real numbers. Prove that:
$$\sum_{i=1}^{n} a_{i}^{2}-\sum_{i=1}^{n} a_{i} a_{i+1} \leqslant\left[\frac{n}{2}\right](M-m)^{2}$$
where $a_{n+1}=a_{1}, M=\max _{1 \leq i \leq n} a_{i}, m=\min _{1 \leq i \leq n} a_{i},[x]$ denotes the greatest integer not excee... | Prove that if $n=2 k$ ($k$ is a positive integer), then
$$2\left(\sum_{i=1}^{n} a_{i}^{2}-\sum_{i=1}^{n} a_{i} a_{i+1}\right)=\sum_{i=1}^{n}\left(a_{i}-a_{i+1}\right)^{2} \leqslant n \times(M-m)^{2}$$
Thus,
$$\sum_{i=1}^{n} a_{i}^{2}-\sum_{i=1}^{n} a_{i} a_{i+1} \leqslant \frac{n}{2}(M-m)^{2}=\left[\frac{n}{2}\right](... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,372 |
Example 7 (1) If $x, y, z$ are positive integers not all equal, find the minimum value of $(x+y+z)^{3}-27 x y z$;
(2) If $x, y, z$ are positive integers all different, find the minimum value of $(x+y+z)^{3}-27 x y z$. | Solution (1)
$$\begin{aligned}
& (x+y+z)^{3}-27 x y z \\
= & x^{3}+y^{3}+z^{3}+3\left(x^{2} y+y^{2} z+z^{2} x\right)+3\left(x y^{2}+y z^{2}+z x^{2}\right)+6 x y z \\
& -27 x y z \\
= & (x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)+3\left(x^{2} y+y^{2} z+\right. \\
& \left.z^{2} x+x y^{2}+y z^{2}+z x^{2}-6 x y z\rig... | 54 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,373 |
Example 8 Let $a_{1}, a_{2}, \cdots$ be an infinite sequence of real numbers, satisfying: there exists a real number $c$, for all $i$ we have $0 \leqslant a_{i} \leqslant c$, and $\left|a_{i}-a_{j}\right| \geqslant \frac{1}{i+j}$ holds (for all $i \neq j$), prove that: $c \geqslant 1$.
---
The translation preserves t... | For a fixed $n \geqslant 2$, let the first $n$ terms of the sequence be sorted as
$$0 \leqslant a_{\sigma(1)} < a_{\sigma(2)} < \cdots < a_{\sigma(n)} \leqslant c$$
where $\sigma(1), \sigma(2), \sigma(3), \cdots, \sigma(n)$ is a permutation of $1, 2, 3, \cdots, n$.
Thus, $c \geqslant a_{\sigma(n)} - a_{\sigma(1)}$
$$\... | c \geqslant 1 | Inequalities | proof | Yes | Yes | inequalities | false | 738,374 |
Example 9 Let $a \leqslant b<c$ be the side lengths of a right triangle $ABC$, find the largest constant $M$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant \frac{M}{a+b+c}$ always holds. | When $a=b=\frac{\sqrt{2}}{2} c$, i.e., $\triangle A B C$ is an isosceles right triangle, we have
$$M \leqslant 2+3 \sqrt{2}$$
Below we prove that
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant \frac{2+3 \sqrt{2}}{a+b+c}$$,
which is equivalent to
$$a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \geqslant(2+3 \sqrt{2}) a b c$$
alwa... | 2+3 \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,375 |
Example 10 The real number $a$ makes the inequality
$$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} \geqslant a\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}\right)$$
hold for any real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$. Find the maximum value of $a$. (2010 Shanghai High School Mathematics Competiti... | Solve for the maximum value of $a$ which is $\frac{2 \sqrt{3}}{3}$.
Because when $x_{1}=1, x_{2}=\sqrt{3}, x_{3}=2, x_{4}=\sqrt{3}, x_{5}=1$, we get $a \leqslant \frac{2}{\sqrt{3}}$.
Furthermore, when $a=\frac{2}{\sqrt{3}}$, the inequality always holds.
In fact,
$$\begin{aligned}
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}... | \frac{2 \sqrt{3}}{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,376 |
Example 11 Given that $a, b, c$ are positive numbers, $a+b+c=10$, and $a \leqslant 2b, b \leqslant 2c, c \leqslant 2a$, find the minimum value of $abc$.
untranslated text remains the same as requested. | Let $\quad x=2 b-a, y=2 c-b, z=2 a-c$,
then $\quad x+y+z=10, x \geqslant 0, y \geqslant 0, z \geqslant 0$,
and $\quad a=\frac{x+2 y+4 z}{7}, b=\frac{y+2 z+4 x}{7}, c=\frac{z+2 x+4 y}{7}$,
thus $a b c=\frac{1}{343} \cdot[(x+2 y+4 z)(y+2 z+4 x)(z+2 x+4 y)]$,
and
$$\begin{aligned}
& (x+2 y+4 z)(y+2 z+4 x)(z+2 x+4 y) \\
=... | \frac{8000}{343} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,377 |
Example 12 Let $a, b, c, d \in \mathbf{R}^{+}, abcd=1$, and let $T=a(b+c+d)+b(c+d)+cd$.
(1) Find the minimum value of $a^{2}+b^{2}+T$;
(2) Find the minimum value of $a^{2}+b^{2}+c^{2}+T$. | (1) $a^{2}+b^{2}+T=a^{2}+b^{2}+(a+b)(c+d)+a b+c d$
$$\begin{array}{l}
\geqslant 2 a b+2 \sqrt{a b} \cdot 2 \sqrt{c d}+a b+c d \\
=4+3 a b+c d \geqslant 4+2 \cdot \sqrt{3 a b c d} \\
=4+2 \sqrt{3}
\end{array}$$
When $a=b, c=d, 3 a b=c d$, i.e., $a=b=\left(\frac{1}{3}\right)^{\frac{1}{t}}, c=d=3^{\frac{1}{4}}$, the equa... | 4+2 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,378 |
2 For any $x \in \mathbf{R}$ and $n \in \mathbf{N}_{+}$, prove:
(1) $\left|\sum_{k=1}^{n} \frac{\sin k x}{k}\right| \leqslant 2 \sqrt{\pi}$;
(2) $\sum_{k=1}^{n} \frac{|\sin k x|}{k} \geqslant|\sin n x|$. | 2. (1) Let the left-hand side of the inequality be denoted as $f(x)$. It is evident that the inequality holds when $x=0$ or $\pi$. Since $|f(x)|$ is an even function and has a period of $\pi$, it suffices to prove the inequality for $x \in (0, \pi)$.
For any fixed $x \in (0, \pi)$, take $m \in \mathbf{N}$ such that $m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,381 |
3 Let $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 2)$ all be positive numbers, prove:
$$\frac{x_{1}^{2}}{x_{1}^{2}+x_{2} x_{3}}+\frac{x_{2}^{2}}{x_{2}^{2}+x_{3} x_{4}}+\cdots+\frac{x_{n-1}^{2}}{x_{n-1}^{2}+x_{n} x_{1}}+\frac{x_{n}^{2}}{x_{n}^{2}+x_{1} x_{2}} \leqslant n-1 .$$ | 3. Let $y_{i}=\frac{x_{i+1} x_{i+2}}{x_{i}^{2}}\left(i=1,2, \cdots, n, x_{n+1}=x_{1}, x_{n+2}=x_{2}\right)$, then the original inequality is equivalent to
$$\frac{1}{1+y_{1}}+\frac{1}{1+y_{2}}+\cdots+\frac{1}{1+y_{n}} \leqslant n-1$$
From the definition of $y_{i}$, we know that $y_{1} y_{2} \cdots y_{n}=1$.
If there a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,382 |
4 Let $x_{1}, x_{2}, \cdots, x_{n}$ be non-negative real numbers, and satisfy:
$$\sum_{i=1}^{n} x_{i}^{2}+\sum_{1 \leq i<j \leqslant n}\left(x_{i} x_{j}\right)^{2}=\frac{n(n+1)}{2}$$
(1) Find the maximum value of $\sum_{i=1}^{n} x_{i}$;
(2) Find all positive integers $n$, such that $\sum_{i=1}^{n} x_{i} \geqslant \sqrt... | 4. (1) $\left(\sum_{i=1}^{n} x_{i}\right)^{2}=\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}$, if $n>2$, assume $x_{1} x_{2}>2$, then
$$x_{1}+x_{2}+x_{3} \geqslant x_{1}+x_{2} \geqslant 2 \sqrt{x_{1} x_{2}}>2 \sqrt{2}>\sqrt{6}$$
When $n \geqslant 4$, let $x_{1}=x_{2}=x, x_{3}=x_{4}=\cdots=x_... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,383 |
$\mathbf{5}$ Given $a \geqslant b \geqslant c>0$, and $a+b+c=3$. Prove:
$$\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \geqslant 3+Q$$
where $Q=|(a-1)(b-1)(c-1)|$. | 5. (1) When $a \geqslant 1 \geqslant b \geqslant c$, $\theta=2+a b c-a b-b c-c a$.
Since $\frac{a}{c}+a c+\frac{b}{a}+a b+\frac{c}{b}+b c \geqslant 2 a+2 b+2 c=6$, and $a+b+c \geqslant 3 \sqrt[3]{a b c}$, $a b c \leqslant 1$, hence the left side $\geqslant 6+a b c-1-a b-b c-a c=3+\theta$.
(2) When $a \geqslant b \geqs... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,384 |
6 Given $a, b, c \in \mathbf{R}^{+}$, and $a b c \leqslant 1$, prove:
$$\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \geqslant Q+a+b+c \text {, }$$
where $Q=|(a-1)(b-1)(c-1)|$. | 6. (1) When $a \leqslant 1, b \leqslant 1, c \leqslant 1$, we prove that the left side $\geqslant 3 \geqslant a+b+c+\theta$. This is equivalent to $a b+b c+c a \leqslant 2+a b c$.
Using $(1-a)(1-b) \geqslant 0$, we get $2-a-b \geqslant 1-a b \geqslant c(1-a b)$, hence
$$2+a b c \geqslant a+b+c \geqslant a b+b c+c a$$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,385 |
8 There are 5 positive numbers that satisfy the conditions:
(1) One of the numbers is $\frac{1}{2}$;
(2) From these 5 numbers, any 2 numbers taken, in the remaining 3 numbers there must be one number, which, together with the sum of the two numbers taken, equals 1.
Find these 5 numbers. | 8. Let these 5 numbers be $0<x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}$.
First, $x_{4}<\frac{1}{2}$, otherwise $x_{4} \geqslant \frac{1}{2}, x_{5} \geqslant \frac{1}{2}$, then $x_{4}+x_{5} \geqslant 1$, which is a contradiction! Hence $x_{5}=\frac{1}{2}$.
Take $x_{1} 、 x_{2}$, there exists ... | \left(\frac{1}{6}, \frac{1}{3}, \frac{1}{3}, \frac{1}{3}, \frac{1}{2}\right) \text{ or } \left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{2}\right) | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 738,387 |
9 Given $x \geqslant 0, y \geqslant 0, z \geqslant 0, x+y+z=1$, find the maximum value of $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2} z^{2}$. | 9. Suppose $x \geqslant y \geqslant z$, then
$$\begin{aligned}
& x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2} z^{2} \\
= & x^{2} \cdot\left[(1-x)^{2}-2 y z\right]+y^{2} z^{2}+x^{2} y^{2} z^{2} \\
= & x^{2}(1-x)^{2}-2 x^{2} y z+y^{2} z^{2}+x^{2} y^{2} z^{2} \\
\leqslant & \frac{1}{16}-\left(x^{2} y z-y^{2} z^{2}\righ... | \frac{1}{16} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,388 |
10 Let $x, y, z \geqslant 0$, and $x+y+z=1$, find the maximum and minimum values of $x^{2} y+y^{2} z+z^{2} x$.
| 10. First, $x^{2} y+y^{2} z+z^{2} x \geqslant 0$, when $x=1, y=z=0$ the equality holds, so the minimum value is 0. Without loss of generality, assume $x=\max \{x, y, z\}$, then
(1) When $x \geqslant y \geqslant z$, we have $x^{2} y+y^{2} z+z^{2} x \leqslant x^{2} y+y^{2} z+z^{2} x+z[x y+$ $(x-y)(y-z)]=(x+z)^{2} y=(1-y)... | \frac{4}{27} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,389 |
11 Given that $a, b, c$ are the lengths of the three sides of $\triangle ABC$, prove:
$$\left|\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}\right|<\frac{1}{22}$$ | 11. Let $a \geqslant b \geqslant c$, and set $b=c+x, a=b+y=c+x+y$, where $x>0, y>0$.
$$\begin{aligned}
\text { LHS } & =\frac{y x(x+y)}{(2 x+y+2 c)(x+2 c)(x+y+2 c)} \\
& 0$, then
$$\begin{aligned}
& 2 x^{3}+13 x^{2} y+27 x y^{2}+18 y^{3} \\
= & 2 x^{2} \cdot k y+13 x^{2} y+22 x y^{2}+\frac{5}{k} x^{2} y+\frac{18}{k^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,390 |
Example 25 Let $x, y, z \geqslant 0$, and $x+y+z=1$, prove:
$$0 \leqslant y z+z x+x y-2 x y z \leqslant \frac{7}{27}$$ | Prove that from the second transformation of Schur's Inequality, we get
$$\left(\sum_{\mathrm{oc}} x\right)^{3}-4\left(\sum_{\mathrm{oc}} x\right)\left(\sum_{\mathrm{oc}} y z\right)+9 x y z \geqslant 0,$$
Given the condition $\sum_{o x} x=1$, we have
$$\begin{array}{c}
1-4 \cdot \sum_{\text {or }} y z+9 x y z \geqslan... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,391 |
12 Find the maximum constant $k$, such that $\frac{k a b c}{a+b+c} \leqslant(a+b)^{2}+(a+b+4 c)^{2}$ holds for all positive real numbers $a$, $b$, $c$. | 12. Let $a=b=2c$, then $k \leqslant 100$. Also, due to
$$\begin{aligned}
& \frac{a+b+c}{a b c} \cdot\left[(a+b)^{2}+(a+b+4 c)^{2}\right] \\
\geqslant & \frac{a+b+c}{a b c} \cdot\left[(a+b)^{2}+(a+2 c+b+2 c)^{2}\right] \\
\geqslant & \frac{a+b+c}{a b c} \cdot\left[4 a b+(2 \sqrt{2 a c}+2 \sqrt{2 b c})^{2}\right] \\
= & ... | 100 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,392 |
13 (1) Prove: For any real numbers $p, q$, we have $p^{2}+q^{2}+1>p(q+1)$;
(2) Find the largest real number $b$, such that for any real numbers $p, q$, we have $p^{2}+q^{2}+1>b p(q+1)$;
(3) Find the largest real number $c$, such that for any integers $p, q$, we have $p^{2}+q^{2}+1>c p(q+1)$. | 13. (1) $p^{2}+q^{2}+1>p(q+1)$ is equivalent to $\left(q-\frac{p}{2}\right)^{2}+\left(\frac{p}{2}-1\right)^{2}+\frac{p^{2}}{2}>0$, which is clearly true.
(2) Let $p=\sqrt{2}, q=1$, then $b \leqslant \sqrt{2}$. We now prove that the inequality $p^{2}+q^{2}+1 \geqslant \sqrt{2} p(q+1)$ holds for all real numbers $p, q$.
... | \sqrt{2} | Inequalities | proof | Yes | Yes | inequalities | false | 738,393 |
14 (1) Let $a$, $b$, $c$ be the side lengths of $\triangle ABC$, and $n \geqslant 2$ be an integer. Prove that:
$$\frac{\sqrt[n]{a^{n}+b^{n}}+\sqrt[n]{b^{n}+c^{n}}+\sqrt[n]{c^{n}+a^{n}}}{a+b+c}<1+\frac{\sqrt[n]{2}}{2}$$
(2) Let $a$, $b$, $c$ be the side lengths of a triangle. Find the smallest positive real number $k$ ... | 14. (1) Let's assume $a \leqslant b \leqslant c$, then $a+b>c$. The original inequality is equivalent to
\[
\sqrt[n]{a^{n}+b^{n}}+\sqrt[n]{b^{n}+c^{n}}+\sqrt[n]{c^{n}+a^{n}}\sqrt[n]{2} \cdot c=\sqrt[n]{c^{n}+c^{n}} \geqslant \sqrt{b^{n}+c^{n}},
\]
and it is not difficult to prove that
\[
\sqrt[n]{a^{n}+b^{n}} \leqsla... | proof | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,394 |
Example 26 Let $x, y, z \in \mathbf{R}^{+}$, and $x+y+z=x y z$, prove that:
$$x^{2}+y^{2}+z^{2}-2(x y+y z+z x)+9 \geqslant 0 .$$ | Prove that because $x+y+z=x y z$, (3) is equivalent to
$$\begin{array}{c}
{\left[x^{2}+y^{2}+z^{2}-2(x y+y z+z x)\right](x+y+z)+9 x y z \geqslant 0} \\
\Leftrightarrow x^{3}+y^{3}+z^{3}-\left(x^{2} y+y^{2} z+z^{2} x+x y^{2}+y z^{2}+z x^{2}\right)+3 x y z \geqslant 0
\end{array}$$
That is,
$$\sum_{\mathrm{oc}} x^{3}-\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,395 |
Example 27 Let $a, b, c \in \mathbf{R}^{+}$, prove:
$$\sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c})+(a+b+c)^{2} \geqslant 4 \sqrt{3 a b c(a+b+c)} .$$ | Prove that by Schur's inequality (in (2), let $r=2$), we get
$$\sum_{\infty} x^{2}(x-y)(x-z) \geqslant 0, x, y, z \in \mathbf{R}^{+}$$
Therefore,
$$\sum_{9 c} x^{4}+x y z \sum_{c x} x \geqslant \sum_{c x} x^{3}(y+z)$$
Since $\sum_{9 c} x^{3}(y+z)=2 \sum_{c y c} y^{2} z^{2}+\sum_{c y c} y z(y-z)^{2} \geqslant 2 \sum_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,396 |
Example 4 Let $a, b, c \in \mathbf{R}^{+}$, prove: $a^{2 a} b^{2 b} c^{2 c} \geqslant a^{b+c} b^{c+a} c^{a+b}$. | Proof: Since the inequality is symmetric with respect to $a, b, c$, without loss of generality, assume $a \geqslant b \geqslant c$, thus
$$\frac{a^{2 a} b^{2 b} c^{2 c}}{a^{b+c} b^{c+a} c^{a+b}}=\left(\frac{a}{b}\right)^{a-b}\left(\frac{b}{c}\right)^{b-c}\left(\frac{a}{c}\right)^{a-c} \geqslant 1,$$
Therefore,
$$a^{2 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,397 |
Hölder's Inequality: Let $w_{1}, w_{2}, \cdots, w_{n}$ be positive real numbers, $w_{1}+w_{2}+\cdots+w_{n}=1$, for any positive real numbers $a_{i j}$, we have
$$\begin{aligned}
& \left(a_{11}+a_{12}+\cdots+a_{1 m}\right)^{w_{1}}\left(a_{21}+a_{22}+\cdots+a_{2 m}\right)^{w_{2}} \cdots\left(a_{n 1}+a_{n 2}+\cdots+a_{n n... | Let \( A_{\alpha}=\sum_{j=1}^{m} a_{\alpha j} \, (\alpha=1,2, \cdots, n) \), then (1) is
i.e.,
$$\begin{array}{c}
\left(A_{1}^{w_{1}} A_{2}^{w_{2}} \cdots A_{n}^{w_{n}}\right)^{-1} \sum_{j=1}^{m} a_{1 j}^{w_{1}} a_{2 j}^{w_{2}} \cdots a_{n j}^{w_{n}} \leqslant 1 \\
\sum_{j=1}^{m}\left(\frac{a_{1 j}}{A_{1}}\right)^{w_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,398 |
Example 28 Let $a, b \in \mathbf{R}^{+}$.
(1) Find the minimum value of $S=\frac{(a+1)^{2}}{b}+\frac{(b+3)^{2}}{a}$;
(2) Find the minimum value of $T=\frac{(a+1)^{3}}{b^{2}}+\frac{(b+3)^{3}}{a^{2}}$. | (1) By Cauchy-Schwarz inequality, we have
So
$$\begin{array}{l}
S \cdot (b+a) \geqslant (a+1+b+3)^{2} \\
S \geqslant \frac{(a+b+4)^{2}}{a+b}=(a+b)+\frac{16}{a+b}+8 \\
\quad \geqslant 2 \sqrt{16}+8=16
\end{array}$$
When $a=\frac{7}{3}, b=\frac{5}{3}$, equality holds.
Thus, the minimum value of $S$ is 16.
(2) By (3) (H... | 27 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,399 |
Example 29 Let $a, b, c \in \mathbf{R}^{+}$, prove:
$$\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}$$ | Prove that by the arithmetic mean inequality, we have
$$\frac{(a+b)+(b+c)+(c+a)}{3} \geqslant \sqrt[3]{(a+b)(b+c)(c+a)},$$
Therefore,
$$\frac{a+b+c}{3} \geqslant \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}$$
By (2) (Hölder's inequality), we have
$$\begin{aligned}
\frac{(a+b)(b+c)(c+a)}{8} & =\frac{1}{27}\left(\frac{a+b}{2}+b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,400 |
Example 30 Let $a, b, c$ be positive real numbers, prove that:
$$\left(a^{5}-a^{2}+3\right)\left(b^{5}-b^{2}+3\right)\left(c^{5}-c^{2}+3\right) \geqslant(a+b+c)^{3} .$$ | Prove that for $x \in \mathbf{R}^{+}, x^{2}-1$ and $x^{3}-1$ have the same sign, so
that is,
$$\begin{array}{l}
\left(x^{2}-1\right)\left(x^{3}-1\right) \geqslant 0 \\
x^{5}-x^{2}+3 \geqslant x^{3}+2
\end{array}$$
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... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,401 |
Example 31 (Power Mean Inequality) Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, $\alpha>\beta>0$, then
$$\left(\frac{1}{n} \sum_{i=1}^{n} a_{i}^{\beta}\right)^{\frac{1}{\beta}} \leqslant\left(\frac{1}{n} \sum_{i=1}^{n} a_{i}^{\alpha}\right)^{\frac{1}{\alpha}} .$$ | Prove that in (6) (Hölder's inequality), by setting \(x^{i}=1, i=1,2, \cdots, n\), 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
$$\frac{1}{n} \sum_{i=1}^{n} y_{i} \leqslant\left(... | \left(\frac{1}{n} \sum_{i=1}^{n} a_{i}^{\beta}\right)^{\frac{1}{\beta}} \leqslant\left(\frac{1}{n} \sum_{i=1}^{n} a_{i}^{\alpha}\right)^{\frac{1}{\alpha}} | Inequalities | proof | Yes | Yes | inequalities | false | 738,402 |
1 Let $x, y, z \in \mathbf{R}$, prove:
$$\begin{array}{c}
\left(x^{2}+y^{2}+z^{2}\right)\left[\left(x^{2}+y^{2}+z^{2}\right)^{2}-(x y+y z+z x)^{2}\right] \\
\geqslant(x+y+z)^{2}\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y+y z+z x)\right]^{2} .
\end{array}$$ | 1. Let $a=x+y+z, b=xy+yz+zx$, then $x^{2}+y^{2}+z^{2}=a^{2}-2b$, so the left side - right side of the original expression $=2b^{2}(a^{2}-3b)=b^{2} \cdot\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] \geqslant 0$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,403 |
2 Let $m, n \in \mathbf{N}^{+}, m>n$, prove: $\left(1+\frac{1}{n}\right)^{n}<\left(1+\frac{1}{m}\right)^{m}$. | 2. We only need to prove: $\left(1+\frac{1}{n}\right)^{n}<\left(1+\frac{1}{n+1}\right)^{n+1}$.
When $n=1$, the above inequality obviously holds. When $n \geqslant 2$, by the binomial theorem,
$$\begin{aligned}
\left(1+\frac{1}{n}\right)^{n}=\sum_{k=2}^{n} & \frac{1}{k!} \cdot\left(1-\frac{1}{n}\right)\left(1-\frac{2}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,404 |
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