problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
9. 133 Let $x, y$ be two distinct positive numbers, and denote
$$R=\sqrt{\frac{x^{2}+y^{2}}{2}}, A=\frac{x+y}{2}, G=\sqrt{x y}, H=\frac{2 x y}{x+y} .$$ | [Solution] Clearly $G<A$. Since $R=\sqrt{2 A^{2}-G^{2}}$, we have
$$R-A=\sqrt{2 A^{2}-G^{2}}-A=\frac{(A+G)(A-G)}{\sqrt{2 A^{2}-G^{2}}+A} .$$
From $G<A$, it immediately follows that
$$A+G<A+\sqrt{2 A^{2}-G^{2}}$$
Thus, we have $\quad R-A<A-G$.
From $G^{2}<2 A^{2}-G^{2}$, we know
$$G \sqrt{2 A^{2}-G^{2}}=\sqrt{G^{2}\le... | G-H<R-A<A-G | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,061 |
9・135 Let $0<a<1, x^{2}+y=0$, prove:
$$\log _{a}\left(a^{x}+a^{y}\right) \leqslant \log _{a}^{2}+\frac{1}{8} .$$ | [Proof] Since $0<a<1, a^{x}+a^{y} \geqslant 2 a^{\frac{x+y}{2}}$, thus
$$\log _{a}\left(a^{x}+a^{y}\right) \leqslant \log _{a} 2+\frac{x+y}{2}$$
Using $x^{2}+y=0$ we get
$$\frac{x+y}{2}=\frac{1}{2} x(1-x)=-\frac{1}{2}\left(x-\frac{1}{2}\right)^{2}+\frac{1}{8} \leqslant \frac{1}{8}$$
Therefore
$$\log _{a}\left(a^{x}+a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,063 |
9.136 Let real numbers $a, b, c$ satisfy
$$a b c>0, a+b+c>0 .$$
Prove: For any natural number $n$ we have
$$a^{n}+b^{n}+c^{n}>0$$ | [Proof] From $a b c>0$, we know that either $a, b, c$ are all positive, or $a, b, c$ consist of one positive and two negative numbers. When $a, b, c$ are all positive, it is obvious that $a^{n}+b^{n}+c^{n}>0$. For the latter case, assume without loss of generality that $a>0, b<0, c<0$, and $a>|b|+|c|$, so
$$a>|b|+|c| \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,064 |
9-137 Given that $a, b, c$ are all positive numbers, prove:
$$a^{a} b^{b} c^{c} \geqslant(a b c)^{\frac{a+b+c}{3}}$$ | [Proof] Since
$$\begin{aligned}
& 3(a \log a + b \log b + c \log c) - (a + b + c)(\log a + \log b + \log c) \\
= & (a - b)(\log a - \log b) + (a - c)(\log a - \log c) + (b - c)(\log b - \log c) \geqslant 0,
\end{aligned}$$
we have \(a \log a + b \log b + c \log c \geqslant \frac{1}{3}(a + b + c)(\log a + \log b + \log... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,065 |
9・138 Given $a>b>c$, prove:
$$a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)>0 .$$ | [Proof]
$$\begin{aligned}
& a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b) \\
= & a^{2}(b-c)+b^{2}(c-b+b-a)+c^{2}(a-b) \\
= & \left(a^{2}-b^{2}\right)(b-c)+\left(c^{2}-b^{2}\right)(a-b) \\
= & (a-b)(b-c)(a+b-c-b) \\
= & (a-b)(b-c)(a-c)>0 .
\end{aligned}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,066 |
1. 138 Prove: There exist infinitely many natural numbers $n$, such that the numbers $1,2, \cdots, 3 n$ can be arranged into the following table:
$$\begin{array}{l}
a_{1}, a_{2}, \cdots, a_{n} \\
b_{1}, b_{2}, \cdots, b_{n} \\
c_{1}, c_{2}, \cdots, c_{n}
\end{array}$$
satisfying the following two conditions:
(1) $a_{1... | Let the set of natural numbers $n$ that satisfy the above two conditions be denoted as $s$. If $n \in s$, by conditions (1) and (2), we know: there exist natural numbers $s$ and $t$, such that
$$\left\{\begin{array}{l}
\frac{3 n(3 n+1)}{2}=6 s n \\
\frac{3 n(3 n+1)}{2}=18 t
\end{array}\right.$$
which implies $\left\{\... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,069 |
9-141 Let $a, b, c$ be the lengths of the three sides of a triangle, prove that
$$\frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b} \geqslant\left(\frac{2}{3}\right)^{n-2} s^{n-1}$$
where $s=\frac{1}{2}(a+b+c), n \geqslant 1$. | [Proof] By the rearrangement inequality, we have
$$\frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b} \geqslant \frac{1}{3}\left(a^{n}+b^{n}+c^{n}\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) .$$
Since \( n \geqslant 1 \), it follows that
$$a^{n}+b^{n}+c^{n} \geqslant \frac{1}{3^{n-1}}(a+b+c)^{n}=\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,070 |
9 - 142 Let $a, b, c$ be the lengths of the three sides of a triangle. Prove:
$$a^{2}(b+c-a)+b^{2}(c+a-b)+c^{2}(a+b-c) \leqslant 3 a b c \text {. }$$ | [Proof] Let $x+y=a, y+z=b, z+x=c$. Since $a, b, c$ are the lengths of the sides of a triangle, $x, y, z$ are all positive numbers and the original inequality is equivalent to
$$\begin{array}{l}
\quad 2 z(x+y)^{2}+2 x(y+z)^{2}+2 y(z+x)^{2} \leqslant 3(x+y)(y+z)(z+x) \text {. } \\
\text { The left side }=2 z\left(x^{2}+y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,071 |
9・143 Let $a, b, c$ all be positive numbers, prove that:
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{1}{2}(a+b+c)$$ | [Proof] Let $s=a+b+c$, then
$$\begin{aligned}
\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} & =\frac{a^{2}}{s-a}+\frac{b^{2}}{s-b}+\frac{c^{2}}{s-c} \\
& =\frac{a^{2}-s^{2}+s^{2}}{s-a}+\frac{b^{2}-s^{2}+s^{2}}{s-b}+\frac{c^{2}-s^{2}+s^{2}}{s-c} \\
& =-4 s+s^{2}\left(\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,072 |
9・144 Let $0 \leqslant a \leqslant 1,0 \leqslant b \leqslant 1,0 \leqslant c \leqslant 1$, prove:
$$\begin{array}{l}
\quad \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \\
\leqslant 1 .
\end{array}$$ | [Proof] Without loss of generality, let $a \leqslant b \leqslant c$, we only need to prove
$$\begin{array}{l}
\frac{a+b+c}{a+b+1}+(1-a)(1-b)(1-c) \leqslant 1, \\
\text { i.e., } \quad(1-c)\left[(1-a)(1-b)-\frac{1}{a+b+1}\right] \leqslant 0 .
\end{array}$$
Since $0 \leqslant a \leqslant 1,0 \leqslant b \leqslant 1$, w... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,073 |
9-145 Let $a, b, c$ be non-negative real numbers satisfying $a+b+c=1$, prove that: $(1+a)(1+b)(1+c) \geqslant 8(1-a)(1-b)(1-c)$.
| [Proof] From $1+a=1-b+1-c \geqslant 2 \sqrt{(1-b)(1-c)}$, $1+b=1-a+1-c \geqslant 2 \sqrt{(1-a)(1-c)}$, $1+c=1-a+1-b \geqslant 2 \sqrt{(1-a)(1-b)}$, therefore $\quad(1+a)(1+b)(1+c) \geqslant 8(1-a)(1-b)(1-c)$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,074 |
9. 146 Prove: Given any 3 distinct positive numbers, they can be denoted as $a$, $b$, $c$ such that
$$\frac{a}{b}+\frac{b}{c}>\frac{a}{c}+\frac{c}{a}$$ | [Proof] Given three positive numbers denoted as $x_{1}, x_{2}, x_{3}$ and assuming $x_{1}\frac{x_{3}}{x_{1}}+\frac{x_{1}}{x_{3}}$, then let $x_{3}=a, x_{2}=b, x_{1}=c$ we get $\frac{a}{b}+\frac{b}{c}$ $>\frac{a}{c}+\frac{c}{a}$
Otherwise, if $\frac{x_{3}}{x_{2}}+\frac{x_{2}}{x_{1}} \leqslant \frac{x_{3}}{x_{1}}+\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,075 |
9. 147 Let $a, b, c$ be positive numbers, prove that:
$$\begin{aligned}
& \sqrt{a b(a+b)}+\sqrt{b c(b+c)}+\sqrt{c a(c+a)} \\
> & \sqrt{(a+b)(b+c)(c+a)} .
\end{aligned}$$ | [Proof] The square of the left side of the original inequality is
$$\begin{array}{l}
a b(a+b)+b c(b+c)+c a(c+a)+2 b \sqrt{a c(a+b)(b+c)} \\
+2 a \sqrt{b c(a+b)(c+a)}+2 c \sqrt{a b(b+c)(c+a)}
\end{array}$$
Since $b \sqrt{a c(a+b)(b+c)}>b \sqrt{a c \cdot a c}=a b c$,
$a \sqrt{b c(a+b)(c+a)}>a b c$,
$c \sqrt{a b(b+c)(c+a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,076 |
9. 148 Given non-negative real numbers $a, b, c$ satisfying $a+b+c \leqslant 3$, prove: $\frac{a}{1+a^{2}}+\frac{b}{1+b^{2}}+\frac{c}{1+c^{2}} \leqslant \frac{3}{2} \leqslant \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$ | [Proof] Clearly, $\frac{a}{1+a^{2}} \leqslant \frac{1}{2}, \frac{b}{1+b^{2}} \leqslant \frac{1}{2}, \frac{c}{1+c^{2}} \leqslant \frac{1}{2}$, thus the left half of the original inequality holds. Now we prove the right half of the inequality. Let
$$x=1+a, y=1+b, z=1+c,$$
then $x+y+z \leqslant 6$. Therefore,
$$\begin{al... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,077 |
$\begin{array}{c}9 \cdot 149 \text { Let } 0<x<1,0<y<1,0<z<1 \text {, prove: } \\ x(1-y)+y(1-z)+z(1-x)<1 .\end{array}$ | ［Proof］Let $s=x(1-y)+y(1-z)+z(1-x)$, then $s=x(1-y-z)+y(1-z)+z$.
If $1-y-z>0$, then
$$s<1-y-z+y(1-z)+z=1-y z<1$$
If $1-y-z \leqslant 0$, then
$$s \leqslant y(1-z)+z<1-z+z=1 .$$
In either case, we have $s<1$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,078 |
9・150 Let $0 \leqslant x, y, z \leqslant 1$, prove:
$$2\left(x^{3}+y^{3}+z^{3}\right)-\left(x^{2} y+y^{2} z+z^{2} x\right) \leqslant 3$$ | [Proof] Since $x^{2} y \geqslant \min \left\{x^{3}, y^{3}\right\}$, we have
$$x^{3}+y^{3}-x^{2} y \leqslant \max \left\{x^{3}, y^{3}\right\} \leqslant 1$$
Similarly, we get $\quad y^{3}+z^{3}-y^{2} z \leqslant \max \left\{y^{3}, z^{3}\right\} \leqslant 1$,
$$z^{3}+x^{3}-z^{2} x \leqslant \max \left\{z^{3}, x^{3}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,079 |
1. 139 A math lottery is now being issued, where 10 numbers from the first 100 natural numbers are filled in on 1 lottery ticket. During the draw, 10 numbers are removed from $1,2, \cdots, 100$. If the 10 numbers on the lottery ticket are all among the remaining 90 numbers, then the ticket wins. Prove:
（1）If 13 lottery... | [Proof] (1) After purchasing 13 lottery tickets, we can fill in the numbers as follows: on the first ticket, fill in $1-10$; on the second ticket, fill in $1-5$ and $11-15$; on the third ticket, fill in $6-15$; on the fourth ticket, fill in 16-25; on the fifth ticket, fill in $16-20$ and $26-30$; on the sixth ticket, f... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,080 |
9. 151 For positive numbers $a, b, c$, prove:
$$a^{3}+b^{3}+c^{3}+3 a b c \geqslant a b(a+b)+b c(b+c)+c a(c+a) .$$ | [Proof] Without loss of generality, let $c=\min \{a, b, c\}$. Since
$$\begin{array}{c}
a^{3}+b^{3}+2 a b c-a b(a+b)-c\left(a^{2}+b^{2}\right)=(a+b-c)(a-b)^{2} \\
c^{3}+a b c-c^{2}(a+b)=c(a-c)(b-c)
\end{array}$$
the inequality to be proved is
$$\text { LHS }- \text { RHS }=(a+b-c)(a-b)^{2}+c(a-c)(b-c) \geqslant 0 \text... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,081 |
9・152 Let $a, b, c$ all be greater than 1, prove:
$$2\left(\frac{\log _{b} a}{a+b}+\frac{\log _{b} b}{b+c}+\frac{\log _{a} c}{c+a}\right) \geqslant \frac{9}{a+b+c} .$$ | [Proof] Since $a>1, b>1, c>1$, we have
$$\log _{b} a \cdot \log _{c} b \cdot \log _{a} c=1$$
Using the AM-GM inequality, we get
$$\frac{\log _{b} a}{a+b}+\frac{\log _{b} b}{b+c}+\frac{\log _{a} c}{c+a} \geqslant 3 \sqrt[3]{\frac{1}{(a+b)(b+c)(c+a)}}$$
Using the AM-GM inequality again, we obtain
$$3 \sqrt[3]{\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,082 |
9. 153 Prove: If the product of three real numbers is 1, and their sum is greater than the sum of their reciprocals, then exactly one of these numbers is greater than 1. | [Proof] Suppose $x, y, z$ satisfy the given conditions, then
$$\begin{array}{l}
x y z=1 \\
x+y+z-\frac{1}{x}-\frac{1}{y}-\frac{1}{z}>0
\end{array}$$
Thus,
$$\begin{aligned}
& (x-1)(y-1)(z-1) \\
= & x y z-x y-y z-z x+x+y+z-1 \\
= & x+y+z-x y z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \\
= & x+y+z-\left(\frac{1}{... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,083 |
9・154 Let $a=\frac{m^{m+1}+n^{n+1}}{m^{m}+n^{n}}$, where $m, n$ are positive integers. Prove: $a^{m}+a^{n} \geqslant m^{m}+n^{n}$. | [Proof] For positive integer $p$, since
$$a^{p}-p^{p}=(a-p)\left(a^{p-1}+a^{p-2} p+\cdots+a p^{p-2}+p^{p-1}\right)$$
Therefore, regardless of the size of $p$, we always have
$$a^{p}-p^{p} \geqslant(a-p) p^{p}$$
From this, we can derive
$$\begin{aligned}
a^{m}+a^{n}-m^{m}-n^{n} & =\left(a^{m}-m^{m}\right)+\left(a^{n}-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,084 |
9. 156 Let real numbers $x_{1}, x_{2}, x_{3}$ be such that the sum of any two is greater than the third. Prove:
$$\frac{2}{3}\left(\sum_{i=1}^{3} x_{i}\right)\left(\sum_{i=1}^{3} x_{i}^{2}\right)>\sum_{i=1}^{3} x_{i}^{3}+x_{1} x_{2} x_{3}$$ | $[$ Proof $]$ Let $\left\{\begin{array}{l}2 y_{1}=x_{2}+x_{3}-x_{1} \\ 2 y_{2}=x_{3}+x_{1}-x_{2} \\ 2 y_{3}=x_{1}+x_{2}-x_{3}\end{array}\right.$, then $y_{1}>0, y_{2}>0, y_{3}>0$, and $x_{1}=y_{2}+y_{3}, x_{2}=y_{3}+y_{1}$, $x_{3}=y_{1}+y_{2}$. Therefore,
$$\begin{array}{l}
\sum_{i=1}^{3} x_{i}=2 \sum_{i=1}^{3} y_{i} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,086 |
9-157 Let $a, b, c$ be non-negative real numbers, prove that:
$$\begin{array}{l}
a^{4}+b^{4}+c^{4}-2\left(a^{2} b^{2}+a^{2} c^{2}+b^{2} c^{2}\right)+a^{2} b c+b^{2} a c+c^{2} a b \\
\geqslant 0
\end{array}$$ | [Proof] By calculation, it is easy to see that the left side of the original inequality is
$$(a+b+c)[a b c-(a+b-c)(b+c-a)(c+a-b)] .$$
Since $a, b, c \geqslant 0$, at most one of the three numbers $a+b-c, b+c-a, c+a-b$ can be negative. If one of the above three numbers is negative, the original inequality obviously hol... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,087 |
9-158 Let $a, b, c$ be non-negative real numbers, prove that:
$$\frac{1}{3}(a+b+c)^{2} \geqslant a \sqrt{b c}+b \sqrt{c a}+c \sqrt{a b} .$$ | [Proof] By the AM-GM inequality, we have
$$\begin{aligned}
a \sqrt{b c}+b \sqrt{c a}+c \sqrt{a b}= & \sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c}) \\
& \leqslant\left(\frac{a+b+c}{3}\right)^{\frac{3}{2}}(\sqrt{a}+\sqrt{b}+\sqrt{c}) .
\end{aligned}$$
By the Cauchy-Schwarz inequality, we know
$$\sqrt{a}+\sqrt{b}+\sqrt{c} \le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,088 |
9・160 Prove that for any positive numbers $a, b, c$ the following inequality holds:
$$\sqrt{a^{2}-a b+b^{2}}+\sqrt{b^{2}-b c+c^{2}} \geqslant \sqrt{a^{2}+a c+c^{2}}$$
and prove that equality holds if and only if $\frac{1}{b}=\frac{1}{a}+\frac{1}{c}$. | [Proof] It suffices to prove
$$\sqrt{a^{2}-a b+b^{2}} \geqslant\left|\sqrt{a^{2}+a c+c^{2}}-\sqrt{b^{2}-b c+c^{2}}\right|$$
Obviously, (1) $\Leftrightarrow$
$$\begin{aligned}
a^{2}-a b+b^{2} \geqslant & a^{2}+a c+c^{2}+b^{2}-b c+c^{2}-2 \sqrt{a^{2}+a c+c^{2}} \\
& \sqrt{b^{2}-b c+c^{2}}
\end{aligned}$$
That is, $4\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,090 |
1. 140 Let $k_{1}<k_{2}<k_{3}<\cdots$ be positive integers, none of which are consecutive, and for $m=1,2,3, \cdots, S_{m}=k_{1}+k_{2}+\cdots+k_{m}$. Prove that for every positive integer $n$, the interval $\left[S_{n}, S_{n+1}\right)$ contains at least one perfect square. | [Proof] The necessary and sufficient condition for the interval $\left[S_{n}, S_{n+1}\right)$ to contain at least one perfect square is that $\left[\sqrt{S_{n}}, \sqrt{S_{n+1}}\right)$ contains at least one integer.
Therefore, to prove this problem, it is only necessary to prove that for each $n \in N$, we have
$$\beg... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,091 |
9-164 Let $a, b, c$ be non-negative real numbers, prove that:
$$a+b+c \geqslant \frac{a(bc+c+1)}{ca+a+1}+\frac{b(ca+a+1)}{ab+b+1}+\frac{c(ab+b+1)}{bc+c+1} .$$ | [Proof] The original inequality is equivalent to
$$\begin{aligned}
a+b+c \geqslant & -\frac{a b c}{c a+a+1}+\frac{a b c}{a b+b+1}+\frac{a b c}{b c+c+1}+3 \\
& -\frac{1}{c a+a+1}-\frac{1}{a b+b+1}-\frac{1}{b c+c+1}
\end{aligned}$$
which is
$$\begin{aligned}
3 \leqslant & b-\frac{a b c}{c a+a+1}+\frac{1}{c a+a+1}+c-\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,094 |
9. 168 Let $f_{a}(x, y)=x^{2}+a x y+y^{2}$, where $0 \leqslant a \leqslant 2$. For any $(x, y)$, let
$$\overline{f_{a}}(x, y)=\min _{m, n \in \mathbb{Z}} f_{a}(x-m, y-n)$$
(1) Prove that $\overline{f_{1}}(x, y)<\frac{1}{2}$;
(2) Prove that $\overline{f_{1}}(x, y) \leqslant \frac{1}{3}$, and find all $(x, y)$ such that ... | [Solution] We only need to solve problem (3) and find all $(x, y)$ that satisfy the equality in (3). For any $(x, y)$, there clearly exist $m^{\prime}, n^{\prime} \in \mathbb{Z}$ such that $\left|x^{\prime}=x-m^{\prime}\right| \leqslant \frac{1}{2},\left|y^{\prime}\right|=$ $\left|y-n^{\prime}\right| \leqslant \frac{1}... | c=\frac{1}{a+2} | Algebra | proof | Yes | Yes | inequalities | false | 735,098 |
9-169 Let $x, y, z$ be positive real numbers, and $x y z=1$. Prove:
$$\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geqslant \frac{3}{4} .$$ | ［Proof］The original inequality is equivalent to
$$x^{4}+x^{3}+y^{4}+y^{3}+z^{4}+z^{3} \geqslant \frac{3}{4}(1+x)(1+y)(1+z)$$
For any positive numbers $u, v, w$, we have
$$u^{3}+v^{3}+w^{3} \geqslant 3 u v w$$
Therefore, it suffices to prove
$$x^{4}+x^{3}+y^{4}+y^{3}+z^{4}+z^{3} \geqslant \frac{1}{4}\left[(x+1)^{3}+(y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,099 |
1- 141 Proof: There are infinitely many pairs of natural numbers $a, b$, satisfying: (1) $a, b$ (both represented in decimal) have the same number of digits;
(2) $a, b$ are both perfect squares;
(3) Writing $b$ after $a$ produces a perfect square.
(For example, $a=16, b=81, 1681=41^{2}$) | [Solution] Let $a=\left(5 \times 10^{n-1}-1\right)^{2}, b=\left(10^{n}-1\right)^{2}$, then $a, b$ are both $2n$-digit numbers, and
$$\begin{aligned}
& \left(5 \times 10^{n-1}-1\right)^{2} \times 10^{2 n}+\left(10^{n}-1\right)^{2} \\
= & \left(5 \times 10^{2 n-1}\right)^{2}-\left(10^{n}-1\right) \times 10^{2 n}+\left(10... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,101 |
9. 171 Let $a, b, c$ be positive real numbers satisfying $abc=1$. Prove that:
$$\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} .$$ | [Proof] Let
$$A=\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}$$
Then, by $\frac{1}{a}=b c$, we have
$$A=\frac{b^{2} c^{2}}{a(b+c)}+\frac{c^{2} a^{2}}{b(c+a)}+\frac{a^{2} b^{2}}{c(a+b)}$$
Using the Cauchy-Schwarz inequality and the Arithmetic-Geometric Mean inequality, we get
$$\begin{aligned}
& {[a(b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,102 |
9.172 Let $a, b, c$ be positive real numbers, and satisfy $abc=1$. Prove:
$$\frac{ab}{a^{5}+b^{5}+ab}+\frac{bc}{b^{5}+c^{5}+bc}+\frac{ca}{c^{5}+a^{5}+ca} \leqslant 1,$$
and specify when equality holds. | [Proof] From
$$a^{5}+b^{5}-a^{2} b^{2}(a+b)=\left(a^{2}-b^{2}\right)\left(a^{3}-b^{3}\right) \geqslant 0,$$
we can get
$$a^{5}+b^{5} \geqslant a^{2} b^{2}(a+b)$$
From the given and (1), we have
$$\begin{aligned}
& \frac{a b}{a^{5}+b^{5}+a b}=\frac{a^{2} b^{2} c}{a^{5}+b^{5}+a^{2} b^{2} c} \\
\leqslant & \frac{a^{2} b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,103 |
9. 173 Let $x, y, z$ be real numbers, prove:
$$|x|+|y|+|z| \leqslant |x+y-z|+|x-y+z|+|-x+y+z| \text {. }$$ | [Proof] Since
$$(x+y-z)+(x-y+z)=2 x,$$
then
$$|x+y-z|+|x-y+z| \geqslant 2|x| .$$
Similarly, we have
$$\begin{array}{c}
|x-y+z|+|-x+y+z| \geqslant 2|z|, \\
|x+y-z|+|-x+y+z| \geqslant 2|y|,
\end{array}$$
Adding (1), (2), and (3) yields
$$\begin{aligned}
& 2(|x+y-z|+|x-y+z|+|-x+y+z|) \\
\geqslant & 2(|x|+|y|+|z|)
\end{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,104 |
9. Proposition (*): Let $a, b, c$ be non-negative real numbers. If $a^{4}+b^{4}+c^{4} \leqslant 2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)$, then
$$a^{2}+b^{2}+c^{2} \leqslant 2(a b+b c+c a) .$$
(1) Prove that proposition (*) is correct;
(2) Write the converse of proposition (*), and determine whether the conver... | ［Solution］(1) Let
$$\begin{aligned}
D & =2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right) \\
& =4 a^{2} b^{2}-\left(a^{2}+b^{2}\right)^{2}+2 c^{2}\left(a^{2}+b^{2}\right)-c^{4} \\
& =(2 a b)^{2}-\left(a^{2}+b^{2}-c^{2}\right)^{2}
\end{aligned}$$
Given that $D \geqslant 0$, we have
$$\be... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,105 |
9・175 Let real numbers $a, b, c, d, p, q$ satisfy:
$$a b+c d=2 p q, a c \geqslant p^{2}>0$$
Prove that $\quad b d \leqslant q^{2}$. | [Proof] By contradiction. If $b d>q^{2}$, then
$$\begin{aligned}
4 a b c d & =4(a c)(b d)>4 p^{2} q^{2}=(a b+c d)^{2} \\
& =a^{2} b^{2}+2 a b c d+c^{2} d^{2}
\end{aligned}$$
This leads to
$$(a b-c d)^{2}<0$$
Contradiction! | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,106 |
9. 177 Let \(a, b, c, d\) be positive real numbers, prove that:
$$\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d} \geqslant \frac{64}{a+b+c+d}$$ | [Proof]
$$\begin{aligned}
& \left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\right)(a+b+c+d) \\
= & 1+\frac{a}{b}+\frac{4 a}{c}+\frac{16 a}{d}+\frac{b}{a}+1+\frac{4 b}{c}+\frac{16 b}{d}+ \\
& +\frac{c}{a}+\frac{c}{b}+4+\frac{16 c}{d}+\frac{d}{a}+\frac{d}{b}+\frac{4 d}{c}+16 \\
= & 22+\left(\frac{a}{b}+\frac{b}{a}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,107 |
9・178 Let $f(x)=a_{1} \sin x+a_{2} \sin 2 x+\cdots+a_{n} \sin n x$, where $a_{1}$, $a_{2}, \cdots, a_{n}$ are real numbers, and $n$ is a positive integer. If for all real numbers $x$ we have
$$|f(x)| \leqslant|\sin x|$$
Prove: $\left|a_{1}+2 a_{2}+\cdots+n a_{n}\right| \leqslant 1$. | [Proof] Let $M=\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right|$. For a positive integer $k(1 \leqslant k \leqslant n)$, since
$$\lim _{x \rightarrow 0} \frac{\sin k x}{\sin x}=k$$
for any $\varepsilon>0$, there exists a real number $x$, such that $\sin x \neq 0$, and
$$\left|\frac{\sin k x}{\sin x}-k\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,108 |
9・179 Let $a_{1}, a_{2}, \cdots, a_{n}$ be natural numbers, where $n>2$, and $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$. Prove: For all real numbers $x_{1}, x_{2}, \cdots, x_{n}$ that are not all 0, the inequality
$$\sum_{i=1}^{n} a_{i} x_{i}^{2}+2 \sum_{i=1}^{n-1} x_{i} x_{i+1}>0$$
holds if and only if ... | [Proof] Necessity. If $a_{2}=1$, then $a_{1}=1$. Take
$$x_{1}=1, x_{2}=-1, x_{3}=\frac{1}{a_{3}}, x_{4}=\cdots=x_{n}=0$$
Then
$$\begin{aligned}
\sum_{i=1}^{n} a_{i} x_{i}^{2}+2 \sum_{i=1}^{n-1} x_{i} x_{i+1} & =x_{1}^{2}+x_{2}^{2}+a_{3} x_{3}^{2}+2 x_{1} x_{2}+2 x_{2} x_{3} \\
& =-\frac{1}{a_{3}}<0
\end{aligned}$$
Th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,109 |
9-180 Let $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$ be real numbers. Prove that the necessary and sufficient condition for the inequality
$$\sum_{i=1}^{n} a_{i} x_{i} \leqslant \sum_{i=1}^{n} b_{i} x_{i}$$
to hold for any real numbers satisfying $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$... | [Proof] First, we prove the necessity. Let $x_{1}=x_{2}=\cdots=x_{n}=1$, then
$$\sum_{i=1}^{n} a_{i} \leqslant \sum_{i=1}^{n} b_{i}$$
Next, let $x_{1}=x_{2}=\cdots=x_{n}=-1$, we get
$$-\sum_{i=1}^{n} a_{i} \leqslant-\sum_{i=1}^{n} b_{i}$$
Thus, $\quad \sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}$.
For $1 \leqslant k \le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,110 |
1. Does there exist a perfect square whose digit sum is 1993? | [Solution] It exists. Because
$$\underbrace{99 \cdots 99}_{n \uparrow 9} 7^{2}=\underbrace{99 \cdots 99}_{n \uparrow 9} 4 \underbrace{00 \cdots 00}_{n \uparrow 0} 9,$$
and when $n=220$, the perfect square
$$\underbrace{99 \cdots 99}_{220 \uparrow 9} 4 \underbrace{00 \cdots 0}_{220 \uparrow 0} 9$$
has a digit sum of 1... | 1993 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,111 |
9・181 Let $a, b, A, B$ be given real numbers,
$$f(\theta)=1-a \cos \theta-b \sin \theta-A \cos 2 \theta-B \sin 2 \theta \text {. }$$
If for all real numbers $\theta$ we have $f(\theta) \geqslant 0$, prove:
$$a^{2}+b^{2} \leqslant 2, A^{2}+B^{2} \leqslant 1$$ | [Proof] Take $\beta, r \in [0, 2\pi]$ such that
$$\begin{array}{l}
a \cos \theta + b \sin \theta = \sqrt{a^2 + b^2} \cos (\theta - \beta) \\
A \cos 2\theta + B \sin 2\theta = \sqrt{A^2 + B^2} \cos (2\theta - \gamma)
\end{array}$$
Then for all real numbers $\theta$,
$$\sqrt{A^2 + B^2} \cos (2\theta - \gamma) \leq 1 - \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,112 |
9-182 Let $a, b, c, d$ all be positive numbers, prove that:
$$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \geqslant 2$$ | [Proof] We discuss in two cases.
(i) $\frac{b}{b+c}+\frac{a}{a+b}+\frac{d}{d+a}+\frac{c}{c+d} \leqslant 2$. By the AM-GM inequality, we have
$$\frac{a+b}{b+c}+\frac{d+a}{a+b}+\frac{c+d}{d+a}+\frac{b+c}{c+d} \geqslant 4,$$
which implies that the original inequality holds.
(ii) $\frac{b}{b+c}+\frac{a}{a+b}+\frac{d}{d+a}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,113 |
9-183 Let $a, b, c, d$ be non-negative real numbers satisfying
$$a b+b c+c d+d a=1$$
Prove:
$$\frac{a^{3}}{b+c+d}+\frac{b^{3}}{a+c+d}+\frac{c^{3}}{a+b+d}+\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3} .$$ | [Proof] Let $a=x_{1}, b=x_{2}, c=x_{3}, d=x_{4}, s=x_{1}+x_{2}+x_{3}+x_{4}$, the original inequality can be written as
$$\sum_{i=1}^{4} \frac{x_{i}^{3}}{s-x_{i}} \geqslant \frac{1}{3} \text {. }$$
By Chebyshev's inequality, we have
$$\begin{aligned}
\sum_{i=1}^{4} \frac{x_{i}^{3}}{s-x_{i}} & \geqslant \frac{1}{4}\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,114 |
9・184 Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ all be positive numbers, prove that:
$$\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)^{2}>4\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}+x_{5} x_{1}\right) .$$ | [Proof] By the cyclic symmetry, we may assume
$$x_{1}=\min \left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\},$$
Let $x_{5+k}=x_{k}, k=1,2,3,4,5$. Since
$$\begin{aligned}
& \left(\sum_{k=1}^{5} x_{k}\right)^{2}-4 \sum_{k=1}^{5} x_{k} x_{k+1}=\sum_{k=1}^{5} x_{k}^{2}=2 \sum_{k=1}^{5} x_{k} x_{k+1}+2 \sum_{k=1}^{5} x_{k} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,115 |
9-186 Let $x_{1}, x_{2}, y_{1}, y_{2}, z_{1}, z_{2}$ be real numbers satisfying
$$x_{1}>0, x_{2}>0, x_{1} y_{1}-z_{1}^{2}>0, x_{2} y_{2}-z_{2}^{2}>0$$
Prove that $\frac{8}{\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(z_{1}+z_{2}\right)^{2}} \leqslant \frac{1}{x_{1} y_{1}-z_{1}^{2}}+\frac{1}{x_{2} y_{2}-z_{2}... | [Proof] Let $k_{1}=x_{1} y_{1}-z_{1}^{2}, k_{2}=x_{2} y_{2}-z_{2}^{2}$, then $k_{1}, k_{2}>0$ and
$$\begin{aligned}
x_{1} y_{2}+x_{2} y_{1} & =\frac{x_{1}}{x_{2}} x_{2} y_{2}+\frac{x_{2}}{x_{1}} x_{1} y_{1}=\frac{x_{1}}{x_{2}}\left(k_{2}+z_{2}^{2}\right)+\frac{x_{2}}{x_{1}}\left(k_{1}+z_{1}^{2}\right) \\
& \geqslant 2 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,117 |
9・187 Let positive numbers $b_{1}, b_{2}, \cdots, b_{1989}$ be such that the system of equations
$$x_{r-1}-2 x_{r}+x_{r+1}+b_{r} x_{r}=0, r=1,2, \cdots, 1989$$
has a solution, and $x_{0}=x_{1989}=0, x_{1}, x_{2}, \cdots, x_{1989}$ are not all 0.
Prove: $b_{1}+b_{2}+\cdots+b_{1989} \geqslant \frac{2}{995}$. | [Proof] Let $n=1989, \left|x_{k}\right|=\max _{0 \leqslant i \leqslant n}\left|x_{i}\right|>0$. Since
$$\begin{aligned}
\sum_{r=1}^{k}-r b_{r} x_{r}= & \sum_{r=1}^{k} r\left(x_{r-1}-2 x_{r}+x_{r+1}\right) \\
= & x_{0}+\sum_{r=1}^{k-1}(r+1-2 r+r-1) x_{r}-2 k x_{k}+(k-1) x_{k}+ \\
& k x_{k+1} \\
& =-(k+1) x_{k}+k x_{k+1}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,118 |
9-188 Let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{10}$, prove that:
$$\frac{a_{1}+a_{2}+\cdots+a_{6}}{6} \leqslant \frac{a_{1}+a_{2}+\cdots+a_{10}}{10}$$ | [Proof] Let $\frac{a_{1}+a_{2}+\cdots+a_{6}}{6}=k$, by the assumption we have
then $a_{1}+a_{2}+\cdots+a_{10} \geqslant 6 k+4 k=10 k$,
i.e., $\quad \frac{a_{1}+a_{2}+\cdots+a_{10}}{10} \geqslant k=\frac{a_{1}+a_{2}+\cdots+a_{6}}{6}$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,119 |
9. 189 Let $A, B, C, D$ be four points in space, prove that:
$$|A C|^{2}+|B D|^{2}+|A D|^{2}+|B C|^{2} \geqslant|A B|^{2}+|C D|^{2} .$$ | [Proof] Let $\left(x_{i}, y_{i}, z_{i}\right), i=1,2,3,4$, be the Cartesian coordinates of $A, B, C, D$, respectively, then
$$\begin{array}{l}
|A C|^{2}=\left(x_{1}-x_{3}\right)^{2}+\left(y_{1}-y_{3}\right)^{2}+\left(z_{1}-z_{3}\right)^{2}, \\
|B D|^{2}=\left(x_{2}-x_{4}\right)^{2}+\left(y_{2}-y_{4}\right)^{2}+\left(z_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,120 |
9・190 Given positive numbers $a, b, c, x, y, z$ and $k$ satisfying
$$a+x=b+y=c+z=k$$
Prove: $a y+b z+c x<k^{2}$. | [Proof] Since $k^{3}=(a+x)(b+y)(c+z)$
$$\begin{array}{l}
=a b c+a c y+b c x+c x y+a b z+a y z+b x z+x y z \\
=a b c+x y z+k(a y+b z+c x)
\end{array}$$
Therefore,
$$a y+b z+c x<k^{2} \text {. }$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,121 |
1. Does there exist 100 positive integers such that their sum equals their least common multiple?
143 | ［Solution］Exists.
In fact, we can take the following 102 numbers:
$$1,2,2^{2}, 2^{3}, \cdots, 2^{67}$$
and $3,3 \cdot 2,3 \cdot 2^{3}, 3 \cdot 2^{5}, \cdots, 3 \cdot 2^{65}$.
Their sum is
$$2^{68}-1+3+3 \cdot 2 \cdot \frac{\left(2^{2}\right)^{33}-1}{2^{2}-1}=3 \cdot 2^{67}$$
Their least common multiple is also $3 \cd... | 3 \cdot 2^{67} | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,122 |
9・191 Let $x_{i} \in[0,1], i=1,2,3,4,5,6$. Prove:
$$\left(x_{1}-x_{2}\right)\left(x_{2}-x_{3}\right)\left(x_{3}-x_{4}\right)\left(x_{4}-x_{5}\right)\left(x_{5}-x_{6}\right)\left(x_{6}-x_{1}\right) \leqslant \frac{1}{16}$$ | [Proof] Let $y_{i}=x_{i}-x_{i+1}, i=1,2,3,4,5, y_{6}=x_{6}-x_{1}$. Without loss of generality, assume $y_{1}$, $y_{2}$, $y_{3}$, $y_{4}$, $y_{5}$, $y_{6}$ are all non-zero. Suppose there are $l$ negative $y_{i}$ and the remaining $6-l$ $y_{i}$ are positive. By $\sum_{i=1}^{6} y_{i}=0$, we have $1 \leqslant l \leqslant ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,123 |
9・192 Let $0<p \leqslant a, b, c, d, e \leqslant q$, prove that
$$\begin{aligned}
& (a+b+c+d+e)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}\right) \\
\leqslant & 25+6\left(\sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}\right)^{2}
\end{aligned}$$
and determine when equality holds. | [Proof] Given positive numbers $u, v$, consider the function
$$f(x)=(u+x)\left(v+\frac{1}{x}\right), 0 \leqslant p \leqslant x \leqslant q$$
It can be proven that for any $x \in [p, q]$,
$$f(x) \leqslant \max \{f(p), f(q)\}$$
In fact, without loss of generality, assume $p < q$, and let $\lambda = \frac{q-x}{q-p}$, th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,124 |
9. 194 Prove: For any real numbers $a_{1}, a_{2}, \cdots, a_{1987}$ and any positive numbers $b_{1}, b_{2}, \cdots, b_{1987}$, we have
$$\frac{\left(a_{1}+a_{2}+\cdots+a_{1987}\right)^{2}}{b_{1}+b_{2}+\cdots+b_{1987}} \leqslant \frac{a_{1}^{2}}{b_{1}}+\frac{a_{2}^{2}}{b_{2}}+\cdots+\frac{a_{1987}^{2}}{b_{1987}}$$ | [Proof] Let $s=b_{1}+b_{2}+\cdots+b_{1987}$, then the left side of the original inequality is $s \cdot\left(\frac{b_{1}}{s} \cdot \frac{a_{1}}{b_{1}}+\frac{b_{2}}{s} \cdot \frac{a_{2}}{b_{2}}+\cdots+\frac{b_{1987}}{s} \cdot \frac{a_{1987}}{b_{1987}}\right)^{2}$.
By the convexity of the function $f(x)=x^{2}$, we have
$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,126 |
9- 195 Given in the plane $\vec{a}+\vec{b}+\vec{c}+\vec{d}=\overrightarrow{0}$, prove that $|\vec{a}|+|\vec{b}|+|\vec{c}|+$ $|\vec{d}| \geqslant|\vec{a}+\vec{d}|+|\vec{b}+\vec{d}|+|\vec{c}+\vec{d}|$ | [Proof] Clearly, the left side of (1) is symmetric with respect to $\vec{a}, \vec{b}, \vec{c}, \vec{d}$. From $\vec{a}+\vec{b}+\vec{c}+\vec{d}=\overrightarrow{0}$, we know
$$|\vec{a}+\vec{d}|=|\vec{b}+\vec{c}|,|\vec{b}+\vec{d}|=|\vec{a}+\vec{c}|,|\vec{c}+\vec{d}|=|\vec{a}+\vec{b}|,$$
Thus, the right side of (1) is als... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,127 |
9-196 Given that $a_{1}, a_{2}, \cdots, a_{n}$ are all positive numbers and $a_{1} \cdot a_{2} \cdots a_{n}=1$. Prove: $\left(2+a_{1}\right)\left(2+a_{2}\right) \cdots\left(2+a_{n}\right) \geqslant 3^{n}$. | [Proof] By the AM-GM inequality
$$2+a_{k}=1+1+a_{k} \geqslant 3 a_{k}^{\frac{1}{3}}, k=1,2, \cdots, n$$
Multiplying both sides of the $n$ inequalities, noting that $a_{1} \cdot a_{2} \cdots \cdot a_{n}=1$, we obtain the desired inequality. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,128 |
9・198 Let $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ be permutations of $1, \frac{1}{2}, \cdots, \frac{1}{n}$. If
$$a_{1}+b_{1} \geqslant a_{2}+b_{2} \geqslant \cdots \geqslant a_{n}+b_{n}$$
Prove: $a_{m}+b_{m} \leqslant \frac{4}{m}, m=1,2, \cdots, n$. | [Proof] For any $1 \leqslant m \leqslant n$, among the $m$ numbers $a_{1}-b_{1}, a_{2}-b_{2}, \cdots, a_{m}-b_{m}$, either at least $\frac{m}{2}$ of them are non-negative, or at least $\frac{m}{2}$ of them are non-positive. Without loss of generality, assume that at least $\frac{m}{2}$ of them are non-negative, i.e., t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,130 |
9・199 Let real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $n \geqslant 2$ and
$$0 \leqslant a_{1} \leqslant a_{2} \leqslant 2 a_{1}, a_{2} \leqslant a_{3} \leqslant 2 a_{2}, \cdots, a_{n-1} \leqslant a_{n} \leqslant 2 a_{n-1}$$
Prove: In $S= \pm a_{1} \pm a_{2} \pm \cdots \pm a_{n}$, the signs can be chosen appropr... | [Proof] By induction. When $n=2$, take $S=-a_{1}+a_{2}$. Since $0 \leqslant a_{1} \leqslant a_{2}$ $\leqslant 2 a_{1}$, then $0 \leqslant S \leqslant a_{1}$.
Assume the proposition holds for $n=k$, consider the case for $n=k+1$.
Take any $a_{1}, a_{2}, \cdots, a_{k}, a_{k+1}$ satisfying the given conditions. For $a_{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,131 |
9・200 Given that $x_{i}, y_{i}(i=1,2, \cdots, n)$ are real numbers and
$$\begin{array}{l}
x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n} \\
y_{1} \geqslant y_{2} \geqslant \cdots \geqslant y_{n}
\end{array}$$
Also, $z_{1}, z_{2}, \cdots, z_{n}$ is any permutation of $y_{1}, y_{2}, \cdots, y_{n}$, prove that:
$... | [Proof]
$$\begin{aligned}
\sum_{i=1}^{n}\left(x_{i}-y_{i}\right)^{2} & =\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n} y_{i}^{2}-2 \sum_{i=1}^{n} x_{i} y_{i}, \sum_{i=1}^{n}\left(x_{i}-z_{i}\right)^{2} \\
& =\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n} z_{i}^{2}-2 \sum_{i=1}^{n} x_{i} z_{i} .
\end{aligned}$$
Since $z_{1}, z_{2}, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,132 |
$1 \cdot 15$ Choose a 1962-digit number that is divisible by 9, and let the sum of its digits be $a$, the sum of the digits of $a$ be $b$, and the sum of the digits of $b$ be $c$. What is $c$? | [Solution] Since any number and the sum of its digits have the same remainder when divided by 9,
thus $c \geqslant 9$, and $9 \mid c$.
On the other hand, $a \leqslant 1962 \times 9 < 19999$,
therefore $b < 1 + 4 \times 9 = 37$,
so $c \leqslant 11$. Hence, $c=9$. | 9 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,133 |
1-144 (1) Does there exist 4 different natural numbers such that the sum of any three of them is a prime number?
(2) What about 5? | [Solution] (1) Exists. For example, the four numbers $1,3,7,9$ meet the condition.
(2) Does not exist. Because: The remainder of any natural number divided by 3 can only be $0,1,2$, three possibilities. For any 5 natural numbers, when divided by 3, if the remainders include all three $0,1,2$, then the sum of the three ... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,134 |
9・202 Given non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$. If $x_{1}+x_{2}+\cdots+x_{n}=$ $n$, prove:
$$\begin{array}{c}
\frac{x_{1}}{1+x_{1}^{2}}+\frac{x_{2}}{1+x_{2}^{2}}+\cdots+\frac{x_{n}}{1+x_{n}^{2}} \leqslant \frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+ \\
\frac{1}{1+x_{n}}
\end{array}$$ | [Proof] For $1 \leqslant k \leqslant n$ we have
$$\frac{x_{k}}{1+x_{k}^{2}}-\frac{1}{1+x_{k}}=\frac{x_{k}-1}{\left(1+x_{k}^{2}\right)\left(1+x_{k}\right)}$$
If $x_{k} \geqslant 1$, then $\left(1+x_{k}^{2}\right)\left(1+x_{k}\right) \geqslant 4$. If $x_{k}<1$, then $\left(1+x_{k}^{2}\right)\left(1+x_{k}\right)<$
4. Thu... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,135 |
9・203 Given real numbers $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 2)$ satisfy:
$$\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=a, \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=\frac{a^{2}}{n-1},
\end{array}$$
where $a>0$, prove:
$$0 \leqslant x_{i} \leqslant \frac{2 a}{n}, i=1,2, \cdots, n$$ | [Proof] First, we prove $x_{i} \geqslant 0, i=1,2, \cdots, n$. If not, then there exists $i \in\{1,2, \cdots, n\}$, such that $x_{i}<0$. From $x_{1}+x_{2}+\cdots+x_{n}=a>0$, it is known that there must exist $j \in\{1,2, \cdots, n\}$, such that $x_{j}>0$. Without loss of generality, assume $x_{1}>0$. Then
$$\begin{alig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,136 |
9.205 For any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, prove:
$$\frac{a_{1}}{a_{2}+a_{3}}+\frac{a_{2}}{a_{3}+a_{4}}+\cdots+\frac{a_{n-2}}{a_{n-1}+a_{n}}+\frac{a_{n-1}}{a_{n}+a_{1}}+\frac{a_{n}}{a_{1}+a_{2}}>\frac{n}{4}$$ | [Proof] Let $a_{n+k}=a_{k}, k=1,2,3, \cdots, n$, then the left side of the inequality can be written as $S=\sum_{k=1}^{n} \frac{a_{k}}{a_{k+1}+a_{k+2}}$. Without loss of generality, assume
$$a_{1}=\max _{1 \leqslant k \leqslant n} .$$
Let
$$\begin{array}{l}
n_{1}=1 ; \\
n_{2}=\left\{\begin{array}{l}
2, a_{2} \geqslant... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,138 |
9-206 Let $a_{1}, a_{2}, \cdots, a_{n}$ be real numbers. If the sum of any two of them is non-negative, then for any non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying
$$x_{1}+x_{2}+\cdots+x_{n}=1,$$
the following inequality holds:
$$a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n} \geqslant a_{1} x_{1}^{2}+... | [Proof] Proposition Proof:
$$\begin{aligned}
& a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n} \\
= & \left(a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}\right)\left(x_{1}+x_{2}+\cdots+x_{n}\right) \\
= & \sum_{i, j=1}^{n} a_{i} x_{i} x_{j} \\
= & \sum_{i=1}^{n} a_{i} x_{i}^{2}+\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}+a_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,139 |
$9 \cdot 207$ Let $x_{1}, x_{2}, \cdots, x_{n}$ be non-negative real numbers, and denote $x_{1}+x_{2}+\cdots+x_{n}=a.$
Prove: $x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{n-1} x_{n} \leqslant \frac{a^{2}}{4}$. | ［Proof］ When $n=2,3$, it is obvious that
$$\begin{array}{l}
x_{1} x_{2} \leqslant \frac{\left(x_{1}+x_{2}\right)^{2}}{4}=\frac{a^{2}}{4} \\
x_{1} x_{2}+x_{2} x_{3}=x_{2}\left(x_{1}+x_{3}\right) \leqslant \frac{\left(x_{1}+x_{2}+x_{3}\right)^{2}}{4}=\frac{a^{2}}{4}
\end{array}$$
When $n \geqslant 4$, a stronger inequal... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,140 |
9.208 Let $0 \leqslant t \leqslant 1,1 \geqslant x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}>0$, prove:
$$\left(1+x_{1}+x_{2}+\cdots+x_{n}\right)^{t} \leqslant 1+x_{1}^{t}+2^{t-1} x_{2}^{t}+\cdots+n^{t-1} x_{n}^{t} \text {. }$$ | [Proof] By induction. When $n=1$, given $0 \leqslant t \leqslant 1, 0 \leqslant x_{1} \leqslant 1$, then $\left(1+x_{1}\right)^{t} \leqslant 1+x_{1} \leqslant 1+x_{1}^{t}$.
Assume the inequality holds for $n=k$, then
$$\begin{aligned}
& \left(1+x_{1}+\cdots+x_{k}+x_{k+1}\right)^{t} \\
= & \left(1+\frac{x_{k+1}}{1+x_{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,141 |
9・209 Let real numbers $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ satisfy
$$\begin{array}{l}
a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}>0 \\
b_{1} \geqslant a_{1} \\
b_{1} b_{2} \geqslant a_{1} a_{2} \\
b_{1} b_{2} b_{3} \geqslant a_{1} a_{2} a_{3} \\
\cdots \cdots \\
b_{1} b_{2} \cdots... | [Proof] Let $k_{i}=\frac{b_{i}}{a_{i}}, i=1,2, \cdots, n$, then
$$k_{1} \geqslant 1, k_{1} k_{2} \geqslant 1, \cdots, k_{1} k_{2} \cdots k_{n} \geqslant 1$$
The inequality to be proved is equivalent to
$$a_{1}\left(k_{1}-1\right)+a_{2}\left(k_{2}-1\right)+\cdots+a_{n}\left(k_{n}-1\right) \geqslant 0$$
Let $d_{m}=\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,142 |
9・210 Let real numbers $a_{0}, a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{n}=0$ and
$$a_{k}=c+\sum_{i=k}^{n-1} a_{i-k}\left(a_{i}+a_{i+1}\right), k=0,1,2, \cdots, n-1 .$$
Prove: $c \leqslant \frac{1}{4 n}$. | [Proof] Let $s_{k}=\sum_{i=0}^{k} a_{i}, k=0,1,2, \cdots, n$, then
$$\begin{aligned}
s_{n}= & \sum_{k=0}^{n} a_{k}=\sum_{k=0}^{n-1} a_{k}=n c+\sum_{k=0}^{n-1} \sum_{i=k}^{n-1} a_{i-k}\left(a_{i}+a_{i+1}\right) \\
= & n c+\sum_{i=0}^{n-1} \sum_{k=0}^{i} a_{i-k}\left(a_{i}+a_{i+1}\right) \\
= & n c+\sum_{i=0}^{n-1}\left(... | c \leqslant \frac{1}{4 n} | Algebra | proof | Yes | Yes | inequalities | false | 735,143 |
1- 145 A six-digit number has its first digit as 5. Is it always possible to add 6 more digits after it so that the resulting twelve-digit number is a perfect square? | [Solution] The answer is negative.
In fact, if it were always possible, then the $10^{5}$ six-digit numbers starting with 5 could be extended to at least $10^{5}$ twelve-digit perfect squares $n^{2}$.
On the other hand, these perfect squares satisfy
$$\begin{array}{l}
5 \times 10^{11} \leqslant n^{2}<6 \times 10^{11} ... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,144 |
$9 \cdot 211$ Let $a_{1}, a_{2}, \cdots, a_{n}$ all be positive numbers $(n \geqslant 2), k \geqslant 1$, prove:
$$\begin{array}{l}
\left(\frac{a_{1}}{a_{2}+\cdots+a_{n}}\right)^{k}+\left(\frac{a_{2}}{a_{3}+\cdots+a_{n}+a_{1}}\right)^{k}+\cdots+\left(\frac{a_{n}}{a_{1}+\cdots+a_{n-1}}\right)^{k} \\
\geqslant \frac{n}{(... | [Proof] Let $s=a_{1}+\cdots+a_{n}$, the inequality to be proved can be written as
$$\sum_{i=1}^{n}\left(\frac{a_{i}}{S-a_{i}}\right)^{k} \geqslant \frac{n}{(n-1)^{k}}$$
By the AM-GM inequality, we have
$$\begin{aligned}
\sum_{i=1}^{n} \frac{a_{i}}{S-a_{i}} & =S \sum_{i=1}^{n} \frac{1}{S-a_{i}}-n \\
& \geqslant n S \sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,145 |
$9 \cdot 212$ For any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, prove:
$$\sum_{k=1}^{n} \sqrt[k]{a_{1} \cdots a_{k}} \leqslant e \sum_{k=1}^{n} a_{k}$$
where $e=\lim _{k \rightarrow \infty}\left(1+\frac{1}{k}\right)^{k}$. | [Proof] Given that $\left(1+\frac{1}{k}\right)^{k}$ monotonically increases and converges to $e$, thus for any $i \in \mathbb{N}$, we have $i\left(1+\frac{1}{i}\right)^{i} \leqslant i e$; Let $b_{i}=i\left(1+\frac{1}{i}\right)^{i}$, then
$$\frac{b_{i}}{i} \leqslant e$$
From $b_{1} b_{2} \cdots b_{k}=2 \cdot \frac{3^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,146 |
9. 213 Given $5 n$ real numbers $r_{i}, s_{i}, t_{i}, u_{i}, v_{i}$ all greater than $1(1 \leqslant i \leqslant n)$, let $R=\frac{1}{n} \sum_{i=1}^{n} r_{i}, S=\frac{1}{n} \sum_{i=1}^{n} s_{i}, T=\frac{1}{n} \sum_{i=1}^{n} t_{i}, U=\frac{1}{n} \sum_{i=1}^{n} u_{i}, V=\frac{1}{n}$ $\sum_{i=1}^{n} v_{i}$, prove:
$$\prod_... | [Proof] First, prove that for any $n$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$ greater than 1, we have
$$\prod_{i=1}^{n} \frac{x_{i}+1}{x_{i}-1} \geqslant\left(\frac{A+1}{A-1}\right)^{n}$$
where $A=\sqrt[n]{x_{1} x_{2} \cdots \cdots x_{n}}$.
In fact, let $x_{i}=\max \left(x_{1}, x_{2}, \cdots, x_{n}\right), x_{j}=\m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,147 |
9・214 Let $x_{1}, x_{2}, \cdots, x_{n}$ all be positive numbers $(n \geqslant 2)$ and
$$\sum_{i=1}^{n} x_{i}=1$$
Prove:
$$\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1-x_{i}}} \geqslant \frac{1}{\sqrt{n-1}} \sum_{i=1}^{n} \sqrt{x_{i}}$$ | [Proof] By symmetry, we can assume $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$, thus we have
$$\frac{1}{\sqrt{1-x_{1}}} \leqslant \frac{1}{\sqrt{1-x_{2}}} \leqslant \cdots \leqslant \frac{1}{\sqrt{1-x_{n}}} .$$
By Chebyshev's inequality, we get
$$\begin{aligned}
\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1-x_{i}}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,148 |
$$\begin{array}{l}
9 \cdot 215 \text { Let }-1 \leqslant x_{i} \leqslant 1, i=1,2, \cdots, n \text { and } \\
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=0 .
\end{array}$$
Prove: $x_{1}+x_{2}+\cdots+x_{n} \leqslant \frac{n}{3}$. | [Proof] Since when $x \geqslant -1$, $(x+1)(2x-1)^2 \geqslant 0$, that is,
$$4x^3 - 3x + 1 \geqslant 0$$
Therefore, $4 \sum_{k=1}^{n} x_k^3 - 3 \sum_{k=1}^{n} x_k + n = -3 \sum_{k=1}^{n} x_k + n \geqslant 0$. This immediately shows that the inequality to be proven holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,149 |
9・216 Let $a_{1}, a_{2}, \cdots, a_{n}$ be real numbers, and for a natural number $k(1 \leqslant k \leqslant n)$, let
$$S_{k}=\sum_{1 \leqslant i_{1}<\cdots<i_{k} \leqslant n} a_{i_{1}} a_{i_{2}} \cdots a_{i_{k}},$$
Prove:
$$S_{k} S_{n-k} \geqslant\left(C_{n}^{k}\right)^{2} a_{1} a_{2} \cdots a_{n}$$ | [Proof] By the AM-GM inequality $S_{k} \geqslant C_{n}^{k}\left(\prod_{1 \leqslant i_{1}<\cdots<i_{k} \leqslant n} a_{i_{1}} a_{i_{2}} \cdots a_{i_{k}}\right) \frac{1}{C_{n}^{k}}$
$$=C_{n}^{k}\left(a_{1} a_{2} \cdots a_{n}\right)^{\frac{C_{n-1}^{k-1}}{C_{n}^{k}}}$$
Since $C_{n}^{k}=C_{n}^{n-k}, C_{n-1}^{k-1}+C_{n-1}^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,150 |
9・217 Let $x_{1}, x_{2}, \cdots, x_{n}$ all be positive numbers, and let $s=x_{1}+x_{2}+\cdots+x_{n}$. Prove:
$$\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \leqslant 1+s+\frac{s^{2}}{2!}+\cdots+\frac{s^{n}}{n!}$$ | [Proof] By the AM-GM inequality, we have
$$\begin{aligned}
\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) & \leqslant\left[\frac{1}{n} \sum_{i=1}^{n}\left(1+x_{i}^{\prime}\right)\right]^{n} \\
& =\left(1+\frac{s}{n}\right)^{n}
\end{aligned}$$
Using the binomial expansion, we get
$$\left(1+\frac{s}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,151 |
9・218 Let $a_{1}, a_{2}, \cdots, a_{n}$ have the same sign, and let $S=\sum_{i=1}^{n} a_{i}$. Prove that:
$$\sum_{i=1}^{n} \frac{a_{i}}{2 S-a_{i}} \geqslant \frac{n}{2 n-1}$$ | [Proof] Without loss of generality, let $a_{1}, a_{2}, \cdots, a_{n}$ all be positive, then
$$\sum_{i=1}^{n} \frac{a_{i}}{2 s-a_{i}}=\sum_{i=1}^{n} \frac{a_{i}-2 s+2 s}{2 s-a_{i}}=-n+2 s \sum_{i=1}^{n} \frac{1}{2 s-a_{i}}$$
By the AM-GM inequality, we have
$$\begin{aligned}
\sum_{i=1}^{n} \frac{1}{2 s-a_{i}} & \geqsla... | \frac{n}{2 n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 735,152 |
9・219 Let $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 3)$ be non-negative real numbers, and $x_{1}+x_{2}+\cdots+x_{n}=1$.
Prove: $x_{1}^{2} x_{2}+x_{2}^{2} x_{3}+\cdots+x_{n-1}^{2} x_{n}+x_{n}^{2} x_{1} \leqslant \frac{4}{27}$. | [Proof] By induction. When $n=3$, without loss of generality, assume $x_{1} \geqslant x_{2}, x_{1} \geqslant x_{3}$. If $x_{2} \geqslant x_{3}$, then by the AM-GM inequality we have
$$\begin{aligned}
x_{1}^{2} x_{2}+x_{2}^{2} x_{3}+x_{3}^{2} x_{1} & \leqslant x_{1}^{2} x_{2}+x_{1} x_{2} x_{3}+x_{1} x_{2} x_{3} \\
& =x_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,153 |
$9 \cdot 220$ Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{2 n-1} \geqslant 0$, prove:
$$a_{1}^{2}-a_{2}^{2}+a_{3}^{2}-\cdots+a_{2 n-1}^{2} \geqslant\left(a_{1}-a_{2}+a_{3}-\cdots+a_{2 n-1}\right)^{2} .$$ | [Proof] By induction, when $n=1$, the proposition is obviously true. Let $n=2$, since
$$\begin{array}{l}
a_{1}^{2}-a_{2}^{2}+a_{3}^{2}-\left(a_{1}-a_{2}+a_{3}\right)^{2} \\
=\left(a_{1}-a_{2}\right)\left(a_{1}+a_{2}-a_{1}+a_{2}-2 a_{3}\right) \\
=2\left(a_{1}-a_{2}\right)\left(a_{2}-a_{3}\right) \geqslant 0 .
\end{arra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,154 |
$1 \cdot 146$ Simplify $\left(\frac{1 \cdot 2 \cdot 4+2 \cdot 4 \cdot 8+\cdots+n \cdot 2 n \cdot 4 n}{1 \cdot 3 \cdot 9+2 \cdot 6 \cdot 18+\cdots+n \cdot 3 n \cdot 9 n}\right)^{\frac{1}{3}}$ | [Solution] Since for $1 \leqslant k \leqslant n$ we have
$$\begin{aligned}
k \cdot 2 k \cdot 4 k & =8 k^{3} \\
k \cdot 3 k \cdot 9 k & =27 k^{3}
\end{aligned}$$
Therefore, the original expression $=\left[\frac{8\left(1^{3}+2^{3}+\cdots+n^{3}\right)}{27\left(1^{3}+2^{3}+\cdots+n^{3}\right)}\right]^{\frac{1}{3}}$
$$=\fr... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,155 |
9・221 Let $0<p \leqslant a_{i} \leqslant q(i=1,2, \cdots, n), b_{1}, b_{2} \cdots b_{n}$ be a permutation of $a_{1}$, $a_{2}, \cdots, a_{n}$, prove that:
$$n \leqslant \frac{a_{1}}{b_{1}}+\frac{a_{2}}{b_{2}}+\cdots+\frac{a_{n}}{b_{n}} \leqslant n+\left[\frac{n}{2}\right]\left(\sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}\right... | [Proof] By the AM-GM inequality, it is obvious that
$$\frac{a_{1}}{b_{1}}+\frac{a_{2}}{b_{2}}+\cdots+\frac{a_{n}}{b_{n}} \geqslant n$$
Thus, we only need to prove the inequality on the right. Without loss of generality, assume
$$a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$$
According to the rearrangement i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,156 |
$9 \cdot 222$ Let $x_{1}, x_{2}, \cdots, x_{n}$ be real numbers $(n \geqslant 3)$, and let
$$p=\sum_{i=1}^{n} x_{i}, q=\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}$$
Prove:
( i ) $\frac{(n-1) p^{2}}{n}-2 q \geqslant 0$,
(ii) $\left|x_{i}-\frac{p}{n}\right| \leqslant \frac{n-1}{n} \sqrt{p^{2}-\frac{2 n}{n-1} q}, i=1,... | [Proof] (i) Since
$$\begin{aligned}
(n-1) p^{2}-2 n q=(n & -1)\left(\sum_{i=1}^{n} x_{i}\right)^{2}-2 n \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \\
& =(n-1) \sum_{i=1}^{n} x_{i}^{2}-2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \\
& =\sum_{1 \leqslant i<j \leqslant n}\left(x_{i}-x_{j}\right)^{2},
\end{aligned}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,157 |
9. 223 Let $a_{i} \in[-1,1], a_{i} a_{i+1} \neq-1, i=1,2, \cdots, n$, where $a_{n+1}=a_{1}$. Prove:
$$\sum_{i=1}^{n} \frac{1}{1+a_{i} a_{i+1}} \geqslant \sum_{i=1}^{n} \frac{1}{1+a_{i}^{2}}$$ | [Proof] Since
$$\begin{array}{l}
\frac{1}{1+a_{i} a_{i+1}}-\frac{1}{1+a_{i}^{2}}=\frac{a_{i}\left(a_{i}-a_{i+1}\right)}{\left(1+a_{i} a_{i+1}\right)\left(1+a_{i}^{2}\right)} \\
\frac{1}{1+a_{i} a_{i+1}}-\frac{1}{1+a_{i+1}^{2}}=\frac{a_{i+1}\left(a_{i+1}-a_{i}\right)}{\left(1+a_{i} a_{i+1}\right)\left(1+a_{i+1}^{2}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,158 |
9・224 Let $0 \leqslant x_{i} \leqslant 1, i=1,2, \cdots, n, n \geqslant 2$. Prove that there exists $i$ such that:
$$\begin{aligned}
1 \leqslant i \leqslant & n-1 \text { and } \\
& x_{i}\left(1-x_{i+1}\right) \geqslant \frac{1}{4} x_{1}\left(1-x_{n}\right) ..
\end{aligned}$$ | [Proof]
$$\begin{array}{l}
\text { Let } x_{k}=a=\max \left\{x_{1}, x_{2}, \cdots, x_{n}\right\} \text {, } \\
x_{t}=b=\min \left\{x_{1}, x_{2}, \cdots, x_{n}\right\} .
\end{array}$$
If $x_{2} \leqslant \frac{1+b}{2}$, then
$$\begin{aligned}
x_{1}\left(1-x_{2}\right) & \geqslant x_{1}\left(1-\frac{1+b}{2}\right)=\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,159 |
9・225 Given positive numbers $x_{1}, x_{2}, \cdots, x_{n}$ and $y_{1}, y_{2}, \cdots, y_{n}$ satisfying
(1) $x_{1}>x_{2}>\cdots>x_{n}, y_{1}>y_{2}>\cdots>y_{n}$,
(2) $x_{1}>y_{1}, x_{1}+x_{2}>y_{1}+y_{2}, \cdots, x_{1}+x_{2}+\cdots+x_{n}$ $>y_{1}+y_{2}+\cdots+y_{n}$.
Prove that for any natural number $k$,
$$x_{1}^{k}+x... | [Proof] Let $S_{i}=x_{1}+x_{2}+\cdots+x_{i}, i=1,2, \cdots, n, S_{0}=0$,
$$T_{i}=y_{1}+y_{2}+\cdots+y_{i}, i=1,2, \cdots, n, T_{0}=0$$
For any positive numbers $a_{1}>a_{2}>\cdots>a_{n}$, we have
$$\begin{aligned}
\sum_{k=1}^{n} a_{k} x_{k} & =\sum_{k=1}^{n} a_{k}\left(S_{k}-S_{k-1}\right)=\sum_{k=1}^{n} a_{k} S_{k}-\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,160 |
9・226 Let $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 2)$ be $n$ distinct real numbers, $S=$ $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}, M=\min _{1 \leqslant i<j \leqslant n}\left(a_{j}-a_{i}\right)^{2}$. Prove that:
$$\frac{S}{M} \geqslant \frac{n\left(n^{2}-1\right)}{12}$$ | [Proof] Without loss of generality, let $a_{1}<a_{2}<\cdots<a_{n}$, then
$$\left(a_{i}-a_{j}\right)^{2} \geqslant M|i-j|^{2}$$
(1)
Let $A=n S-\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}$, then
$$A=\sum_{i, j=1}^{n} a_{i}^{2}-\sum_{i, j=1}^{n} a_{i} a_{j}=\sum_{i, j=1}^{n} a_{i}\left(a_{i}-a_{j}\right)$$
Also, $A=\sum_{i... | \frac{S}{M} \geqslant \frac{n\left(n^{2}-1\right)}{12} | Inequalities | proof | Yes | Yes | inequalities | false | 735,161 |
9.227 Let $x_{1}, x_{2}, \cdots, x_{n}$ be non-negative real numbers, $a$ be the minimum of them, and denote $x_{n+1}=x_{1}$. Prove:
$$\sum_{j=1}^{n} \frac{1+x_{j}}{1+x_{j+1}} \leqslant n+\frac{1}{(1+a)^{2}} \sum_{j=1}^{n}\left(x_{j}-a\right)^{2},$$
where equality holds if and only if $x_{1}=x_{2}=\cdots=x_{n}$. | [Proof] Use induction. When $n=1$, the inequality to be proved is obviously true. Assume that the conclusion of the problem holds when $n=k$. When $n=k+1$, by cyclic symmetry, we may assume without loss of generality that $x_{k+1}$ is the largest. Then, by the induction hypothesis, we have
$$\sum_{j=1}^{k-1} \frac{1+x_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,162 |
9・228 Given two sets of real numbers $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$. Arrange $\left\{a_{k}\right\}$ in ascending order and $\left\{b_{k}\right\}$ in descending order to get
$$\begin{array}{l}
a_{i_{1}} \leqslant a_{i_{2}} \leqslant \cdots \leqslant a_{i_{n}} \\
b_{k_{1}} \geqslant b_{k_... | [Proof] Let $a_{i_{l}}+b_{k_{l}}=\max \left\{a_{i_{1}}+b_{k_{1}}, \cdots, a_{i_{n}}+b_{k_{n}}\right\}$, since $i_{l}, i_{l+1}, \cdots$ $i_{n}$, are $n-l+1$ distinct numbers in $\{1,2, \cdots, n\}$, and $b_{k_{1}} \geqslant b_{k_{2}} \geqslant \cdots \geqslant b_{k_{l}}$, so $\square$
$$\max \left\{b_{i_{l}}, b_{i_{l+1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,163 |
9.230 Prove: For any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, the inequality
$$\frac{1}{a_{1}}+\frac{2}{a_{1}+a_{2}}+\cdots+\frac{n}{a_{1}+\cdots+a_{n}}<4\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)$$ | [Proof] Without loss of generality, let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, and denote
$S_{n}=\frac{1}{a_{1}}+\frac{2}{a_{1}+a_{2}}+\cdots+\frac{n}{a_{1}+\cdots+a_{n}}$. When $n<4$, the inequality to be proved is obviously true. We will use induction to prove that when $n \geqslant 4$,
$$S_{n} \le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,165 |
$\begin{array}{l}\text { 1・147 Simplify } \\ \quad(2+1)\left(2^{2}+1\right)\left(2^{4}+1\right)\left(2^{8}+1\right) \cdots \cdots\left(2^{256}+1\right) .\end{array}$ | [Solution] $\begin{aligned} \text { Original expression } & =(2-1)(2+1)\left(2^{2}+1\right)\left(2^{4}+1\right) \cdots\left(2^{256}+1\right) \\ & =2^{512}-1 .\end{aligned}$ | 2^{512}-1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,166 |
9.231 Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ real numbers, and let
$$\begin{array}{l}
b_{k}=\frac{1}{k}\left(a_{1}+a_{2}+\cdots+a_{k}\right), k=1,2, \cdots, n, \\
C=\left(a_{1}-b_{1}\right)^{2}+\left(a_{2}-b_{2}\right)^{2}+\cdots+\left(a_{n}-b_{n}\right)^{2}, \\
D=\left(a_{1}-b_{n}\right)^{2}+\left(a_{2}-b_{n}\right)... | [Proof] By induction, when $n=1$, the proposition is obviously true. Assume the proposition holds for $n=k$. Take any $k+1$ real numbers $a_{1}, a_{2}, \cdots, a_{k+1}$. Let $C_{k}=\sum_{i=1}^{k}\left(a_{i}-b_{i}\right)^{2}$, $C_{k+1}=\sum_{i=1}^{k+1}\left(a_{i}-b_{i}\right)^{2}$, $D_{k}=\sum_{i=1}^{k}\left(a_{i}-b_{k}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,167 |
9・232 If $x_{1}, x_{2}, \cdots, x_{n} \in[a, b]$, where $0<a<b$, prove:
$$\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}\right) \leqslant \frac{(a+b)^{2}}{4 a b} n^{2} .$$ | [Proof] From $0<a \leqslant x_{i} \leqslant b, i=1,2, \cdots, n$ we can get
that is
$$\left(\sqrt{x_{i}}-\frac{b}{\sqrt{x_{i}}}\right)\left(\sqrt{x_{i}}-\frac{a}{\sqrt{x_{i}}}\right) \leqslant 0$$
thus
$$x_{i}+\frac{a b}{x_{i}} \leqslant a+b, i=1,2, \cdots, n$$
Furthermore,
$$\begin{array}{l}
\sum_{i=1}^{n} x_{i}+a ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,168 |
9.233 Let $n$ be a natural number greater than 2. Prove that the following inequality holds for any real numbers $a_{1}, a_{2}, \cdots, a_{n}$ if and only if $n=3$ or $n=5$:
$$\sum_{i=1}^{n}\left(a_{i}-a_{1}\right) \cdots\left(a_{i}-a_{i-1}\right)\left(a_{i}-a_{i+1}\right) \cdots\left(a_{i}-a_{n}\right) \geqslant 0 .$$ | [Proof] When $n=3$, without loss of generality, let $a_{1} \leqslant a_{2} \leqslant a_{3}$. The original inequality
$$\begin{aligned}
\text { LHS }= & \left(a_{1}-a_{2}\right)\left(a_{1}-a_{3}\right)+\left(a_{2}-a_{1}\right)\left(a_{2}-a_{3}\right)+\left(a_{3}-\right. \\
& \left.a_{1}\right)\left(a_{3}-a_{2}\right) \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,169 |
9・235 Let $a_{1}, a_{2}, \cdots, a_{n}$ be real numbers greater than 1 and
$$\left|a_{k+1}-a_{k}\right|<1, k=1,2, \cdots, n-1$$
Prove:
$$\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}+\frac{a_{n}}{a_{1}}<2 n-1 .$$ | [Proof] For $k=1,2, \cdots, n-1$, if $a_{k}<a_{k+1}$, then from $a_{k}1$ we know $\frac{a_{k}}{a_{k+1}}<1+\frac{1}{a_{k+1}}<2$. Suppose there are $l$ values of $k$ such that
$$1 \leqslant k \leqslant n-1 \text { and } a_{k}<a_{k+1} \text {, }$$
and the remaining $n-1-l$ values of $k$ have $a_{k} \geqslant a_{k+1}$. Si... | 2 n-1 | Inequalities | proof | Yes | Yes | inequalities | false | 735,171 |
9. 236 There are several positive numbers written on the blackboard. It is known that the sum of their pairwise products is 1. Prove: one of these numbers can be erased so that the sum of the remaining positive numbers is less than $\sqrt{2}$.
untranslated text remains the same as requested. | [Proof] Let the positive numbers written on the blackboard be $x_{1}, x_{2}, \cdots, x_{n}$ and
$$x_{n}=\max \left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$$
By assumption,
Thus,
$$\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2}=x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}+2$$
$$\begin{aligned}
& \left(x_{1}+x_{2}+\cdots+x_{n-1}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,172 |
9.237 Prove: regardless of how the "+" and "-" signs are chosen for the odd powers of $x$ on the left side of the following expression, we always have
$$x^{2 n} \pm x^{2 n-1}+x^{2 n-2} \pm x^{2 n-3}+\cdots+x^{2} \pm x+1>\frac{1}{2}$$ | [Proof] It is clear that it suffices to prove for any $x \geqslant 0$
$$f(x)=x^{2 n}-x^{2 n-1}+x^{2 n-2}-x^{2 n-3}+\cdots+x^{2}-x+1>\frac{1}{2}$$
Indeed, since $f(x)=\frac{1+x^{2 n+1}}{1+x}$, if $x \geqslant 1$ or $x=0$, then $f(x) \geqslant 1$. If $0 < x < 1$, then $f(x) > \frac{1}{1+x} \geqslant \frac{1}{2}$
In any... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,173 |
9. 238 Let $a_{1}, a_{2}, \cdots a_{n}$ all be positive numbers, and their sum is 1. Prove:
$$\frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n-1}^{2}}{a_{n-1}+a_{n}}+\frac{a_{n}^{2}}{a_{n}+a_{1}} \geqslant \frac{1}{2} .$$ | [Proof] Since
$$\frac{a_{1}^{2}-a_{2}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}-a_{3}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n-1}^{2}-a_{n}^{2}}{a_{n-1}+a_{n}}+\frac{a_{n}^{2}-a_{1}^{2}}{a_{n}+a_{1}}=0$$
Therefore,
$$\begin{aligned}
& \frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n}^{2}}{a_{n}+a_{1}} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,174 |
$9 \cdot 239$ Let $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ be positive real numbers, and $\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} b_{k}$.
Prove:
$$\sum_{k=1}^{n} \frac{a_{k}^{2}}{a_{k}+b_{k}} \geqslant \frac{1}{2} \sum_{k=1}^{n} a_{k}$$ | [Proof] Since
and
$$\begin{aligned}
\sum_{k=1}^{n} \frac{a_{k}^{2}}{a_{k}+b_{k}} & =\sum_{k=1}^{n} \frac{a_{k}^{2}+a_{k} b_{k}-a_{k} b_{k}}{a_{k}+b_{k}} \\
& =\sum_{k=1}^{n} a_{k}-\sum_{k=1}^{n} \frac{a_{k} b_{k}}{a_{k}+b_{k}}, \\
\sum_{k=1}^{n} \frac{a_{k} b_{k}}{a_{k}+b_{k}} & \leqslant \frac{1}{4} \sum_{k=1}^{n} \f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,175 |
9・240 Let $a_{1}, a_{2}, \cdots, a_{n}$ all be positive numbers, and for any $1 \leqslant k \leqslant n$, we have $a_{1} a_{2} \cdots a_{k} \geqslant 1$. Prove:
$$\frac{1}{1+a_{1}}+\frac{2}{\left(1+a_{1}\right)\left(1+a_{2}\right)}+\cdots+\frac{n}{\left(1+a_{1}\right) \cdots\left(1+a_{n}\right)}<2 .$$ | [Proof] For any $1 \leqslant k \leqslant n$, since
$$\begin{array}{l}
1+a_{1} \geqslant 2 \sqrt{a_{1}} \\
1+a_{2} \geqslant 2 \sqrt{a_{2}} \\
\cdots \cdots \cdot \\
1+a_{k} \geqslant 2 \sqrt{a_{k}}
\end{array}$$
Therefore, $\left(1+a_{1}\right) \cdots\left(1+a_{k}\right) \geqslant 2^{k} \sqrt{a_{1} \cdots a_{k}} \geqs... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,176 |
$1 \cdot 148 \quad$ Calculate $\sqrt{31 \times 30 \times 29 \times 28+1}$. | ［Solution］Since
$$\begin{aligned}
& (n-1) n(n+1)(n+2)+1 \\
= & \left(n^{2}+n\right)\left(n^{2}+n-2\right)+1 \\
= & \left(n^{2}+n\right)^{2}-2\left(n^{2}+n\right)+1 \\
= & \left(n^{2}+n-1\right)^{2},
\end{aligned}$$
therefore, by setting $n=29$ in the above expression, we get
$$28 \times 29 \times 30 \times 31+1=\left(... | 869 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,177 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.