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9. 10 Prove: The following system of inequalities has no real solutions:
$$\left\{\begin{array}{l}
|x|>|y-z+t| \\
|y|>|x-z+t| \\
|z|>|x-y+t| \\
|t|>|x-y+z|
\end{array}\right.$$ | [Proof] By contradiction. Let real numbers $x, y, z, t$ satisfy the above inequalities. Then, squaring both sides of the inequalities, we get
$$\left\{\begin{array}{l}
(x+y-z+t)(x-y+z-t)>0, \\
(y+x-z+t)(y-x+z-t)>0, \\
(z+x-y+t)(z-x+y-t)>0 \\
(t+x-y+z)(t-x+y-z)>0
\end{array}\right.$$
Thus, we have
$$\begin{aligned}
& -... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,942 |
9. 11 For any real numbers $a, b, c$, prove that there exists a real number $x$ such that
$$a \cos x + b \cos 3x + c \cos 9x \geqslant \frac{1}{2}(|a| + |b| + |c|).$$ | [Proof] According to the positive and negative situations of $a, b, c$, we discuss 8 cases as follows: when $a \geqslant 0, b \geqslant 0, c \geqslant 0$, we can take $x=0$; when $a \leqslant 0, b \leqslant 0, c \leqslant 0$, we can take $x=\pi$; when $a \geqslant 0, b \leqslant 0, c \leqslant 0$, we can take $x=\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,943 |
9. 12 (a) Prove: From any 3 positive numbers, two numbers $x, y$ can always be selected such that
$$0<\frac{x-y}{1+x y} \leqslant 1$$
( $b$ ) Can two numbers be selected from any 4 real numbers to satisfy (1). | [Proof] $(a)$ For any three positive numbers $x, y, z$, let
$$\alpha=\operatorname{arctg} x, \beta=\operatorname{arctg} y, \gamma=\operatorname{arctg} z$$
Since $\alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$, there exist two among $\alpha, \beta, \gamma$, say $\alpha, \beta$ such that
thus $0<\operatorname{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,944 |
9・13 Solve the inequality
$$\sqrt{\frac{\pi}{4}-\operatorname{arctg} \frac{|x|+|y|}{\pi}}+\operatorname{tg}^{2} x+1 \leqslant \sqrt{2}|\operatorname{tg} x|(\sin x+\cos x)$$ | [Solution] Let $x, y$ satisfy the inequality. From $\sin x + \cos x \leqslant \sqrt{2}$, we have
$$\begin{aligned}
\tan^2 x + 1 & \leqslant \sqrt{\frac{\pi}{4} - \arctan \frac{|x| + |y|}{\pi}} + \tan^2 x + 1 \\
& \leqslant \sqrt{2}|\tan x|(\sin x + \cos x) \\
& \leqslant 2|\tan x|
\end{aligned}$$
Since $\tan^2 x + 1 \... | (x, y) \in \left\{\left(\frac{\pi}{4}, \frac{3\pi}{4}\right), \left(\frac{\pi}{4}, -\frac{3\pi}{4}\right)\right\} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,945 |
9. 15 Determine all positive real number tuples $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)$ that satisfy the following inequalities:
$$\begin{array}{l}
\left(x_{1}^{2}-x_{3} x_{5}\right)\left(x_{2}^{2}-x_{3} x_{5}\right) \leqslant 0 \\
\left(x_{2}^{2}-x_{4} x_{1}\right)\left(x_{3}^{2}-x_{4} x_{1}\right) \leqslant ... | [Solution] Let $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)$ be the required set of solutions and $x_{1}=\min \left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}, x_{2}=\min \left\{x_{2}, x_{3}, x_{4}, x_{5}\right\}$. Since $x_{1}^{2} \leqslant x_{3} x_{5}, x_{2}^{2} \leqslant x_{3} x_{5}$ and $\left(x_{1}^{2}-x_{3} x_... | x_{1}=x_{2}=x_{3}=x_{4}=x_{5} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,947 |
9. 16 Given 7 real numbers, prove that there must be two numbers, denoted as $x, y$, satisfying
$$0<\frac{x-y}{1+x y} \leqslant \frac{1}{\sqrt{3}}$$ | ［Proof］Let the given 7 numbers be
Let
$$\begin{array}{c}
a_{1}<a_{2}<a_{3}<a_{4}<a_{5}<a_{6}<a_{7} \\
\theta_{i}=\operatorname{arctg} a_{i}, i=1,2, \cdots, 7
\end{array}$$
Then $-\frac{\pi}{2}<\theta_{1}<\theta_{2}<\cdots<\theta_{7}<\frac{\pi}{2}$.
Thus, there must exist $1 \leqslant i \leqslant 6$, such that
$$0<\th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,948 |
9・17 Let non-negative real numbers $a, b, c$ satisfy
$$a^{4}+b^{4}+c^{4} \leqslant 2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right) .$$
(i) Prove that each of $a, b, c$ is not greater than the sum of the other two.
(ii) Prove that $a^{2}+b^{2}+c^{2} \leqslant 2(a b+b c+c a)$.
(iii) Can the original inequality be deriv... | [Proof] (i) Since
$$\begin{aligned}
0 & \leqslant 2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)-a^{4}-b^{4}-c^{4} \\
& =b^{2}\left(2 a^{2}+2 c^{2}\right)-b^{4}-\left(c^{2}-a^{2}\right)^{2} \\
& =b^{2}(c+a)^{2}+b^{2}(c-a)^{2}-b^{4}-(c-a)^{2}(c+a)^{2} \\
& =\left[b^{2}-(c-a)^{2}\right]\left[(c+a)^{2}-b^{2}\right] \\
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,949 |
1-126 Question: When and only when do real numbers $x_{0}, x_{1}, \cdots, x_{n}(n \geqslant 2)$ satisfy what conditions so that there exist real numbers $y_{0}, y_{1}, \cdots, y_{n}$, such that $z_{0}^{2}=z_{1}^{2}+z_{2}^{2}+\cdots+z_{n}^{2}$ holds. Where $z_{k}=x_{k}+i y_{k}, i$ is the imaginary unit, $k=0,1,2, \cdots... | ［Solution］It is easy to know that (1) is equivalent to
$$\left\{\begin{array}{l}
\sum_{k=1}^{n} x_{k}^{2}-x_{0}^{2}=\sum_{k=1}^{n} y_{k}^{2}-y_{0}^{2} \\
\sum_{k=1}^{n} x_{k} y_{k}=x_{0} y_{0}
\end{array}\right.$$
If there exist real numbers $y_{0}, y_{1}, \cdots, y_{n}$ such that (2) holds, then
$$x_{0}^{2} y_{0}^{2}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,950 |
9・19 Let real numbers $x_{1}, x_{2}, \cdots, x_{100}$ have a sum of 1 and
$$\left|x_{k+1}-x_{k}\right|<\frac{1}{50}, k=1,2, \cdots, 99$$
Prove: It is possible to select 50 numbers from these 100 numbers such that the absolute difference between their sum and $\frac{1}{2}$ is no greater than $\frac{1}{100}$. | [Proof] If $\left|x_{1}+x_{3}+x_{5}+\cdots+x_{99}-\frac{1}{2}\right| \leqslant \frac{1}{100}$, then $x_{1}, x_{3}$, $x_{5}, \cdots, x_{99}$ are the required numbers. Otherwise, assume
$$x_{1}+x_{3}+x_{5}+\cdots+x_{99}>\frac{1}{2}+\frac{1}{100}$$.
First, swap $x_{1}$ with $x_{2}$, then swap $x_{3}$ with $x_{4}$, and the... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 734,951 |
9. 20 Given 12 real numbers $a_{1}, a_{2}, a_{3}, \cdots, a_{12}$ satisfy
$$\left\{\begin{array}{l}
a_{2}\left(a_{1}-a_{2}+a_{3}\right)<0, \\
a_{3}\left(a_{2}-a_{3}+a_{4}\right)<0, \\
\cdots \cdots \cdots \cdots \\
a_{11}\left(a_{10}-a_{11}+a_{12}\right)<0
\end{array}\right.$$
Prove: At least 3 positive numbers and 3 ... | [Proof] By contradiction. Suppose the conclusion to be proved does not hold. Without loss of generality, assume that among $a_{1}, a_{2}, \cdots, a_{12}$, there are at most two negative numbers, then there exists $1 \leqslant k \leqslant 9$ such that $a_{k}, a_{k+1}, a_{k+2}, a_{k+3}$ are all non-negative real numbers.... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,952 |
9・21 Let real numbers $a_{1}, a_{2}, \cdots, a_{100}$ satisfy
$$\left\{\begin{array}{l}
a_{1}-3 a_{2}+2 a_{3} \geqslant 0 \\
a_{2}-3 a_{3}+2 a_{4} \geqslant 0 \\
\cdots \cdots \cdots \cdots \\
a_{99}-3 a_{100}+2 a_{1} \geqslant 0 \\
a_{100}-3 a_{1}+2 a_{2} \geqslant 0
\end{array}\right.$$
Prove that $a_{1}=a_{2}=\cdot... | [Proof] Let $s_{1}=a_{1}-3 a_{2}+2 a_{3}, s_{2}=a_{2}-3 a_{3}+2 a_{4}, \cdots, s_{99}=a_{99}$ $-3 a_{100}+2 a_{1}, s_{100}=a_{100}-3 a_{1}+2 a_{2}$. Clearly, we have
$$s_{1}+s_{2}+\cdots+s_{99}+s_{100}=0$$
Also, $s_{i} \geqslant 0, i=1,2, \cdots, 100$, so $s_{i}=0, i=1,2, \cdots, 100$, from which we can derive $a_{1}-... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,953 |
9. 23 Find all real numbers $\alpha$ such that
$$\cos \alpha, \cos 2 \alpha, \cos 4 \alpha, \cdots, \cos 2^{n} \alpha, \cdots$$
are all negative. | [Solution] Let $\alpha$ satisfy the requirements of the problem, then from $\cos 4 \alpha<0$ and $\cos 2 \alpha<0$ we get
$$-\frac{\sqrt{2}}{2}<\cos 2 \alpha<0$$
Furthermore, by $\cos \alpha<0$ we know $\cos \alpha<-\frac{\sqrt{2-\sqrt{2}}}{2}<-\frac{1}{4}$. This immediately gives us
$$\cos 2^{n} \alpha<-\frac{1}{4}, ... | \alpha= \pm \frac{2 \pi}{3}+2 k \pi, k \in \mathbb{Z} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,955 |
9.24 Prove: For any real numbers $a, b$, there exist $x$ and $y$ in $[0,1]$, such that
$$|x y-a x-b y| \geqslant \frac{1}{3}$$
and ask whether the above proposition still holds if $\frac{1}{3}$ is replaced by $\frac{1}{2}$, or 0.33334. | [Proof] By contradiction, assume the proposition is not true, then there exist real numbers $a, b$, such that for any $x, y$ in $[0,1]$ we have
$$|x y-a x-b y|>\frac{1}{3},$$
which leads to a contradiction! Therefore, the proposition holds.
The following proof: If $\frac{1}{3}$ in the original proposition is replaced ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,956 |
9. 25 Prove: Any vectors $\vec{a}, \vec{b}, \vec{c}$ cannot simultaneously satisfy $\sqrt{3}|\vec{a}|<| \vec{b}-$ $\vec{c}|, \sqrt{3}| \vec{b}|<| \vec{c}-\vec{a}|, \sqrt{3}| \vec{c}|<| \vec{a}-\vec{b} |$. | [Proof] By contradiction. If vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy the above inequalities, then squaring both sides yields
$$\left\{\begin{array}{l}
3|\vec{a}|^{2}<|\vec{b}|^{2}+|\vec{c}|^{2}-2 \vec{b} \cdot \vec{c}, \\
3|\vec{b}|^{2}<|\vec{c}|^{2}+|\vec{a}|^{2}-2 \vec{c} \cdot \vec{a}, \\
3|\vec{c}|^{2}<|\vec{a}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,957 |
9. 26 Given a real number $a$, find the largest real number $\lambda$ such that any rectangle in the plane with sides parallel to the $x$-axis and $y$-axis, if it contains the figure $G$ defined by the following inequalities:
$$\left\{\begin{array}{l}
y \leqslant -x^{2}, \\
y \geqslant x^{2}-2 x+a
\end{array}\right.$$
... | [Solution] Clearly, the $x$-coordinates of the points in $G$ satisfy
$$2 x^{2}-2 x+a \leqslant 0 \text {. }$$
Thus, when $a>\frac{1}{2}$, $G$ is an empty set. When $a=\frac{1}{2}$, $x=\frac{1}{2}$, from which we immediately get that $G$ is a single point $\left(\frac{1}{2},-\frac{1}{4}\right)$. Therefore, when $a \geq... | \lambda=\left\{\begin{array}{ll}
0, & a \geqslant \frac{1}{2} . \\
1-2 a, & 0<a<\frac{1}{2} . \\
(1-a) \sqrt{1-2 a}, & a<0 .
\end{array}\right.} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,958 |
9・27 (i) Let positive numbers $a, b, c$ satisfy
$$\left(a^{2}+b^{2}+c^{2}\right)^{2}>2\left(a^{4}+b^{4}+c^{4}\right)$$
Prove that $a, b, c$ are the lengths of the sides of some triangle.
(ii) Let positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)^{2}>(n-1)\left(a... | ［Proof］(i) From the assumption, we have
$$2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}-a^{4}-b^{4}-c^{4}>0$$
Factoring the left side of the above equation, we get
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)>0$$
By symmetry, without loss of generality, let $a \geqslant b \geqslant c$, then we have
$$b+c-a>0$$
This shows that $a, b, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,959 |
9.28 Prove: The set of real numbers satisfying the inequality
$$\sum_{k=1}^{70} \frac{k}{x-k} \geqslant \frac{5}{4}$$
is the union of pairwise disjoint intervals, and the sum of the lengths of these intervals is 1988. | [Proof] Let $R(x)=\sum_{k=1}^{70} \frac{k}{x-k}$. Clearly, when $x>70$, then $R(x)>0$ and it is strictly decreasing.
Since $x>70$ and as $x \rightarrow 70$, $R(x) \rightarrow+\infty$, and as $x \rightarrow+\infty$, $R(x) \rightarrow 0$, there exists a unique $x_{70}$ such that $x_{70}>70$, $R\left(x_{70}\right)=\frac{... | 1988 | Inequalities | proof | Yes | Yes | inequalities | false | 734,961 |
9 - 29 Let real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$. For an integer $k \geqslant 2$, prove: there exist integers $a_{1}, a_{2}, \cdots, a_{n}$, not all 0, such that $\left|a_{i}\right| \leqslant k-1$, $i=1,2, \cdots, n$, and
$$\left|a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n}... | Let $S=\left\{\left(\lambda_{1}, \lambda_{2}, \cdots, \lambda_{n}\right) \mid \lambda_{i} \in \mathbb{Z}, 0 \leqslant \lambda_{i} \leqslant k-1, i=1\right.$, $2, \cdots, n\}$. For any $\left(\lambda_{1}, \lambda_{2}, \cdots, \lambda_{n}\right) \in S$, by the Cauchy-Schwarz inequality, we have
$$0 \leqslant \sum_{i=1}^{... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,962 |
$9 \cdot 31$ Let $0 \leqslant p_{i} \leqslant 1, i=1,2, \cdots, n$, prove: there exists $x \in[0,1]$ satisfying $\sum_{i=1}^{n} \frac{1}{\left|x-p_{i}\right|} \leqslant 8 n\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2 n-1}\right)$. | [Proof] Consider $2n$ open intervals $I_{k}=\left(\frac{k}{2 n}, \frac{k+1}{2 n}\right), k=0,1, \cdots, 2 n-1$. Clearly, we can select $n$ of them such that none of the selected open intervals $I_{k}$ contains any $p_{i}$. Let $x_{j}(j=1,2, \cdots, n)$ denote the midpoints of the selected open intervals, and let $d_{i ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,963 |
9. 32 Prove: In the open interval $(0,1)$, there must exist four pairs of distinct positive numbers $(a, b)(a \neq b)$, satisfying
$$\sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}>\frac{a}{2 b}+\frac{b}{2 a}-a b-\frac{1}{8 a b} .$$ | [Proof] Since $a \in(0,1), b \in(0,1)$, we can set
$$a=\cos \alpha, b=\cos \beta$$
where $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$. Clearly, using the above expressions, we get
$$\begin{aligned}
a b+\sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)} & =\cos \alpha \cos \beta+\sin \alpha \sin \beta \\
& =\cos (\alp... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,964 |
9.34 Find the range of real number $a$ such that the inequality
$$\sin ^{6} x+\cos ^{6} x+2 a \sin x \cos x \geqslant 0$$
holds for all real numbers $x$. | ［Solution］ Let $f(x)=\sin ^{6} x+\cos ^{6} x+2 a \sin x \cos x$. Since
$$\begin{aligned}
1 & =\left(\sin ^{2} x+\cos ^{2} x\right)^{3} \\
& =\sin ^{6} x+3 \sin ^{4} x \cos ^{2} x+3 \sin ^{2} x \cos ^{4} x+\cos ^{6} x \\
& =\sin ^{6} x+\cos ^{6} x+3 \sin ^{2} x \cos ^{2} x \\
& =\sin ^{6} x+\cos ^{6} x+\frac{3}{4}+\sin ... | |a| \leqslant \frac{1}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,966 |
9.35 Find the largest positive integer $n$, such that there exists a unique integer $k$ satisfying
$$\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$$ | ［Solution］From the assumption, we have $\frac{15}{8}>\frac{n+k}{n}>\frac{13}{7}$, thus $48 n<56 k<49 n$.
Therefore, the problem is reduced to finding the largest open interval $(48 n, 49 n)$, such that this interval contains only one multiple of 56. Since the length of such an interval is $n$, it contains $n-1$ integer... | 112 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,967 |
9・36 Find the smallest real number $c$ such that for any positive sequence $\left\{x_{n}\right\}$, if $x_{1}+x_{2}+\cdots+x_{n} \leqslant x_{n+1}, n=1,2,3, \cdots$, then
$$\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}} \leqslant c \sqrt{x_{1}+x_{2}+\cdots+x_{n}}, n=1,2,3, \cdots$$ | ［Solution］ Let $x_{n}=2^{n}$, then
$$x_{1}+x_{2}+\cdots+x_{n}=2^{n+1}-2 \leqslant 2^{n+1}=x_{n+1}, n=1,2, \cdots$$
For any natural number $n$,
$$\frac{\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}}}{\sqrt{x_{1}+x_{2}+\cdots+x_{n}}}=\frac{\sqrt{2^{n+1}}-\sqrt{2}}{(\sqrt{2}-1) \sqrt{2^{n+1}-2}}$$
Since
$$\lim _{n \right... | \sqrt{2}+1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,968 |
$9 \cdot 37$ Find the largest real number $\alpha$ such that for any positive integers $m$ and $n$ satisfying $\frac{m}{n}<\sqrt{7}$, we have
$$\frac{\alpha}{n^{2}} \leqslant 7-\frac{m^{2}}{n^{2}}$$ | ［Solution］ Since
$$\begin{array}{l}
\frac{m}{n}0 \\
\frac{\alpha}{n^{2}} \leqslant 7-\frac{m^{2}}{n^{2}} \Leftrightarrow \alpha \leqslant 7 n^{2}-m^{2}
\end{array}$$
Thus, the maximum real number sought is
$$\alpha=\min \left\{7 n^{2}-m^{2} ; n, m \in N \text {, and } 7 n^{2}-m^{2}>0\right\} \text {. }$$
Since $7 n^{... | 3 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,969 |
1. 128 For every positive integer $n$, if the number of 1s in the binary representation of $n$ is even, then let $a_{n}=0$, otherwise let $a_{n}=1$. Prove: there do not exist positive integers $k$ and $m$ such that $a_{k+j}=a_{k+m+j}=a_{k+2 m+j}, 0 \leqslant j \leqslant m-1$. | ［Proof］Obviously, $a_{2 n}=a_{n}$,
$$a_{2 n+1}=1-a_{2 n}=1-a_{n}$$
If there exist positive integers $k$ and $m$ that satisfy the conditions, then we can assume $m$ is the smallest $m$ among all $(k, m)$ that satisfy the conditions.
If $m$ is odd, let's assume
$$a_{k}=a_{k+m}=a_{k+2 m}=0$$
(If $a_{k}=1$, the case can ... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,970 |
9.38 Find real numbers $A, B, C$ such that for any real numbers $x, y, z$,
$$A(x-y)(x-z)+B(y-z)(y-x)+C(z-x)(z-y) \geqslant 0$$
is satisfied. | [Solution] In (1), let $x=y \neq z$, we get $C(z-x)^{2} \geqslant 0$, thus $C \geqslant 0$. By symmetry, we have
$$A \geqslant 0, B \geqslant 0, C \geqslant 0$$
Let $s=x-y, t=y-z$, then $x-z=s+t$, and (1) is equivalent to
$$A s(s+t)-B s t+C+(s+t) \geqslant 0$$
i.e., $\square$
$$A s^{2}+(A-B+C) s t+C t^{2} \geqslant 0... | A^{2}+B^{2}+C^{2} \leqslant 2(A B+B C+C A) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,971 |
9-39 Let $a \leqslant b<c$ be the side lengths of a right triangle. Find the maximum constant $M$ such that
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant \frac{M}{a+b+c}$$ | [Solution] Let $I=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$, then $I=3+\frac{b}{a}+\frac{a}{b}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}$.
Since
$$\begin{array}{l}
\frac{b}{a}+\frac{a}{b} \geqslant 2 \\
\frac{a}{c}+\frac{c}{a}=\frac{2 a}{c}+\frac{c}{a}-\frac{a}{c} \geqslant 2 \sqrt{2}-\frac{a}{c} ... | 5+3\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,972 |
$9 \cdot 40$ Find the smallest real number $a$, such that for any non-negative real numbers $x$, $y$, $z$ whose sum is 1, we have
$$a\left(x^{2}+y^{2}+z^{2}\right)+x y z \geqslant \frac{a}{3}+\frac{1}{27} .$$ | Let $I=a\left(x^{2}+y^{2}+z^{2}\right)+x y z$. Without loss of generality, assume $x \leqslant y \leqslant z$, and let $x=\frac{1}{3}+\delta_{1}, y=\frac{1}{3}+\delta_{2}, z=\frac{1}{3}+\delta_{3}$, then
$$-\frac{1}{3} \leqslant \delta_{1} \leqslant 0,0 \leqslant \delta_{3} \leqslant \frac{2}{3}, \delta_{1}+\delta_{2}+... | \frac{2}{9} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,973 |
9. 41 Let $a_{1}, a_{2}, \cdots, a_{n}$ be given real numbers, not all zero. If real numbers $r_{1}$, $r_{2}, \cdots, r_{n}$ satisfy the inequality
$$\sum_{k=1}^{n} r_{k}\left(x_{k}-a_{k}\right) \leqslant\left(\sum_{k=1}^{n} x_{k}^{2}\right)^{\frac{1}{2}}-\left(\sum_{k=1}^{n} a_{k}^{2}\right)^{\frac{1}{2}}$$
for any r... | [Solution] Let $x_{1}=x_{2}=\cdots=x_{n}=0$, by the assumption we get
$$\sum_{k=1}^{n} r_{k} a_{k} \geqslant\left(\sum_{k=1}^{n} a_{k}^{2}\right)^{\frac{1}{2}}$$
Let $x_{k}=2 a_{k}(k=1,2, \cdots, n)$, we also get
$$\sum_{k=1}^{n} r_{k} a_{k} \leqslant\left(\sum_{k=1}^{n} a_{k}^{2}\right)^{\frac{1}{2}}$$
(1) and (2) gi... | r_{i}=\frac{a_{i}}{\sqrt{\sum_{k=1}^{n} a_{k}^{2}}}, i=1,2, \cdots, n | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,974 |
9.46 Let real numbers $a, b, c, d$ satisfy $a^{2}+b^{2}+c^{2}+d^{2} \leqslant 1$, find
$$S=(a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4}+(b+d)^{4}+(c+d)^{4}$$
the maximum value. | [Solution] $S \leqslant(a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4}+(b+d)^{4}$
$$\begin{aligned}
& +(c+d)^{4}+(a-b)^{4}+(a-c)^{4}+(a-d)^{4}+(b- \\
& c)^{4}+(b-d)^{4}+(c-d)^{4} \\
= & 6\left(a^{4}+b^{4}+c^{4}+d^{4}+2 a^{2} b^{2}+2 a^{2} c^{2}+2 a^{2} d^{2}+2 b^{2} c^{2}\right. \\
& \left.+2 b^{2} d^{2}+2 c^{2} d^{2}\right) \... | 6 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,976 |
9.47 Find the maximum value of the following expression:
$$x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}} .$$ | ［Solution］ By Cauchy-Schwarz inequality, we have
$$\left|x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right|^{2} \leqslant\left(x^{2}+y^{2}\right)\left(2-x^{2}-y^{2}\right) .$$
Using the AM-GM inequality, we get
$$\left|x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right| \leqslant \frac{x^{2}+y^{2}+2-x^{2}-y^{2}}{2}=1 .$$
If $x=\frac{1}{2... | 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,977 |
9.48 A pile of stones has a total weight of 100 kilograms, with each stone weighing no more than 2 kilograms. By taking some of these stones in various ways and calculating the difference between the sum of their weights and 10 kilograms. Among all these differences, the minimum absolute value is denoted as $d$. Among ... | [Solution] Take any pile of stones that satisfies the conditions of the problem, and let their weights be $x_{1}, x_{2}$, $\cdots, x_{n}$, without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$. According to the definition of $d$, there exists a natural number $k$ such that
$$x_{1}+... | \frac{10}{11} | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,978 |
9.49 (1) Given real numbers $a_{1}, a_{2}, b_{1}, b_{2}$, and positive numbers $p_{1}, p_{2}, q_{1}, q_{2}$. Prove that in the $2 \times 2$ matrix
$$\left(\begin{array}{ll}
\frac{a_{1}+b_{1}}{p_{1}+q_{1}} & \frac{a_{1}+b_{2}}{p_{1}+q_{2}} \\
\frac{a_{2}+b_{1}}{p_{2}+q_{1}} & \frac{a_{2}+b_{2}}{p_{2}+q_{2}}
\end{array}\... | ［Proof］It suffices to prove (2). Let
$$f(x)=\max _{1 \leqslant i \leqslant m}\left(p_{i} x-a_{i}\right), g(x)=\max _{1 \leqslant i \leqslant n}\left(-q_{j} x+b_{j}\right)$$
Since $p_{1}, p_{2}, \cdots, p_{m}, q_{1}, q_{2}, \cdots, q_{n}$ are all positive, $f(x)$ is strictly increasing and $\lim _{x \rightarrow+\infty}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,979 |
1- 129 For a given positive integer $n$ greater than 1, do there exist $2n$ distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$, that simultaneously satisfy the following two conditions:
(1) $a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n}$;
(2) $n-1>\sum_{i=1}^{n} \frac{a_{i}-b_{i}}... | [Solution] There exist $2n$ numbers that satisfy the proposition requirements.
In fact, we can set
$$\left\{\begin{array}{l}
a_{i}=2 M i \quad i=1,2, \cdots, n-1, M \geqslant 8000 n, M \in \mathbb{N}; \\
b_{i}=2 i, \quad i=1,2, \cdots, n-1 ; \\
a_{n}=(M-1)^{2} n(n-1) ; \\
b_{n}=M(M-1) n(n-1) .
\end{array}\right.$$
Cle... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,980 |
$9 \cdot 51$ Let $m, n \in N$, find the minimum value of $\left|12^{m}-5^{n}\right|$. | [Solution] Clearly, $|12^{m}-5^{n}|$ is an odd number, and it is not divisible by 3 or 5. From $12^{m}-5^{n} \equiv -1 \pmod{4}$, we know that $12^{m}-5^{n}$ cannot be 1.
If $5^{n}-12^{m}=1$, then by
$$5^{n}-12^{m} \equiv 5^{n} \equiv (-1)^{n} \pmod{3}$$
we can deduce that $n$ is even, let $n=2k$. Also, since
$$5^{n}-... | 7 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,981 |
9・53 Let $x, y, z$ be positive numbers and $x^{2}+y^{2}+z^{2}=1$, find
$$\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}$$
the minimum value. | [Solution] Let $x=x_{1}, y=x_{2}, z=x_{3}$. By the Cauchy-Schwarz inequality,
$$\begin{aligned}
1=\left(\sum_{i=1}^{3} x_{i}^{2}\right)^{2} & =\left(\sum_{i=1}^{3} \sqrt{\frac{x_{i}}{1-x_{i}^{2}}} \cdot x_{i}^{\frac{3}{2}} \sqrt{1-x_{i}^{2}}\right)^{2} \\
& \leqslant\left(\sum_{i=1}^{3} \frac{x_{i}}{1-x_{i}^{2}}\right)... | \frac{3 \sqrt{3}}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,982 |
$9 \cdot 55$ Let $n \geqslant 2$, find the maximum and minimum value of the product $x_{1} x_{2} \cdots x_{n}$ under the conditions $x_{i} \geqslant \frac{1}{n}, i=1,2, \cdots$, $n$, and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$. | [Solution] First, find the maximum value. By the AM-GM inequality, we have
$$\sqrt[n]{x_{1}^{2} \cdot x_{2}^{2} \cdot \cdots \cdot x_{n}^{2}} \leqslant \frac{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}{n}=\frac{1}{n}$$
and the equality holds when $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{\sqrt{n}}>\frac{1}{n}(n \geqslant 2)$. Ther... | \frac{\sqrt{n^{2}-n+1}}{n^{n}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,984 |
9. 56 Find
$$\begin{aligned}
A & =\sqrt{\left(1264-z_{1}-\cdots-z_{n}\right)^{2}+x_{n}^{2}+y_{n}^{2}}+ \\
& \sqrt{z_{n}^{2}+x_{n-1}^{2}+y_{n-1}^{2}}+\cdots+\sqrt{z_{2}^{2}+x_{1}^{2}+y_{1}^{2}}+ \\
& \sqrt{z_{1}^{2}+\left(948-x_{1}-\cdots-x_{n}\right)^{2}+\left(1185-y_{1}-\cdots-y_{n}\right)^{2}}
\end{aligned}$$
the mi... | [Solution] Consider the following points in a three-dimensional space with a Cartesian coordinate system:
$$\begin{array}{l}
B=(0,0,1264) \\
M_{n}=\left(x_{n}, y_{n}, z_{1}+\cdots+z_{n}\right) \\
M_{n-1}=\left(x_{n-1}+x_{n}, y_{n-1}+y_{n}, z_{1}+\cdots+z_{n-1}\right) \\
\cdots \cdots \cdots \cdots \\
M_{2}=\left(x_{2}+... | 1975 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,985 |
$9 \cdot 59$ Let $x_{i} \geqslant 0, i=1,2, \cdots, n, n \geqslant 2$ and $\sum_{i=1}^{n} x_{i}=1$, find the maximum value of $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)$. | [Solution] Let $z=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) ; x_{i} \geqslant 0, i=1,2, \cdots, n\right.$ and $\sum_{i=1}^{n} x_{i}=$ $1\}, F(v)=\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)$. When $n=2$, it is obvious that
$$\max _{v \in:} F(v)=\frac{1}{4}$$
and it is achieved when $x_{... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,986 |
$9 \cdot 60$ Let $x_{1}, x_{2}, x_{3}, x_{4}$ all be positive numbers, and $x_{1}+x_{2}+x_{3}+x_{4}=\pi$, find the minimum value of the expression
$$\left(2 \sin ^{2} x_{1}+\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right)\left(2 \sin ^{2} x_{3}+\frac{1}{\sin ^{2} x_{3}}\right)\le... | [Solution] By the AM-GM inequality, we have
$$2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}=2 \sin ^{2} x_{i}+\frac{1}{2 \sin ^{2} x_{i}}+\frac{1}{2 \sin ^{2} x_{i}} \geqslant 3 \sqrt[3]{\frac{1}{2 \sin ^{2} x_{i}}}$$
Therefore, $\prod_{i=1}^{4}\left(2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}\right) \geqslant 81\left(4... | 81 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,987 |
1. Does there exist a natural number $n$ such that the first four digits of $n!$ are 1993? | ［Solution］Exists.
Let $m=1000100000$, when $k<99999$, if
$$(m+k)!=\overline{a b c d \cdots}$$
then $\square$
$$\begin{array}{l}
(m+k+1)!=(m+k)!\times(m+k+1) \\
=\overline{a b c d} \cdots \times 10001 \cdots \\
=\overline{a b c x \cdots} \text {, where } x=d \text { or } d+1 \text {. }
\end{array}$$
Thus, if $m!=\over... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,988 |
9.61 Given a natural number $n \geqslant 2$. Find the smallest positive number $\lambda$, such that for any positive numbers $a_{1}$, $a_{2}, \cdots, a_{n}$, and any $n$ numbers $b_{1}, b_{2}, \cdots, b_{n}$ in $\left[0, \frac{1}{2}\right]$, if
$$a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n}=1$$
then $\quad a_{1} ... | [Solution] By Cauchy-Schwarz inequality, we have
$$1=\sum_{i=1}^{n} b_{i}=\sum_{i=1}^{n} \frac{\sqrt{b_{i}}}{\sqrt{a_{i}}} \sqrt{a_{i} b_{i}} \leqslant\left(\sum_{i=1}^{n} \frac{b_{i}}{a_{i}}\right)^{\frac{1}{2}}\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{\frac{1}{2}}$$
Thus,
$$\frac{1}{\sum_{i=1}^{n} a_{i} b_{i}} \leqsl... | \frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,989 |
9.62 Find the smallest positive number $\lambda$, such that for any triangle with side lengths $a, b, c$, if $a \geqslant \frac{b+c}{3}$, then
$$a c+b c-c^{2} \leqslant \lambda\left(a^{2}+b^{2}+3 c^{2}+2 a b-4 b c\right) .$$ | [Solution] It is easy to know that
$$\begin{aligned}
a^{2}+b^{2}+3 c^{2}+2 a b-4 b c & =(a+b-c)^{2}+2 c^{2}+2 a c-2 b c \\
& =(a+b-c)^{2}+2 c(a+c-b) .
\end{aligned}$$
Let \( I=\frac{(a+b-c)^{2}+2 c(a+c-b)}{2 c(a+b-c)}=\frac{a+b-c}{2 c}+\frac{a+c-b}{a+b-c} \).
Since \( a \geqslant \frac{1}{3}(b+c) \), we have \( a \geq... | \frac{2 \sqrt{2}+1}{7} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,990 |
9.64 Given a natural number $n$ greater than 1. Let $a, b, c, d$ be natural numbers satisfying
$$\frac{a}{b}+\frac{c}{d}<1, a+c \leqslant n$$
Find the maximum value of $\frac{a}{b}+\frac{c}{d}$. | [Solution] First, let $a+c=n$. If $b, d \geqslant n+1$, then
$$\frac{a}{b}+\frac{c}{d} \leqslant \frac{n}{n+1}$$
It is also clear that $\frac{n}{n+1}0$, when $b>\bar{b}$, $f(b+1)-f(b)<0$. Therefore, when $b$ equals the smallest integer not less than $b_{1}$, $f(b)$ is maximized. From $(2 n-1)^{2}+12(n+1)=4 n^{2}+8 n+1... | 1-\frac{1}{\left[\frac{2 n}{3}+\frac{7}{6}\right]\left(\left[\frac{2 n}{3}+\frac{7}{6}\right]\left(n-\left[\frac{2 n}{3}+\frac{1}{6}\right]\right)+1\right)} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,992 |
9.65 Let $u, v$ be positive real numbers, and for a given positive integer $n$, find the necessary and sufficient conditions that $u, v$ must satisfy so that there exist real numbers $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n} \geqslant 0$ satisfying
$$\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n}=u \\
a_{1}^{2... | [Solution] If $a_{1}, a_{2}, \cdots, a_{n}$ satisfying the conditions exist, by the Cauchy-Schwarz inequality, we have
$$\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2} \leqslant n\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)$$
It is also clear that
$$\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2} \geqslant a_{1}^{2}+a_{2}^{... | \frac{k u+\sqrt{k\left[(k+1) v-u^{2}\right]}}{k(k+1)} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,993 |
$9 \cdot 67$ Let $n$ be a fixed integer, $n \geqslant 2$.
(a) Determine the smallest constant $c$, such that the inequality
$$\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant c\left(\sum_{1 \leqslant i \leqslant n} x_{i}\right)^{4}$$
holds for all non-negative real numbers $x_{1... | [Solution] Since the inequality is homogeneous and symmetric, we may assume without loss of generality that $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n} \geqslant 0$, and
$$\sum_{i=1}^{n} x_{i}=1$$
Thus, we only need to consider
$$F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{1 \leqslant i < j \leqslant n... | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,995 |
9.68 Find the range of real number $a$ such that for any real number $x$ and any $\theta \in \left[0, \frac{\pi}{2}\right]$, the following inequality always holds:
$$(x+3+2 \sin \theta \cos \theta)^{2}+(x+a \sin \theta+a \cos \theta)^{2} \geqslant \frac{1}{8}$$ | [Solution] It is easy to know that the original inequality is equivalent to
$$(3+2 \sin \theta \cos \theta-a \sin \theta-a \cos \theta)^{2} \geqslant \frac{1}{4}$$
for any $\theta \in\left[0, \frac{\pi}{2}\right]$.
From (1), we get
$$a \geqslant \frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta}$$... | a \geqslant \frac{7}{2} \text{ or } a \leqslant \sqrt{6} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,996 |
$9 \cdot 69$ (1) Prove: For $n>2$, there is
$$3-\frac{2}{(n-1)!}<\frac{2^{3}}{2!}+\frac{3^{3}}{3!}+\cdots+\frac{n^{3}}{n!}<3 .$$
(2) Let $a, b, c$ be constants. For $n>2$, there is
$$b-\frac{c}{(n-2)!}<\frac{2^{3}-a}{2!}+\frac{3^{3}-a}{3!}+\cdots+\frac{n^{3}-a}{n!}<b .$$ | [Solution] (1) For $n \geqslant 2$, let
$$d_{n}=3-\left(\frac{2}{2!}+\frac{7}{3!}+\cdots+\frac{n^{2}-2}{n!}\right)$$
We first prove that $d_{n}=\frac{n+2}{n!}, n \geqslant 2$.
Indeed, $d_{2}=3-\frac{2}{2!}=\frac{6-2}{2!}=\frac{2+2}{2!}$.
Assume $d_{k}=\frac{k+2}{k!}$.
Then $\quad d_{k+1}=d_{k}-\frac{(k+1)^{2}-2}{(k+1)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,997 |
$1 \cdot 131$ Can $9 \times 9$ grid be filled with natural numbers 1 to 8, one number per cell, so that the sum of the 9 numbers in each $3 \times 3$ square is the same? | [Solution] First, fill out a $9 \times 9$ grid table $T$:
\begin{tabular}{c|c|c|c|c|c|c|c|c}
\hline 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline 3 & 4 & 5 & 6 & 7 & 8 & 0 & 1 & 2 \\
\hline 6 & 7 & 8 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline 3 & 4 & 5 & 6 & 7 & 8 & 0 & 1 & 2 \\
\hline ... | 369 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 734,999 |
9・72 Prove the inequality
$$\frac{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}>\frac{1}{4}$$
where the number of square roots in the numerator is one more than in the denominator. | [Proof] Let the square root value in the denominator be $a$. It is easy to prove
$$1\frac{1}{4}$$
From $\frac{2-\sqrt{2+a}}{2-a}=\frac{1}{2+\sqrt{2+a}}$ and (1), we can immediately obtain that (2) holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,000 |
9.74 Try to compare the size of the following pairs of numbers:
(1) $2^{2^{2^{2}}}$ n times and $3^{3^{3^{3^{3}}}}$ n-1 times.
(2) $3^{3^{3^{3^{*}}}}$ n times and $4^{4^{4^{4^{4}}}}$ n-1 times. | [Solution] Let
$$\begin{array}{l}
A_{n}=2^{2^{2^{2^{\cdot}}}} n \text { times, } B_{n}=3^{3^{3^{3^{3^{*}}}}} n \text { times, } \\
C_{n}=4^{4^{4^{4^{4^{4}}}}} n \text { times. }
\end{array}$$
(1) When $n=2$, $A_{2}=4>3=B_{1}$. Using induction, it is easy to prove that when $n \geqslant 3$, $A_{n}2 C_{n-1}$. In fact, if... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,001 |
9・75 Let $a_{1}, a_{2}, \cdots, a_{1985}$ be a permutation of $1,2, \cdots, 1985$, and multiply each number $a_{k}$ by its index $k$. Prove that the maximum value of these 1985 products is not less than $993^{2}$. | [Proof] There are exactly 993 numbers $a_{k}$ that are not less than 993. Among these numbers, there must be one, denoted as $a_{n}$, whose subscript $n \geqslant 993$, hence
$$n a_{n} \geqslant 993^{2} .$$
Thus, the conclusion to be proven holds. | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,002 |
9・77 Let $a, b, c, d, e$ and $f$ be natural numbers and
$$\frac{a}{b}>\frac{c}{d}>\frac{e}{f}$$
If $a f-b e=1$, prove: $d \geqslant b+f$. | [Proof] From the assumption, we have
$$\begin{aligned}
d & =d(a f-b e)=a d f-b c f+b c f-h e d \\
& =f(a d-b c)+b(c f-e d) .
\end{aligned}$$
Given that \(a d-b c>0\), \(c f-e d>0\), and \(a, b, c, d, e, f\) are all natural numbers, we have
$$a d-b c \geqslant 1, c f-e d \geqslant 1$$
Thus, we can conclude
$$d \geqsla... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,003 |
9. 78 For natural numbers $n$ and $k$ greater than 1, prove:
$$\sum_{i=2}^{k^{k}} \frac{1}{i}>k \sum_{i=2}^{n} \frac{1}{i}$$ | [Proof] Let $S_{m}=\frac{1}{n^{m-1}+1}+\frac{1}{n^{m-1}+2}+\cdots+\frac{1}{n^{m}}$,
then $S=\sum_{i=2}^{n^{k}} \frac{1}{i}=\sum_{m=1}^{k} S_{m}$.
Furthermore, let
then
$$\begin{aligned}
S_{m, l} & =\frac{1}{l n^{m-1}+1}+\frac{1}{l n^{m-1}+2}+\cdots+\frac{1}{(l+1) n^{m-1}} \\
S_{m} & =\sum_{l=1}^{n-1} S_{m, l}
\end{ali... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,004 |
$9 \cdot 79$ Let $p_{1}, p_{2}, \cdots, p_{n}$ be different natural numbers, and all greater than 1, prove that
$$\left(1-\frac{1}{p_{1}^{2}}\right)\left(1-\frac{1}{p_{2}^{2}}\right) \cdots\left(1-\frac{1}{p_{n}^{2}}\right)>\frac{1}{2} .$$ | ［Proof］Let $m=\max \left\{p_{1}, p_{2}, \cdots, p_{n}\right\}$, then
$$\begin{array}{l}
\left(1-\frac{1}{p_{1}^{2}}\right)\left(1-\frac{1}{p_{2}^{2}}\right) \cdots\left(1-\frac{1}{p_{n}^{2}}\right) \geqslant\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right) \cdots\left(1-\frac{1}{m^{2}}\right) \\
=\frac{2^{2}... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,005 |
1. 132 Starting from 1 and writing natural numbers in ascending order up to a natural number $n(n>1)$, we get a number: Can the resulting number be read the same way from left to right as from right to left? Prove your conclusion. | ［Solution］ Not possible.
If there exists a number $N$ written from 1 to $n$ that meets the conditions
$$N=123 \cdots 321$$
Let the number of digits in $N$ be $m$, clearly $m>18$.
Let $A$ be the number formed by the first $\left[\frac{m}{2}\right]$ digits of $N$, and $B$ be the number formed by the last $\left[\frac{m}... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,007 |
9.81 Let \(a_{1}, a_{2}, \cdots, a_{n}\) be \(n\) distinct positive integers, none of whose decimal representations contain the digit 9. Prove that:
$$\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<30 .$$ | [Proof] Arrange all positive integers in decimal representation without the digit 9 as $b_{1}, b_{2}$, $b_{3}, \cdots$. It is easy to see that
$$\sum_{k=1}^{\infty} \frac{1}{b_{k}} \leqslant c+\frac{9}{10} c+\left(\frac{9}{10}\right)^{2} c+\left(\frac{9}{10}\right)^{3} c+\cdots$$
where $c=1+\frac{1}{2}+\frac{1}{3}+\cd... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,008 |
9. 82 Let $m$ and $n$ be two coprime natural numbers and $n < m$. Try to compare the following two numbers:
$$\sum_{k=1}^{n}\left[k \cdot \frac{m}{n}\right] \text { and } \sum_{k=1}^{m}\left[k \cdot \frac{n}{m}\right]$$
in size. | [Solution] Since
$$\frac{k m}{n}=\left[\frac{k m}{n}\right]+\frac{a_{k}}{n}, k=1,2, \cdots, n$$
where $a_{k}$ is an integer and $0 \leqslant a_{k} \leqslant n-1$. Also, since $m, n$ are coprime, $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ is a permutation of $\{0,1,2, \cdots, n-1\}$. Therefore,
$$\begin{aligned}
\sum... | m+\frac{1}{2}(m-1)(n-1) > n+\frac{1}{2}(m-1)(n-1) | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,009 |
9.83 In an $m \times m$ grid, each small square is filled with a non-negative integer. If a small square is filled with 0, then the sum of the numbers in all the small squares in the row and column containing this small square is not less than $m$. Prove that the sum of all the numbers in the grid is not less than $\fr... | [Proof] Consider the sum of the numbers filled in each row and each column, and let the smallest of these $2m$ numbers be denoted as $p$. Without loss of generality, assume $p$ is the sum of the numbers in the first row. If $p \geqslant \frac{m}{2}$, then the conclusion to be proven is obviously true. Suppose $p < \fra... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,010 |
9.84 For any positive integers $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$, prove:
$$\begin{aligned}
& \left(a_{1} b_{2}+a_{1} b_{3}+a_{2} b_{1}+a_{2} b_{3}+a_{3} b_{1}+a_{3} b_{2}\right)^{2} \\
\geqslant & 4\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}\right)\left(b_{1} b_{2}+b_{2} b_{3}+b_{3} b_{1}\right)
\end{aligned}$$
... | [Proof] Let the quadratic function be
$$\begin{aligned}
f(x)= & \left(b_{1} x-a_{1}\right)\left(b_{2} x-a_{2}\right)+\left(b_{2} x-a_{2}\right)\left(b_{3} x-a_{3}\right)+\left(b_{3} x\right. \\
& \left.-a_{3}\right)\left(b_{1} x-a_{1}\right) .
\end{aligned}$$
If the inequality to be proved does not hold, then for any ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,011 |
9. 85 For any natural number $k$, prove that in its decimal representation: the sum of all its digits is no more than 8 times the sum of all the digits of $8 k$. | [Proof] For any $x \in N$, let $S(x)$ denote the sum of all digits in the decimal representation of $x$. It is easy to see that
$$s(a+b) \leqslant s(a)+s(b) \text{, for any } a, b \in N \text{.}$$
Let the decimal representation of $a$ be $\overline{a_{n} a_{n-1} \cdots a_{2} a_{1}}$, then
$$a b=\sum_{k=1}^{n} a_{k} \c... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,012 |
$9 \cdot 86$ Write $n$ natural numbers $(n \geqslant 3)$ on a circle, such that the ratio of the sum of the two adjacent numbers to the number itself is a natural number for each number, and let the sum of all these ratios be $s_{n}$. Prove:
$$2 n \leqslant s_{n}<3 n$$ | [Proof] Label $n$ numbers as $x_{1}, x_{2}, \cdots, x_{n}$, then
$$s_{n}=\frac{x_{1}+x_{3}}{x_{2}}+\cdots+\frac{x_{n-2}+x_{n}}{x_{n-1}}+\frac{x_{n-1}+x_{1}}{x_{n}}+\frac{x_{n}+x_{2}}{x_{1}} .$$
Thus, $s_{n}=\left(\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{1}}\right)+\left(\frac{x_{3}}{x_{2}}+\frac{x_{2}}{x_{3}}\right)+\cdots... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,013 |
$9 \cdot 87$ Using the formula $1^{3}+2^{3}+\cdots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$, prove: For distinct natural numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have
$$\sum_{k=1}^{n}\left(a_{k}^{7}+a_{k}^{5}\right) \geqslant 2\left(\sum_{k=1}^{n} a_{k}^{3}\right)^{2}$$
and ask whether the equality can hold. | [Proof] By induction. It is easy to verify that the inequality holds for $n=1$. Assume the inequality holds for $n=k$. For $n=k+1$, without loss of generality, let $a_{1}<a_{2}<\cdots<a_{k}<a_{k+1}$. Since
$$\left(\sum_{i=1}^{k+1} a_{i}^{3}\right)^{2}=\left(\sum_{i=1}^{k} a_{i}^{3}\right)^{2}+2 a_{k+1}^{3} \sum_{i=1}^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,014 |
9.88 For natural numbers $n \geqslant 2$, prove:
$$\sqrt{2 \sqrt{3 \sqrt{4 \cdots \sqrt{(n-1) \sqrt{n}}}}}<3 .$$ | [Proof] We can prove a more general proposition: for $2 \leqslant m \leqslant n$,
$$\sqrt{m \sqrt{(m+1) \sqrt{\cdots \sqrt{n}}}}<m+1$$
We use backward induction on $m$. When $m=n$, it is clearly true that
$$\sqrt{n}<n+1$$
Assume the proposition holds for $m=k(2<k \leqslant n)$, then for $m=k-1$ we have
$$\sqrt{(k-1) ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,015 |
9.89 Let $n \in N$, prove:
$$\frac{2}{3} n \sqrt{n}<\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}<\frac{4 n+3}{6} \sqrt{n}$$ | ［Proof］ By induction. When $n=1$, it is obvious that
$$\frac{2}{3}(2 k-1) \sqrt{k+1}$$
From this, we can deduce that
$$\frac{2}{3} k \sqrt{k}+\sqrt{k+1}>\frac{2}{3}(k+1) \sqrt{k+1} .$$
A simple calculation also shows that
$$(4 k+3) \sqrt{k}<(4 k+1) \sqrt{k+1}$$
Therefore,
$$\frac{4 k+3}{6} \sqrt{k}<\frac{4(k+1)-3}{6... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,016 |
$9 \cdot 91$ For $0 \leqslant k \leqslant n$, let $a_{n}, b_{n}$ be the number of binomial coefficients $c_{n}^{k}$ that are congruent to 1, 2 modulo 3, respectively. Prove that: $a_{n}>b_{n}$. | [Proof] Let $P(x)$ and $Q(x)$ be two polynomials with integer coefficients. If the coefficients of $P(x) - Q(x)$ are all divisible by 3, then we denote it as
$$P(x) \equiv Q(x) \pmod{3}$$
Returning to the problem we are discussing. If $n=0$, it is obvious that $a_{n}=1, b_{n}=0$. If $n = c \cdot 3^{k}$, where $k$ is a... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,017 |
9・96 Let natural numbers $k, l, m, n$ satisfy $k<l<m<n$, and $l m=k n$, prove:
$$\left(\frac{n-k}{2}\right)^{2} \geqslant k+2$$ | [Proof] Let $l=k+a, m=k+b$, then $a, b$ are natural numbers, and $a2 \sqrt{a b}+1 \geqslant 2 \sqrt{k}+1$
Obviously, when $k>3$, we have
thus $\square$
$$2 \sqrt{k}+1>2 \sqrt{k+2}$$
which means $\square$
$$n-k>2 \sqrt{k+2}$$
$$\left(\frac{n-k}{2}\right)^{2}>k+2$$
If $k=1$ or 2, then $n \geqslant \frac{(k+1)(k+2)}{k... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,021 |
9.97. Let $m \leqslant n$ be natural numbers. Select $m$ different numbers $a_{1}, a_{2}, \cdots, a_{m}$ from $\{1,2, \cdots, n\}$, such that if $a_{i}+a_{j} \leqslant n$, then there exists $a$ satisfying $a_{i}+a_{j}=a_{k}$. Prove:
$$\frac{a_{1}+a_{2}+\cdots+a_{m}}{m} \geqslant \frac{n+1}{2}$$ | [Proof] Without loss of generality, let $a_{1}<a_{2}<\cdots<a_{m}$. We can prove that
$$a_{i}+a_{m-i+1} \geqslant n+1, \quad i=1,2, \cdots, m$$
In fact, if there exists $1 \leqslant i \leqslant m$ such that (1) does not hold, i.e.,
$$a_{i}+a_{m-i+1} \leqslant n$$
Assume without loss of generality that $i \leqslant m-... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,022 |
9.98 Let $A_{1}, A_{2}, \cdots, A_{29}$ be 29 sets consisting of certain natural numbers. For $1 \leqslant i, j \leqslant 29$ and natural number $x$, define
$N_{i}(x)=A_{i}$ the number of elements $\leqslant x$ in $A_{i}$,
$N_{i j}(x)=A_{i} \cap A_{j}$ the number of elements $\leqslant x$ in $A_{i} \cap A_{j}$.
If $N_{... | [Proof] Without loss of generality, assume that the elements of $A_i (1 \leqslant i \leqslant 29)$ are all $\leqslant 1988$. By the assumption, we get
$$N_{i}(1988) \geqslant \frac{1988}{e}=731 \cdot 3 \cdots, i=1,2, \cdots, 29.$$
Thus, we can assume $N_{i}(1988)=732, i=1,2, \cdots, 29$.
For $1 \leqslant i \leqslant 2... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,023 |
9.99 Let $\delta(x)$ be the largest odd divisor of the positive integer $x$. For any positive integer $x$, prove:
$$\left|\sum_{n=1}^{x} \frac{\delta(n)}{n}-\frac{2}{3} x\right|<1$$ | [Proof] Let $S(x)=\sum_{n=1}^{x} \frac{\delta(n)}{n}$, obviously $S(1)=1$. Since
$$\delta(2 m+1)=2 m+1, \delta(2 m)=\delta(m)$$
Therefore,
$$\begin{array}{l}
S(2 x+1)=\sum_{n=1}^{2 x} \frac{\delta(n)}{n}+\frac{\delta(2 x+1)}{2 x+1}=S(2 x)+1 \\
S(2 x)=\sum_{m=1}^{x} \frac{\delta(2 m)}{2 m}+\sum_{m=1}^{x} \frac{\delta(2... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,024 |
9- 100 Given the sequence $\left\{r_{n}\right\}$ satisfies $r_{1}=2$,
$$r_{n}=r_{1} r_{2} \cdots r_{n-1}+1, n=2,3, \cdots$$
If natural numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<1$$
Prove: $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} \leqslant \fr... | ［Proof］ First, it is easy to prove by induction that the sequence $\left\{r_{n}\right\}$ satisfies
$$\frac{1}{r_{1}}+\frac{1}{r_{2}}+\cdots+\frac{1}{r_{n}}=1-\frac{1}{r_{1} r_{2} \cdots r_{n}}, n=1,2, \cdots .$$
The following is a proof of the original inequality by induction. When $n=1$, the original inequality clear... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,025 |
1. 14 Proof: In any sequence of 39 consecutive natural numbers, there exists a natural number whose sum of digits is divisible by 11. | [Proof] Among the first 20 known numbers, there exist two numbers whose last digit in decimal notation is zero.
Among these two numbers, there must be one number whose digit before the last 0 is not equal to 9. Let this number be $N$, and let $S$ be the sum of the digits of $N$. Then
$$N, N+1, \cdots, N+9, N+19$$
are... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,026 |
1. 134 A $n \times n$ matrix (square matrix) is called a silver matrix of order $n$ if its elements are taken from the set
$$S=\{1,2, \cdots, 2 n-1\}$$
and for each $i=1,2, \cdots, n$, the elements in the $i$-th row and the $i$-th column together are exactly all the elements of $S$. Prove:
(a) There does not exist a s... | [Proof] (a) Let $n>1$, and there exists an $n$-order silver matrix $A$. Since all $2n-1$ numbers in $S$ must appear in $A$, and there are only $n$ elements on the main diagonal of $A$, there must be at least one $x \in S$ that is not on the main diagonal of $A$.
Choose such an $x$. For each $i=1,2, \cdots, n$, let the... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,027 |
$9 \cdot 101$ For each positive integer $n>1$, prove:
$$n\left((n+1)^{\frac{2}{n}}-1\right)<\sum_{j=1}^{n} \frac{2 j+1}{j^{2}}<n\left(1-n^{-\frac{2}{n-1}}\right)+4 .$$ | [Proof] Clearly,
$$2 j+1=(j+1)^{2}-j^{2},$$
Therefore, from the above equation and the arithmetic-geometric mean inequality, we have
$$\begin{aligned}
\sum_{j=1}^{n} \frac{2 j+1}{j^{2}} & =\sum_{j=1}^{n}\left[\frac{(j+1)^{2}}{j^{2}}-1\right] \\
& =\sum_{j=1}^{n} \frac{(j+1)^{2}}{j^{2}}-n \\
& =\frac{2^{2}}{1^{2}}+\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,028 |
9. 102 Proof:
$$\frac{1}{1999}<\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \cdots \cdot \frac{1997}{1998}<\frac{1}{44}$$ | [Proof] Let
$$p=\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \cdot \frac{1997}{1998}$$
If we add 1 to each denominator in \( p \), the value clearly becomes smaller. Therefore, we have
$$p>\frac{1}{3} \cdot \frac{3}{5} \cdot \frac{5}{7} \cdots \cdots \cdot \frac{1995}{1997} \cdot \frac{1997}{1999}=\frac{1}{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,029 |
9. 103. For $|x|<1$ and integer $n \geqslant 2$, prove:
$$(1-x)^{n}+(1+x)^{n}<2^{n}$$ | [Proof] By induction. When $n=2$, it is obvious that
$$(1-x)^{n}+(1+x)^{n}=2+2 x^{2}<2^{2}$$
Assume that the inequality holds for $n=k-1$, where $k \geqslant 3$.
When $n=k$, we have
$$\begin{aligned}
& (1-x)^{k}+(1+x)^{k} \\
= & (1-x)^{k-1}+(1+x)^{k-1}+x\left[(1+x)^{k-1}-(1-x)^{k-1}\right]
\end{aligned}$$
By the indu... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,030 |
9・104 Let $f(x)$ be defined on $[0,1]$ and $f(0)=f(1)=0$. If for any different $x_{1}, x_{2} \in[0,1]$, we have $\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|<\left|x_{2}-x_{1}\right|$, prove that $\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|<\frac{1}{2}$. | [Proof] Without loss of generality, let $\quad 0 \leqslant x_{1} < x_{2} \leqslant 1$ and $x_{2} - x_{1} > \frac{1}{2}$. From $f(0)=f(1)=0$, we can get
$$\begin{aligned}
\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right| & =\left|f\left(x_{2}\right)-f(1)+f(0)-f\left(x_{1}\right)\right| \\
& \leqslant\left|f\left(x_{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,031 |
9・105 Let $a, b$ be real numbers such that the inequality
$$a \cos x + b \cos 3x > 1$$
has no solution. Prove: $|b| \leqslant 1$. | [Proof] By contradiction. Suppose $|b|>1$, take
$$\begin{array}{l}
x_{1}=\frac{1}{3} \arccos \frac{1}{b}, \\
x_{2}=\frac{2}{3} \pi+x_{1} .
\end{array}$$
Since $a \cos x+b \cos 3 x>1$ has no solution, we have
$$a \cos x_{1} \leqslant 0, \quad a \cos x_{2} \leqslant 0 .$$
Also, $0<\cos x_{1}<1, \cos x_{2}<0$. Therefore... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,032 |
9. 106 Prove: There exists a unique pair of real numbers $(p, q)=\left(-1, \frac{\sqrt{2}+1}{2}\right)$, such that the inequality
$$\left|\sqrt{1-x^{2}}-p x-q\right| \leqslant \frac{\sqrt{2}-1}{2}$$
holds for all $x \in[0,1]$. | [Proof] For any $x \in [0,1]$, let $x = \sin \theta, \sqrt{1-x^2} = \cos \theta$, from which we can deduce
$$1 \leq \sqrt{1-x^2} + x \leq \sqrt{2}$$
Thus, we have $-\frac{\sqrt{2}-1}{2} \leq \sqrt{1-x^2} + x - \frac{1+\sqrt{2}}{2} \leq \frac{\sqrt{2}-1}{2}$,
which means $\left|\sqrt{1-x^2} + x - \frac{1+\sqrt{2}}{2}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,033 |
9. 108 Prove: For any positive number $t$,
$$t^{4}-t+\frac{1}{2}>0$$ | [Proof] Since
$$\begin{aligned}
t^{4}-t+\frac{1}{2} & =\left(t^{4}-t^{2}+\frac{1}{4}\right)+\left(t^{2}-t+\frac{1}{4}\right) \\
& =\left(t^{2}-\frac{1}{2}\right)^{2}+\left(t-\frac{1}{2}\right)^{2}
\end{aligned}$$
Also, \( t^{2}-\frac{1}{2} \) and \( t-\frac{1}{2} \) cannot both be 0, so
$$t^{4}-t+\frac{1}{2}>0$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,034 |
9・109 Let $0<x<a$, prove:
$$(a-x)^{6}-3 a(a-x)^{5}+\frac{5}{2} a^{2}(a-x)^{4}-\frac{1}{2} a^{4}(a-x)^{2}<0 .$$ | [Proof] Let $y=\frac{1}{a}(a-x)$, then $0<y<1$, and it suffices to prove
$$f(y)=a^{6} y^{6}-3 a^{6} y^{5}+\frac{5}{2} a^{6} y^{4}-\frac{1}{2} a^{6} y^{2}<0$$
Since
$$\begin{aligned}
f(y) & =a^{6} y^{2}\left(y^{4}-3 y^{3}+\frac{5}{2} y^{2}-\frac{1}{2}\right) \\
& =a^{6} y^{2}(y-1)\left(y^{3}-2 y^{2}+\frac{1}{2} y+\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,035 |
1. 135 We call natural numbers similar if they can be written using the same set of digits (for example, 112, 121, 211 are similar because they can all be written using the digit set $1,1,2$). Prove that there exist three similar 1995-digit numbers, whose digits do not contain 0, but the sum of two of these numbers equ... | [Solution] Since 1995 is a multiple of 3, and
$$459+495=954$$
therefore, there is a 1995-digit number
$$\begin{array}{l}
459459459 \cdots 459, \\
495495495 \cdots 495, \\
954954954 \cdots 954,
\end{array}$$
which meets the requirements of the problem. Hence, the proposition is proved. | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,037 |
9. 111 For any natural number $n$ and real number $a$, prove:
$$|a| \cdot |a-1| \cdot \cdots \cdot |a-n| \geqslant \langle a \rangle \cdot \frac{n!}{2^n},$$
where $\langle a \rangle$ denotes the distance from $a$ to the nearest integer. | [Proof] Take a permutation $x_{0}, x_{1}, x_{2}, \cdots, x_{n}$ of $0,1,2, \cdots, n$ such that $\left|a-x_{0}\right| \leqslant\left|a-x_{1}\right| \leqslant\left|a-x_{2}\right| \leqslant \cdots \leqslant\left|a-x_{n}\right|$.
Obviously $\square$
$\langle a\rangle \leqslant\left|a-x_{0}\right|$.
Since for any natural n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,038 |
9・112 Let $x>0, n$ be a positive integer. Prove:
$$[n x] \geqslant[x]+\frac{[2 x]}{2}+\frac{[3 x]}{3}+\cdots+\frac{[n x]}{n}$$ | [Proof] Let $x_{k}=\sum_{m=1}^{k} \frac{[m x]}{m}$, the original inequality can be written as
$$x_{n} \leqslant[n x]$$
We will prove (1) by induction.
When $n=1$, (1) clearly holds.
Assume that (1) holds for $n \leqslant k-1$, consider the case when $n=k$. Since
$$\begin{array}{l}
x_{m}=x_{m-1}+\frac{[m x]}{m}, m=2,3,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,039 |
9・116 Let the function $f:(0,+\infty) \rightarrow(0,+\infty)$ satisfy $f(x y) \leqslant f(x) f(y)$, for any $x>0, y>0$. Prove that for any $x>0, n \in N$, we have
$$f\left(x^{n}\right) \leqslant f(x) f\left(x^{2}\right)^{\frac{1}{2}} \cdots f\left(x^{n}\right)^{\frac{1}{n}}$$ | [Proof] Let $F_{n}(x)=f(x) f\left(x^{2}\right)^{\frac{1}{2}} \cdots f\left(x^{n}\right)^{\frac{1}{n}}$, then
$$F_{n}(x)=F_{n-1}(x) f\left(x^{n}\right)^{\frac{1}{n}}$$
Thus, we have
$$\begin{array}{l}
F_{n}^{n}(x)=F_{n-1}^{n}(x) f\left(x^{n}\right) \\
F_{n-1}^{n-1}(x)=F_{n-2}^{n-1}(x) f\left(x^{n-1}\right) \\
\cdots \c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,043 |
9. 117 Let $0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$, prove that:
$$\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta} \geqslant 9$$
and determine the values of $\alpha, \beta$ for which equality holds. | [Proof] Since $\frac{1}{\sin ^{2} \beta \cos ^{2} \beta} \geqslant 4$, equality holds if and only if $\beta=\frac{\pi}{4}$. By the AM-GM inequality, we have
$$\begin{aligned}
\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta} & \geqslant \frac{1}{\cos ^{2} \alpha}+\frac{2}{\sin ^{2} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,044 |
9・118 (1) Factorize $x^{12}+x^{9}+x^{6}+x^{3}+1$.
(2) For any real number $\theta$, prove: $5+8 \cos \theta+4 \cos 2 \theta+\cos 3 \theta \geqslant 0$. | [Solution] (1) Since $x^{15}-1=\left(x^{3}-1\right)\left(x^{12}+x^{9}+x^{6}+x^{3}+1\right)$,
$$x^{15}-1=\left(x^{5}-1\right)\left(x^{10}+x^{5}+1\right)$$
Therefore, $x^{12}+x^{9}+x^{6}+x^{3}+1=\frac{\left(x^{4}+x^{3}+x^{2}+1\right)\left(x^{10}+x^{5}+1\right)}{x^{2}+x+1}$.
Noting that if $x^{2}+x+1=0$ it implies $x^{10... | (1+\cos \theta)(2 \cos \theta+1)^{2} \geqslant 0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,045 |
9. 119 Let $\alpha, \beta, \gamma$ be the three interior angles of a triangle. Prove:
$$\begin{array}{l}
2\left(\frac{\sin \alpha}{\alpha}+\frac{\sin \beta}{\beta}+\frac{\sin \gamma}{\gamma}\right) \leqslant\left(\frac{1}{\beta}+\frac{1}{\gamma}\right) \sin \alpha+\left(\frac{1}{\gamma}+\frac{1}{\alpha}\right) \sin \be... | [Proof] Without loss of generality, let $\alpha \leqslant \beta \leqslant \gamma$. Since $\alpha, \beta, \gamma$ are the three interior angles of a triangle, it is easy to see that
$$\sin \alpha \leqslant \sin \beta \leqslant \sin \gamma$$
By the rearrangement inequality, we have
$$\begin{array}{l}
\frac{\sin \alpha}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,046 |
9. 120 Let $A, B, C$ be the three interior angles of a triangle. Prove that:
$$-2<\sin 3 A+\sin 3 B+\sin 3 C \leqslant \frac{3}{2} \sqrt{3},$$
and determine when the equality holds. | [Proof] Without loss of generality, let $A \geqslant 60^{\circ}$, then $B+C=180^{\circ}-A \leqslant 120^{\circ}$, thus $0^{\circ} \leqslant \frac{3}{2}|B-C|\cos \frac{3}{2}(B+C)$.
Furthermore, by $\sin \frac{3}{2}(B+C) \geqslant 0$, we get
$$2 \sin \frac{3}{2}(B+C) \cos \frac{3}{2}(B-C) \geqslant 2 \sin \frac{3}{2}(B+C... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,047 |
1. 136 Proof: For any natural number $k>1$, there exists a power of 2 such that at least half of the last $k$ digits are 9 (for example, for $k=2$, there is $2^{12}=$ 4096; for $k=3$, there is $2^{53}=\cdots 992$; etc.). | [Proof] We will prove that for any natural number $k$, in the last $k+2$ digits of $2^{10 \cdot 5^{k}+k+2}$, there are at least $\left[\frac{2}{3}(k+2)\right]$ nines.
In fact, we know that
$$\begin{array}{l}
2^{10}=1024, \\
\begin{aligned}
2^{10} \cdot 5^{k}= & 1024^{5^{k}}=\left(10^{3}+24\right)^{5^{k}} \\
= & 10^{3 ... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,048 |
9. 121 Prove: For each natural number $n$, the inequality
$|\sin 1|+|\sin 2|+\cdots+|\sin (3 n-1)|+|\sin 3 n|>\frac{8}{5} n$ holds. | [Proof] Let $f(x)=|\sin x|+|\sin (x+1)|+|\sin (x+2)|$. Clearly, we only need to prove that for any real number $x$,
$$f(x)>\frac{8}{5}$$
Since $f(x)$ is a periodic function with period $\pi$, it suffices to prove (1) for $x \in [0, \pi]$.
When $0 \leqslant x \leqslant \pi-2$, $f(x)=\sin x+\sin (x+1)+\sin (x+2)$. Sinc... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,049 |
9・122 Let real numbers $\theta_{1}, \theta_{2}, \cdots, \theta_{n}$ satisfy
$$\sin \theta_{1}+\sin \theta_{2}+\cdots+\sin \theta_{n}=0$$
Prove: $\left|\sin \theta_{1}+2 \sin \theta_{2}+\cdots+n \sin \theta_{n}\right| \leqslant\left[\frac{n^{2}}{4}\right]$. | ［Proof］ Let $x_{k}=\sin \theta_{k}, k=1,2, \cdots, n$.
When $n=2 m$,
$$\begin{aligned}
{\left[\frac{n^{2}}{4}\right]-\sum_{k=1}^{n} k x_{k} } & =m^{2}-\sum_{k=1}^{n} k x_{k} \\
& =\sum_{k=1}^{m}(m+k)-\sum_{k=1}^{m} k-\sum_{k=1}^{2 m} k x_{k} \\
& =\sum_{k=1}^{m}(m+k)\left(1-x_{m+k}\right)-\sum_{k=1}^{m} k\left(1+x_{k}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,050 |
9. 123 Let $\alpha, \beta$ be real numbers, and $\cos \alpha \neq \cos \beta, k$ be a positive integer greater than 1. Prove:
$$\left|\frac{\cos k \beta \cos \alpha-\cos k \alpha \cos \beta}{\cos \beta-\cos \alpha}\right|<k^{2}-1$$ | [Proof] Let $x=\frac{1}{2}(\alpha-\beta), y=\frac{1}{2}(\alpha+\beta)$, then
$$\cos k \beta \cos \alpha-\cos k \alpha \cos \beta$$
$$\begin{aligned}
= & \frac{1}{2}[\cos (k \beta+\alpha)+\cos (k \beta-\alpha)-\cos (k \alpha+\beta)-\cos (k \alpha-\beta)] \\
= & \frac{1}{2}[\cos (k \beta+\alpha)-\cos (k \alpha+\beta)]+\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,051 |
9. $125 \alpha, \beta, \gamma$ are the three interior angles of a given triangle. Prove:
$$\csc ^{2} \frac{\alpha}{2}+\csc ^{2} \frac{\beta}{2}+\csc ^{2} \frac{\gamma}{2} \geqslant 12$$
and find the condition for equality. | [Proof] By the arithmetic-geometric mean inequality, we have
$$\csc ^{2} \frac{\alpha}{2}+\csc ^{2} \frac{\beta}{2}+\csc ^{2} \frac{\gamma}{2} \geqslant 3\left(\csc \frac{\alpha}{2} \cdot \csc \frac{\beta}{2} \cdot \csc \frac{\gamma}{2}\right)^{\frac{2}{3}}$$
Equality holds if and only if $\alpha=\beta=\gamma$.
By the... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,053 |
9・126 Let real numbers $x, y$ satisfy $|x|<1,|y|<1$. Prove: $\left|\frac{x-y}{1-x y}\right|<1$.
| [Proof] Since $|x|<1$ and $|y|<1$,
it follows that $x^{2}+y^{2}<1+x^{2} y^{2}$.
Thus, $(x-y)^{2}<(1-x y)^{2}$.
Therefore, $|x-y|<|1-x y|$,
which means $\left|\frac{x-y}{1-x y}\right|<1$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,054 |
9. 127 Prove: For any $x>\sqrt{2}$ and $y>\sqrt{2}$, we have $x^{4}-x^{3} y+x^{2} y^{2}-x y^{3}+y^{4}>x^{2}+y^{2}$. | [Proof] Since $x>\sqrt{2}, y>\sqrt{2}$, then
$$\begin{array}{l}
\frac{\left(x^{2}+y^{2}\right)^{2}}{4} \geqslant x^{2} y^{2} \\
\frac{\left(x^{2}+y^{2}\right)^{2}}{2} \geqslant\left(x^{2}+y^{2}\right) x y \\
\frac{\left(x^{2}+y^{2}\right)^{2}}{4}>\left(x^{2}+y^{2}\right)
\end{array}$$
Thus, $\left(x^{2}+y^{2}\right)^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,055 |
9-128 For any real number $x$, we have $\cos (a \sin x)>\sin (b \cos x)$. Prove: $a^{2}+b^{2}<\frac{\pi^{2}}{4}$. | [Proof] By contradiction, assume $a^{2}+b^{2} \geqslant \frac{\pi^{2}}{4}$. Since
$$a \sin x+b \cos x=\sqrt{a^{2}+b^{2}} \sin (x+\varphi)$$
where $\varphi$ is a fixed real number depending only on $a, b$, such that
$$\cos \varphi=\frac{a}{\sqrt{a^{2}+b^{2}}}, \sin \varphi=\frac{b}{\sqrt{a^{2}+b^{2}}} .$$
Since $\sqrt... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,056 |
9. 130 Let $a>1, b>1$, prove:
$$\frac{a^{2}}{b-1}+\frac{b^{2}}{a-1} \geqslant 8$$ | [Proof] Let $x=a-1, y=b-1$, then $x>0, y>0$, and
Since
$$\begin{aligned}
\frac{a^{2}}{b-1}+\frac{b^{2}}{a-1} & =\frac{(x+1)^{2}}{y}+\frac{(y+1)^{2}}{x} \\
& =\frac{x^{2}}{y}+\frac{2 x}{y}+\frac{1}{y}+\frac{y^{2}}{x}+\frac{2 y}{x}+\frac{1}{x}
\end{aligned}$$
Therefore,
$$\begin{array}{l}
\frac{x^{2}}{y}+\frac{y^{2}}{x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,058 |
9.131 Let $a<b$, prove:
$$a^{3}-3 a \leqslant b^{3}-3 b+4,$$
and determine when the equality holds. | ［Proof］ Let $c=b-a>0$, then
$$\begin{aligned}
b^{3}-3 b+4-a^{3}+3 a & =3 c a^{2}+3 c^{2} a+c^{3}-3 c+4 \\
& =3 c\left(a+\frac{c}{2}\right)^{2}+\frac{1}{4} c^{3}-3 c+4
\end{aligned}$$
Furthermore,
$$\begin{aligned}
\frac{1}{4} c^{3}-3 c+4 & =\frac{1}{4}\left(c^{3}-12 c+16\right)=\frac{1}{4}\left(c^{3}-16 c+4 c+16\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,059 |
9. 132 Prove or disprove the proposition: If $x, y$ are real numbers and
$$\begin{array}{l}
y \geqslant 0, y(y+1) \leqslant(x+1)^{2} \\
y(y-1) \leqslant x^{2}
\end{array}$$ | [Solution] We prove the proposition is true. By contradiction. Suppose
$$y(y-1)>x^{2}$$
Then, from $y \geqslant 0$, we can deduce $y>1$, and further obtain
$$y>\frac{1}{2}+\sqrt{\frac{1}{4}+x^{2}} \text {. }$$
From the assumption $y(y+1) \leqslant(x+1)^{2}$ and $y>1$, we know
$$y \leqslant-\frac{1}{2}+\sqrt{\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,060 |
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