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 |
|---|---|---|---|---|---|---|---|---|---|
3 Given natural numbers $a, b, n > 1, A_{n-1}$ and $A_{n}$ are $a$-ary numbers, $B_{n-1}$ and $B_{n}$ are $b$-ary numbers, $A_{n-1}, A_{n}, B_{n-1}, B_{n}$ are defined as:
$A_{n}=x_{n} x_{n-1} \cdots x_{0}, A_{n-1}=x_{n-1} x_{n-2} \cdots x_{0}$ (written in $a$-ary)
$B_{n}=x_{n} x_{n-1} \cdots x_{0}, B_{n-1}=x_{n-1} x_{... | 3. Since $A_{n}>0, B_{n}>0$, it suffices to prove $A_{n} B_{n-1}-A_{n-1} B_{n}>0$, and
$$\begin{aligned}
& A_{n} B_{n-1}-A_{n-1} B_{n} \\
= & \left(x_{n} \cdot a^{n}+A_{n-1}\right) B_{n-1}-A_{n-1}\left(x_{n} b^{n}+B_{n-1}\right) \\
= & x_{n}\left(a^{n} B_{n-1}-b^{n} A_{n-1}\right) \\
= & x_{n}\left[x_{n-1}\left(a^{n} b... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 738,405 |
4 Let $a, b, c \in \mathbf{R}^{+}$, prove:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant \frac{a^{8}+b^{8}+c^{8}}{a^{3} b^{3} c^{3}} \text {. }$$ | 4. Repeatedly using $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a(a, b, c \in \mathbf{R})$, we have
$$\begin{aligned}
a^{8}+b^{8}+c^{8} & \geqslant a^{4} b^{4}+b^{4} c^{4}+c^{4} a^{4} \geqslant a^{2} b^{4} c^{2}+b^{2} c^{4} a^{2}+c^{2} a^{4} b^{2} \\
& =a^{2} b^{2} c^{2}\left(a^{2}+b^{2}+c^{2}\right) \geqslant a^{2} b^{2} c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,406 |
5 Let real numbers $a_{1}, a_{2}, \cdots, a_{100}$ satisfy:
(1) $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{100} \geqslant 0$;
(2) $a_{1}+a_{2} \leqslant 100$;
(3) $a_{3}+a_{4}+\cdots+a_{100} \leqslant 100$.
Find the maximum value of $a_{1}^{2}+a_{2}^{2}+\cdots+a_{100}^{2}$. | 5. $a_{1}^{2}+a_{2}^{2}+\cdots+a_{100}^{2} \leqslant\left(100-a_{2}\right)^{2}+a_{2}^{2}+a_{3}^{2}+\cdots+a_{100}^{2}$
$$\begin{array}{l}
\leqslant 100^{2}-\left(a_{1}+a_{2}+\cdots+a_{100}\right) a_{2}+2 a_{2}^{2}+a_{3}^{2}+\cdots+a_{100}^{2} \\
=100^{2}-a_{2}\left(a_{1}-a_{2}\right)-a_{3}\left(a_{2}-a_{3}\right)-\cdot... | 100^{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,407 |
Example 5 Let $a, b, c \in \mathbf{R}^{+}, a^{2}+b^{2}+c^{2}=1$, find
$$S=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}-\frac{2\left(a^{3}+b^{3}+c^{3}\right)}{a b c}$$
the minimum value. | When $a=b=c$, $S=3$. Conjecture: $S \geqslant 3$.
In fact,
$$\begin{aligned}
S-3 & =\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}-3-\frac{2\left(a^{3}+b^{3}+c^{3}\right)}{a b c} \\
& =\frac{a^{2}+b^{2}+c^{2}}{a^{2}}+\frac{a^{2}+b^{2}+c^{2}}{b^{2}}+\frac{a^{2}+b^{2}+c^{2}}{c^{2}}-3-2\left(\frac{a^{2}}{b c}+\frac{b^{2}... | 3 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,408 |
6 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 $R=\frac{1}{n} \cdot \sum_{i=1}^{n} r_{i}$, $S=\frac{1}{n} \cdot \sum_{i=1}^{n} s_{i}, T=\frac{1}{n} \cdot \sum_{i=1}^{n} t_{i}, U=\frac{1}{n} \cdot \sum_{i=1}^{n} u_{i}, V=\frac{1}{n} \cdot \sum_{i=1}^{n... | 6. First, prove a lemma:
Lemma: Let $x_{1}, x_{2}, \cdots, x_{n}$ be $n$ real numbers greater than 1, and let $A=\sqrt[n]{x_{1} x_{2} \cdots x_{n}}$. Then, $\prod_{i=1}^{n} \frac{\left(x_{i}+1\right)}{\left(x_{i}-1\right)} \geqslant\left(\frac{A+1}{A-1}\right)^{n}$.
Proof: Without loss of generality, assume $x_{1} \l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,409 |
7 Let $k, n$ be positive integers, $1 \leqslant k < n$; $x_{1}, x_{2}, \cdots, x_{k}$ are $k$ positive numbers, and it is known that their sum equals their product. Prove that: $x_{1}^{n-1}+x_{2}^{n-1}+\cdots+x_{k}^{n-1} \geqslant k n$. | 7. Let $T=x_{1} x_{2} \cdots x_{k}=x_{1}+x_{2}+\cdots+x_{k}$. By the AM-GM inequality $\frac{T}{k} \geqslant T^{\frac{1}{k}}, x_{1}^{n-1}+$ $x_{2}^{n-1}+\cdots+x_{k}^{n-1} \geqslant k \cdot T^{n-1}$. Therefore, it suffices to prove: $T^{n-1} \geqslant n$. And $\frac{T}{k} \geqslant T^{\frac{1}{k}}$ is equivalent to $T^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,410 |
8 If $a, b, c \in \mathbf{R}$, prove:
$$- \left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right) \geqslant(a b+b c+c a)^{3} .$$ | 8. If we can prove: $\frac{27}{64}(a+b)^{2}(b+c)^{2}(c+a)^{2} \geqslant(a b+b c+c a)^{2}$, then the conclusion holds. Let $S_{1}=a+b+c, S_{2}=a b+b c+c a, S_{3}=a b c$. The problem is transformed into proving: $27\left(S_{1} S_{2}-S_{3}\right)^{2} \geqslant 64 S_{2}^{3}$.
We discuss two cases:
(1) If $a, b, c$ are all... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,411 |
9 Prove: For any $c>0$, there exist a positive integer $n$ and a sequence of complex numbers $a_{1}, a_{2}, \cdots, a_{n}$, such that
$$c \cdot \frac{1}{2^{n}} \sum_{\varepsilon_{1}, \varepsilon_{2} \cdots \cdots, \varepsilon_{n}}\left|\varepsilon_{1} a_{1}+\varepsilon_{2} a_{2}+\cdots+\varepsilon_{n} a_{n}\right|<\lef... | 9. Consider $S=$
$$\begin{array}{l}
\sum_{\varepsilon_{1}, \varepsilon_{2}, \cdots, \varepsilon_{n}}\left|\left(a_{1} \varepsilon_{1}+a_{2} \varepsilon_{2}+\cdots+a_{n} \varepsilon_{n}\right)\right|^{2} \\
\quad=\sum_{\varepsilon_{1}, \varepsilon_{2}, \cdots, \varepsilon_{n}}\left(a_{1} \varepsilon_{1}+a_{2} \varepsilo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,412 |
10. Let $a, b$ be positive constants, $\theta \in\left(0, \frac{\pi}{2}\right)$, find the maximum value of $y=a \sqrt{\sin \theta}+b \sqrt{\cos \theta}$. | 10. Let $\sqrt{u}=a \cdot \sqrt{\sin \theta}, \sqrt{v}=b \cdot \sqrt{\cos \theta}$, then the condition transforms to: $\frac{u^{2}}{a^{4}}+\frac{v^{2}}{b^{4}}=1, u$ 、 $v \geqslant 0$, using the Cauchy inequality,
$$\begin{aligned}
y & =\sqrt{u}+\sqrt{v} \leqslant \sqrt{\left(\frac{u}{a^{\frac{4}{3}}}+\frac{v}{b^{\frac{... | \left(a^{\frac{4}{3}}+b^{\frac{4}{3}}\right)^{\frac{1}{4}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,413 |
11 Let $n$ real numbers, their absolute values are all less than or equal to 2, and their sum of cubes is 0. Prove: their sum $\leqslant \frac{2}{3} n$. | 11. Let these $n$ real numbers be $y_{1}, y_{2}, \cdots, y_{n}$. Let $x_{i}=\frac{y_{i}}{2}$, then $\left|x_{i}\right| \leqslant 1, \sum_{i=1}^{n} x_{i}^{3}=0$. To prove: $\sum_{i=1}^{n} x_{i} \leqslant \frac{1}{3} n$.
Using the method of undetermined coefficients: Assume $x_{i} \leqslant \frac{1}{3}+\lambda x_{i}^{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,414 |
12 Given a positive integer $n \geqslant 3$, and real numbers $x_{1}, x_{2}, \cdots, x_{n}$ in the interval $[-1,1]$ that satisfy: $\sum_{k=1}^{n} x_{k}^{5}=0$. Prove that: $\sum_{k=1}^{n} x_{k} \leqslant \frac{8}{15} n$. | 12. Since $x_{k} \in[-1,1]$, then $0 \leqslant\left(1+x_{k}\right)\left(B x_{k}-1\right)^{4}$, where $B$ is a real number to be determined. Expanding, we get: $B^{4} x_{k}^{5}+\left(B^{4}-4 B^{3}\right) x_{k}^{4}+B^{2}(6-4 B) x_{k}^{3}+B(4 B-6) x_{k}^{2}+(1-$ $4 B) x_{k}+1 \geqslant 0$
Let $B=\frac{3}{2}$, from the ab... | \sum_{k=1}^{n} x_{k} \leqslant \frac{8}{15} n | Inequalities | proof | Yes | Yes | inequalities | false | 738,415 |
13 Given $n$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$ with arithmetic mean $a$. Prove:
$$\sum_{k=1}^{n}\left(x_{k}-a\right)^{2} \leqslant \frac{1}{2}\left(\sum_{k=1}^{n}\left|x_{k}-a\right|\right)^{2} .$$ | 13. First, prove the case when $a=0$. At this time, $\sum_{k=1}^{n} x_{k}^{2}=-2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \leqslant 2 \sum_{i \neq j}^{n}\left|x_{i} x_{j}\right|$, thus $2 \sum_{k=1}^{n} x_{k}^{2} \leqslant \sum_{k=1}^{n} x_{k}^{2}+2 \sum_{i \neq j}^{n}\left|x_{i} x_{j}\right|=\left(\sum_{k=1}^{n}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,416 |
14 Let $x, y, z \geqslant 0$, prove:
$$x(y+z-x)^{2}+y(z+x-y)^{2}+z(x+y-z)^{2} \geqslant 3 x y z .$$
Determine the conditions under which equality holds. | 14. Without loss of generality, let $x+y+z=1$, then the original inequality is equivalent to
$$4 x^{3}+4 y^{3}+4 z^{3}-4\left(x^{2}+y^{2}+z^{2}\right)+1 \geqslant 3 x y z$$
Furthermore, since $x^{3}+y^{3}+z^{3}=1+3 x y z-3(x y+y z+z x), x^{2}+y^{2}+z^{2}=1-$ $2(x y+y z+z x)$, the original inequality is equivalent to $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,417 |
Example 6 Let $n$ be a positive integer, and $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers. Prove that:
$$\begin{array}{l}
\quad \frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+\cdots+\frac{1}{a_{n}^{2}}+\frac{1}{\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}} \geqslant \frac{n^{3}+1}{\left(n^{2}+2011\right)^{2}}\left(\fra... | Prove that by Cauchy-Schwarz inequality,
$$\begin{array}{c}
\left(1+1+\cdots+1+\frac{1}{n^{2}}\right)\left(\frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+\cdots+\frac{1}{a_{n}^{2}}+\frac{1}{\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}}\right) \\
\geqslant\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}+\frac{1}{n\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,419 |
16. Let $x, y, z \in \mathbf{R}^{+}$, prove:
$$(x y+y z+z x)\left[\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}\right] \geqslant \frac{9}{4} .$$ | 16. It is evident that this problem is equivalent to proving: $\sum_{\mathrm{oc}} \frac{y z}{x(y+z)^{2}} \geqslant \frac{9}{4(x+y+z)}$.
Without loss of generality, assume $x \geqslant y \geqslant z$, and $x+y+z=1$. Then
$$\begin{aligned}
\sum_{o c} \frac{4 y z}{x(y+z)^{2}} & =\sum_{o c} \frac{(y+z)^{2}-(y-z)^{2}}{x(y+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,420 |
17 Prove: In an acute $\triangle A B C$, we have
$$\begin{array}{c}
\cot ^{3} A+\cot ^{3} B+\cot ^{3} C+6 \cot A \cot B \cot C \\
\geqslant \cot A+\cot B+\cot C
\end{array}$$ | 17. Let $x=\cot A, y=\cot B, z=\cot C$, from $A+B+C=\pi$ we know, $\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$, i.e., $x y + y z + z x = 1$. Therefore, the inequality to be proven is equivalent to
$$x^{3} + y^{3} + z^{3} + 6 x y z \geqslant (x + y + z)(x y + y z + z x),$$
which is $x(x-y)(x-z) + y(y-z)(y-x) + z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,421 |
18. Let \(a, b, c \in \left(0, \frac{\pi}{2}\right)\), prove:
$$\begin{array}{c}
\frac{\sin a \sin (a-b) \sin (a-c)}{\sin (b+c)}+\frac{\sin b \sin (b-c) \sin (b-a)}{\sin (c+a)} \\
+\frac{\sin c \sin (c-a) \sin (c-b)}{\sin (a+b)} \geqslant 0
\end{array}$$ | 18. Since $\sin (x-y) \sin (x+y)=\frac{1}{2}(\cos 2 \beta-\cos 2 \alpha)=\sin ^{2} \alpha-\sin ^{2} \beta$, we have
$$\begin{aligned}
& \sin a \sin (a-b) \sin (a-c) \sin (a+b) \sin (a+c) \\
= & \sin a\left(\sin ^{2} a-\sin ^{2} b\right)\left(\sin ^{2} a-\sin ^{2} c\right)
\end{aligned}$$
Let $x=\sin ^{2} a, y=\sin ^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,422 |
19 Let positive real numbers $a, b, c$ satisfy: $a^{2}+b^{2}+c^{2}+(a+b+c)^{2} \leqslant 4$, prove: $\frac{a b+1}{(a+b)^{2}}+\frac{b c+1}{(b+c)^{2}}+\frac{c a+1}{(c+a)^{2}} \geqslant 3$ | 19. From the given: $a^{2}+b^{2}+c^{2}+ab+bc+ac \leqslant 2$,
Therefore,
$$\begin{aligned}
\sum_{oc} \frac{2ab+2}{(a+b)^{2}} & \geqslant \sum_{oc} \frac{2ab+a^{2}+b^{2}+c^{2}+ab+bc+ac}{(a+b)^{2}} \\
& =\sum_{or} \frac{(a+b)^{2}+(c+a)(c+b)}{(a+b)^{2}} \\
& =3+\sum_{or} \frac{(c+a)(c+b)}{(a+b)^{2}} \geqslant 6
\end{alig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,423 |
20 Let positive real numbers $a, b, c$ satisfy: $abc=1$, prove that for integers $k \geqslant 2$, we have $\frac{a^{k}}{a+b}+\frac{b^{k}}{b+c}+\frac{c^{k}}{c+a} \geqslant \frac{3}{2}$. (2007 China Southeast Mathematical Olympiad) | 20. Since $\frac{a^{k}}{a+b}+\frac{1}{4}(a+b)+\underbrace{\frac{1}{2}+\frac{1}{2}+\cdots+\frac{1}{2}}_{k-2 \uparrow \frac{1}{2}} \geqslant k \cdot \sqrt[k]{\frac{a^{k}}{2^{k}}}=\frac{k}{2} a$, therefore $\quad \frac{a^{k}}{a+b} \geqslant \frac{k}{2} a-\frac{1}{4}(a+b)-\frac{k-2}{2}$.
Similarly, we get $\frac{b^{k}}{b+... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 738,424 |
Example 1 Prove the Lagrange identity:
$$\left(\sum_{i=1}^{n} a_{i}^{2}\right) \cdot\left(\sum_{i=1}^{n} b_{i}^{2}\right)=\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2}+\sum_{1 \leqslant i<j \leqslant n}\left(a_{i} b_{j}-a_{j} b_{i}\right)^{2},$$
and use this identity to show that the Cauchy inequality holds. | $$\begin{aligned}
& \left(\sum_{i=1}^{n} a_{i}^{2}\right) \cdot\left(\sum_{i=1}^{n} b_{i}^{2}\right)-\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2} \\
= & \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i}^{2} b_{j}^{2}-\sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} b_{i} a_{j} b_{j} \\
= & \frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n}\left(a_{i}^{2} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,425 |
Example 2 If $p>s \geqslant r>q, p+q=r+s, a_{1}, a_{2}, \cdots, a_{n}>0$, then
$$\left(\sum_{i=1}^{n} a_{i}^{p}\right) \cdot\left(\sum_{i=1}^{n} a_{i}^{q}\right) \geqslant\left(\sum_{i=1}^{n} a_{i}^{s}\right) \cdot\left(\sum_{i=1}^{n} a_{i}^{r}\right) .$$ | $$\begin{array}{l}
\quad\left(\sum_{i=1}^{n} a_{i}^{p}\right) \cdot\left(\sum_{i=1}^{n} a_{i}^{q}\right)-\left(\sum_{i=1}^{n} a_{i}^{s}\right) \cdot\left(\sum_{i=1}^{n} a_{i}^{r}\right) \\
=\sum_{i=1}^{n} \sum_{j=1}^{n}\left(a_{i}^{p} a_{j}^{q}-a_{i}^{s} a_{j}^{r}\right) \\
\quad=\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,426 |
Example 4 The set of real numbers $\left\{a_{0}, a_{1}, \cdots, a_{n}\right\}$ satisfies the following conditions:
(1) $a_{0}=a_{n}=0$;
(2) For $1 \leqslant k \leqslant n-1, a_{k}=c+\sum_{i=k}^{n-1} a_{i-k}\left(a_{i}+a_{i+1}\right)$.
Prove:
$$c \leqslant \frac{1}{4 n}$$ | Prove that for $\quad s_{k}=\sum_{i=0}^{k} a_{i}, k=1,2, \cdots, n$,
then
$$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)$$
Additional definitions:
$$\begin{array}{l}
a_{-1}=a_{-2}=\cdots=a_{-(n-1)}=0 \\
s_{n}=n c+\sum_{k=0}^{n-1} \sum_{i=0}^... | c \leqslant \frac{1}{4 n} | Algebra | proof | Yes | Yes | inequalities | false | 738,428 |
Example 5 Let $a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}, n \in \mathbf{N}_{+}$. Prove that for $n \geqslant 2$, we have
$$a_{n}^{2}>2\left(\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n}\right)$$ | Proof:
$$\begin{aligned}
a_{n}^{2}-a_{n-1}^{2} & =\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)^{2}-\left(1+\frac{1}{2}+\cdots+\frac{1}{n-1}\right)^{2} \\
& =\frac{1}{n^{2}}+2 \cdot \frac{1}{n}\left(1+\frac{1}{2}+\cdots+\frac{1}{n-1}\right) \\
& =\frac{1}{n^{2}}+\frac{2}{n}\left(a_{n}-\frac{1}{n}\right) \\
& =2 \cdot \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,429 |
Example 7: Prove that for any positive real numbers $a, b, c$, we have
$$\frac{1}{a^{3}+b^{3}+a b c}+\frac{1}{b^{3}+c^{3}+a b c}+\frac{1}{c^{3}+a^{3}+a b c} \leqslant \frac{1}{a b c} .$$ | Prove that since
$$a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right) \geqslant(a+b) a b,$$
we have $\frac{1}{a^{3}+b^{3}+a b c} \leqslant \frac{1}{a b(a+b)+a b c}=\frac{c}{a b c(a+b+c)}$,
Similarly, we get
$$\begin{array}{l}
\frac{1}{b^{3}+c^{3}+a b c} \leqslant \frac{a}{a b c(a+b+c)} \\
\frac{1}{c^{3}+a^{3}+a b c} \leqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,430 |
Example 7 (Chung's Inequality) Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}>0$, and $\sum_{i=1}^{k} a_{i} \leqslant \sum_{i=1}^{k} b_{i}(1 \leqslant$ $k \leqslant n)$, then
(1) $\sum_{i=1}^{n} a_{i}^{2} \leqslant \sum_{i=1}^{n} b_{i}^{2}$;
(2) $\sum_{i=1}^{n} a_{i}^{3} \leqslant \sum_{i=1}^{n} a_{i} b_{i... | Prove (1) by Abel transformation formula,
$$\begin{aligned}
\sum_{i=1}^{n} a_{i}^{2} & =a_{n}\left(\sum_{i=1}^{n} a_{i}\right)+\sum_{k=1}^{n-1}\left(\sum_{i=1}^{k} a_{i}\right)\left(a_{k}-a_{k+1}\right) \\
& \leqslant a_{n}\left(\sum_{i=1}^{n} b_{i}\right)+\sum_{k=1}^{n-1}\left(\sum_{i=1}^{k} b_{i}\right)\left(a_{k}-a_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,432 |
Example 8 Let $a_{1}, a_{2}, \cdots$ be a sequence of positive real numbers, and for all $i, j=1,2, \cdots$ satisfy $a_{i+j} \leqslant a_{i}+a_{j}$. Prove that for positive integers $n$, we have
$$a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n} \geqslant a_{n}$$ | Prove using Abel's transformation method.
Let $s_{i}=a_{1}+a_{2}+\cdots+a_{i}, i=1,2, \cdots, n$.
Agree that $s_{0}=0$, then
$$2 s_{i}=\left(a_{1}+a_{i}\right)+\cdots+\left(a_{i}+a_{1}\right) \geqslant i a_{i+1} .$$
That is
$$s_{i} \geqslant \frac{i}{2} \cdot a_{i+1}$$
Therefore
$$\begin{aligned}
\sum_{i=1}^{n} \frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,433 |
Example 9 (Rearrangement Inequality) Let there be two ordered arrays: $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$ and $b_{1} \leqslant$ $b_{2} \leqslant \cdots \leqslant b_{n}$. Prove:
$$\begin{aligned}
& a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}(\text { ordered sum }) \\
\geqslant & a_{1} b_{j_{1}}+a_{2}... | Prove that
$$\begin{array}{c}
s_{i}=b_{1}+b_{2}+\cdots+b_{i} \\
s_{i}^{\prime}=b_{j_{1}}+b_{j_{2}}+\cdots+b_{j_{i}}(i=1,2, \cdots, n) \\
s_{i} \leqslant s_{i}^{\prime}(i=1,2, \cdots, n-1) \\
s_{n}=s_{n}^{\prime}
\end{array}$$
From the given conditions, it is easy to see
Also, since $a_{i}-a_{i+1} \leqslant 0$, it fol... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,434 |
Example 10 Arrange the numbers $1,2,3, \cdots, 2007$ in any order to get 2007! different sequences. Does there exist 4 sequences:
$$a_{1}, a_{2}, \cdots, a_{2007} ; b_{1}, b_{2}, \cdots, b_{2007} ; c_{1}, c_{2}, \cdots, c_{2007} ; d_{1}, d_{2}, \cdots, d_{2007},$$
such that $a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{2007} b_{... | $$\begin{aligned}
a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{2007} b_{2007} & \leqslant 1 \cdot 1+2 \cdot 2+\cdots+2007 \cdot 2007 \\
& =\frac{2007 \cdot 2008 \cdot 4015}{6}=2696779140,
\end{aligned}$$
$$\begin{aligned}
c_{1} d_{1}+c_{2} d_{2}+\cdots+c_{2007} d_{2007} & \geqslant 1 \cdot 2007+2 \cdot 2006+\cdots+2007 \cdot 1 \\... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 738,435 |
Example 11 Prove: For each positive integer $n$, we have
$$\frac{2 n+1}{3} \sqrt{n} \leqslant \sum_{i=1}^{n} \sqrt{i} \leqslant \frac{4 n+3}{6} \sqrt{n}-\frac{1}{6} \text {. }$$
The equality holds on both sides if and only if $n=1$. | It is easy to verify that when $n=1$, both inequalities hold as equalities. Below, we assume without loss of generality that $n \geqslant 2$. First, we prove the left inequality. Let $a_{i}=1, b_{i}=\sqrt{i}(1 \leqslant i \leqslant n)$, then
$$s_{i}=a_{1}+a_{2}+\cdots+a_{i}=i$$
Using the Abel summation formula, we get... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,436 |
Example 12 Let $\left\{a_{n}\right\}$ be an infinite sequence of positive numbers. If there exists a constant $C$ such that $\sum_{k=1}^{n} \frac{1}{a_{k}} \leqslant C$ for all positive integers $n$.
Prove: There exists a constant $M$ such that $\sum_{k=1}^{n} \frac{k^{2} \cdot a_{k}}{\left(a_{1}+a_{2}+\cdots+a_{k}\rig... | Let \( S_{n}=\sum_{k=1}^{n} \frac{k^{2} \cdot a_{k}}{\left(a_{1}+a_{2}+\cdots+a_{k}\right)^{2}}, A_{n}=a_{1}+a_{2}+\cdots+a_{n}(n \geqslant 1) \), \( A_{0}=0 \). Therefore,
\[
\begin{array}{c}
S_{n}=\sum_{k=1}^{n} \frac{k^{2} \cdot\left(A_{k}-A_{k-1}\right)}{A_{k}^{2}} \leqslant \frac{1}{a_{1}}+\sum_{k=2}^{n} \frac{k^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,437 |
Example 13 Let $n$ be a positive integer, and real numbers $a_{1}, a_{2}, \cdots, a_{n}$ and $r_{1}, r_{2}, \cdots, r_{n}$ satisfy: $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$ and $0 \leqslant r_{1} \leqslant r_{2} \leqslant \cdots \leqslant r_{n}$, prove that:
$$\sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} a_{j} \... | Prove the construction of an $n \times n$ table:
$$A_{1}=\left(\begin{array}{ccc}
a_{1} a_{1} r_{1} & a_{1} a_{2} r_{1} & a_{1} a_{3} r_{1} \cdots a_{1} a_{n} r_{1} \\
a_{2} a_{1} r_{1} & a_{2} a_{2} r_{2} & a_{2} a_{3} r_{2} \cdots a_{2} a_{n} r_{2} \\
a_{3} a_{1} r_{1} & a_{3} a_{2} r_{2} & a_{3} a_{3} r_{3} \cdots a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,438 |
Let $a, b, p, q>0$. Prove:
$$\frac{a^{p+q}+b^{p+q}}{2} \geqslant\left(\frac{a^{p}+b^{p}}{2}\right)\left(\frac{a^{q}+b^{q}}{2}\right)$$ | 1. The original inequality is equivalent to $\left(a^{p}-b^{p}\right)\left(a^{q}-b^{q}\right) \geqslant 0$. Since $a^{p}-b^{p}$ and $a^{q}-b^{q}$ have the same sign or are both zero, the conclusion holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,439 |
2 Let $0<a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} ; 0<b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{n}$, then
$$\frac{\sum_{i=1}^{n} a_{i}^{2} b_{i}}{\sum_{i=1}^{n} a_{i} b_{i}} \geqslant \frac{\sum_{i=1}^{n} a_{i}^{2}}{\sum_{i=1}^{n} a_{i}}$$ | 2. The original inequality is equivalent to $\left(\sum a_{i}^{2} b_{i}\right)\left(\sum a_{i}\right)-\left(\sum a_{i}^{2}\right)\left(\sum a_{i} b_{i}\right) \geqslant 0$, which is $\frac{1}{2} \sum \sum\left(a_{i}-a_{j}\right)\left(b_{i}-b_{j}\right) a_{i} a_{j} \geqslant 0$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,440 |
Example 8 Let $a_{i} \geqslant 1(i=1,2, \cdots, n)$, prove:
$$\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right) \geqslant \frac{2^{n}}{n+1}\left(1+a_{1}+a_{2}+\cdots+a_{n}\right)$$ | Prove
$$\begin{aligned}
& \left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right) \\
= & 2^{n}\left(1+\frac{a_{1}-1}{2}\right)\left(1+\frac{a_{2}-1}{2}\right) \cdots\left(1+\frac{a_{n}-1}{2}\right) .
\end{aligned}$$
Since $a_{i}-1 \geqslant 0$, we have:
$$\begin{aligned}
& \left(1+a_{1}\right)\left(1+a_{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,441 |
3 Let $n$ be a given positive integer, $n \geqslant 3$, and for $n$ given real numbers $a_{1}, a_{2}, \cdots, a_{n}$, let $m$ be the minimum value of $\left|a_{i}-a_{j}\right|(1 \leqslant i<j \leqslant n)$. Find the maximum value of $m$ under the condition $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$. | 3. Let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, then
$$\begin{aligned}
a_{j}-a_{i} & =\left(a_{j}-a_{j-1}\right)+\left(a_{j-1}-a_{j-2}\right)+\cdots+\left(a_{i+1}-a_{i}\right) \\
& \geqslant(j-i) m
\end{aligned}$$
Therefore, $\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2} \geqslant \fr... | \sqrt{\frac{12}{(n-1) n(n+1)}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,442 |
Find all positive integers $a_{1}, a_{2}, \cdots, a_{n}$, such that
$$\frac{99}{100}=\frac{a_{0}}{a_{1}}+\frac{a_{1}}{a_{2}}+\cdots+\frac{a_{n-1}}{a_{n}}$$
where $a_{0}=1$, and $\left(a_{k+1}-1\right) a_{k-1} \geqslant a_{k}^{2}-1(k=1,2, \cdots, n-1)$. | 4. Clearly $a_{k}>a_{k-1}$, and $a_{k} \geqslant 2 . k=1,2, \cdots, n-1$.
From the condition, we have
$$\frac{a_{k-1}}{a_{k}} \leqslant \frac{a_{k-1}}{a_{k}-1}-\frac{a_{k}}{a_{k+1}-1}$$
Summing the above inequality for $k=i+1, i+2, \cdots, n$, we get
$$\frac{a_{i}}{a_{i+1}}+\frac{a_{i+1}}{a_{i+2}}+\cdots+\frac{a_{n-1... | a_{1}=2, a_{2}=5, a_{3}=56, a_{4}=78400 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 738,443 |
55 Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ positive numbers $(n \geqslant 2)$ that are not all equal, and satisfy $\sum_{k=1}^{n} a_{k}^{-2 n}=1$. Prove:
$$\sum_{k=1}^{n} a_{k}^{2 n}-n^{2} \sum_{1 \leq i<j \leq n} \frac{1}{a_{i}^{2 n}+a_{j}^{2 n}}>n^{2 n}-n^{2}$$ | 5. The original inequality is equivalent to $\sum_{k=1}^{n} a_{k}^{2 n} \cdot \sum_{k=1}^{n} \frac{1}{a_{k}^{2 n}} - n^{2} \sum_{i < j} \left( \frac{a_i}{a_j} - \frac{a_j}{a_i} \right)^2$.
By the Lagrange identity, $\sum_{k=1}^{n} a_{k}^{2 n} \cdot \sum_{k=1}^{n} \frac{1}{a_{k}^{2 n}} - n^{2} = \sum_{i < j} \left( \fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,444 |
6 Given $n(\geqslant 2)$ real numbers $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, let $x=\frac{1}{n} \sum_{i=1}^{n} a_{i}, y=\frac{1}{n} \sum_{i=1}^{n} a_{i}^{2}$, prove: $2 \sqrt{y-x^{2}} \leqslant a_{n}-a_{1} \leqslant \sqrt{2 n\left(y-x^{2}\right)}$. | 6.
$$\begin{aligned}
n^{2}\left(y-x^{2}\right)= & n \sum_{j=1}^{n} a_{j}^{2}-\left(\sum_{j=1}^{n} a_{j}\right)^{2} \\
= & (n-1) \sum_{j=1}^{n} a_{j}^{2}-2 \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j} \\
= & (n-1)\left(a_{1}^{2}+a_{n}^{2}\right)-2 a_{1} a_{n}-2 a_{1}\left(a_{2}+a_{3}+\cdots+a_{n-1}\right) \\
& -2 a_{n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,445 |
Consider the inequality
$$\begin{aligned}
& \left|z_{1}^{\prime}-z_{2}^{\prime}\right|+\left|z_{2}^{\prime}-z_{3}^{\prime}\right|+\cdots+\left|z_{n-1}^{\prime}-z_{n}^{\prime}\right| \\
\leqslant & \lambda \cdot\left[\left|z_{1}-z_{2}\right|+\left|z_{2}-z_{3}\right|+\cdots+\left|z_{n-1}-z_{n}\right|\right]
\end{aligned}... | 7.
$$\begin{aligned}
\sum_{k=1}^{n-1}\left|z_{k}^{\prime}-z_{k+1}^{\prime}\right| & =\sum_{k=1}^{n-1}\left|\frac{1}{k} \sum_{j=1}^{k} z_{j}-\frac{1}{k+1} \sum_{j=1}^{k+1} z_{j}\right| \\
& =\sum_{k=1}^{n-1} \frac{1}{k(k+1)}\left|\sum_{j=1}^{k} j\left(z_{j}-z_{j+1}\right)\right| \\
& \leqslant \sum_{k=1}^{n-1}\left(\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,446 |
10 Prove Chebyshev's inequality:
Let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{n}$, then
$$n \cdot \sum_{k=1}^{n} a_{k} b_{k} \geqslant\left(\sum_{k=1}^{n} a_{k}\right) \cdot\left(\sum_{k=1}^{n} b_{k}\right) \geqslant n \sum_{k=1}^{n} a_{k} b_{n-k+1} .$... | 10. Define $b_{n+t}=b_{n}, t=0,1,2, \cdots$. Then $\sum_{i=1}^{n} b_{i+t}=\sum_{i=1}^{n} b_{i}, \sum_{i=1}^{k} b_{i+t} \geqslant \sum_{i=1}^{k} b_{i}$, $a_{k}-a_{k+1} \leqslant 0(1 \leqslant k \leqslant n-1)$.
By the summation by parts formula,
$$\begin{aligned}
\left(\sum_{k=1}^{n} a_{k}\right) \cdot\left(\sum_{k=1}^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,449 |
11 (Generalized form of W. Janous inequality)
Let $a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{R}^{+}, p, q \in \mathbf{R}^{+}$. Denote $S=\left(a_{1}^{p}+a_{2}^{p}+\cdots+a_{n}^{p}\right)^{\frac{1}{p}}$, then for any permutation $i_{1}, i_{2}, \cdots, i_{n}$ of $1,2, \cdots, n$, we have: $\sum_{k=1}^{n} \frac{a_{k}^{q}-a_... | 11. By symmetry, without loss of generality, let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$, then
$$\sum_{j=1}^{k} a_{j}^{q} \geqslant \sum_{j=1}^{k} a_{i_{j}}^{q}(1 \leqslant k \leqslant n-1), \sum_{j=1}^{n} a_{j}^{q}=\sum_{j=1}^{n} a_{i_{j}}^{q}$$
and $\left(S^{p}-a_{k}^{p}\right)^{-1} \geqslant\left(S... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,450 |
12 Let $a_{1}, a_{2}, \cdots$ be a sequence of positive real numbers, satisfying the condition for all $n \geqslant 1$: $\sum_{j=1}^{n} a_{j} \geqslant \sqrt{n}$. Prove that for all $n \geqslant 1$, $\sum_{j=1}^{n} a_{j}^{2} \geqslant \frac{1}{4}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)$. | 12. Let $b_{k}=a_{1}+a_{2}+\cdots+a_{k}-\sqrt{k} \geqslant 0,1 \leqslant k \leqslant n$, then $a_{k}=\left(b_{k}-b_{k-1}\right)+$ $(\sqrt{k}-\sqrt{k-1})$. Therefore,
$$\begin{array}{l}
\sum_{j=1}^{n} a_{j}^{2}= {\left[b_{1}^{2}+\left(b_{2}-b_{1}\right)^{2}+\cdots+\left(b_{n}-b_{n-1}\right)^{2}\right]+\left[1^{2}+(\sqr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,451 |
13 Prove: For any real number $x$, $\sum_{k=1}^{n} \frac{[k x]}{k} \leqslant[n x]$, where $[x]$ denotes the greatest integer less than or equal to $x$. | 13. Let $A_{n}=\sum_{k=1}^{n} \frac{[k x]}{k}$, we will prove by mathematical induction that: $A_{n} \leqslant[n x]$.
When $n=1$, it is obviously true. Assume for $1 \leqslant k \leqslant n-1$, we have $A_{k} \leqslant[k x]$, then by the summation by parts formula,
$$\begin{aligned}
n A_{n} & =\sum_{k=1}^{n} k \cdot \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,453 |
14 Given $a_{k} \geqslant 0, k=1,2, \cdots, n$. Define $A_{k}=\frac{1}{k} \cdot \sum_{i=1}^{k} a_{i}$, prove:
$$\sum_{k=1}^{n} A_{k}^{2} \leqslant 4 \sum_{k=1}^{n} a_{k}^{2}$$ | 14. If we set $\frac{1}{c} \sum_{k=1}^{n} A_{k}^{2} \leqslant \sum_{k=1}^{n} A_{k} \cdot a_{k}$, then we have $\sum_{k=1}^{n} A_{k} \cdot a_{k} \leqslant c \cdot \sum_{k=1}^{n} a_{k}^{2}$. Thus, the problem can be transformed into an Abel method:
$$\begin{aligned}
\sum_{k=1}^{n} A_{k} a_{k} & =\sum_{k=1}^{n} A_{k}\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,454 |
Example 1 Given $0^{\circ} \leqslant \alpha \leqslant 90^{\circ}$, prove:
$$2 \leqslant \sqrt{5-4 \sin \alpha}+\sin \alpha \leqslant \frac{9}{4} \text {. }$$ | Prove that if $x=\sqrt{5-4 \sin \alpha}$, then $\sin \alpha=\frac{5-x^{2}}{4}$. Since $0 \leqslant \sin \alpha \leqslant 1$, it follows that $1 \leqslant x \leqslant \sqrt{5}$. Let
$$\begin{aligned}
y & =\sqrt{5-4 \sin \alpha}+\sin \alpha \\
& =x+\frac{5-x^{2}}{4} \\
& =-\frac{1}{4}(x-2)^{2}+\frac{9}{4}
\end{aligned}$$... | 2 \leqslant \sqrt{5-4 \sin \alpha}+\sin \alpha \leqslant \frac{9}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 738,455 |
Example 2 Given real numbers $x, y$ satisfy $x^{2}+y^{2}-4 x-6 y+9=0$, prove:
$$19 \leqslant x^{2}+y^{2}+12 x+6 y \leqslant 99$$ | Prove that the given conditions can be transformed into
i.e.,
$$\begin{array}{l}
(x-2)^{2}+(y-3)^{2}=4 \\
\left(\frac{x-2}{2}\right)^{2}+\left(\frac{y-3}{2}\right)^{2}=1
\end{array}$$
Let $\frac{x-2}{2}=\cos \theta, \frac{y-3}{2}=\sin \theta$, where $\theta \in[0,2 \pi)$, then
$$\begin{aligned}
& x^{2}+y^{2}+12 x+6 y... | 19 \leqslant x^{2}+y^{2}+12 x+6 y \leqslant 99 | Algebra | proof | Yes | Yes | inequalities | false | 738,456 |
Example 3 Let $a$, $b$, $c$ be the lengths of the three sides of a triangle, prove:
$$a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \geqslant 0 .$$ | Prove that for $a=y+z, b=z+x, c=x+y, x, y, z \in \mathbf{R}^{+}$, the inequality to be proved is equivalent to
$$\begin{array}{cc}
& (y+z)^{2}(z+x)(y-x)+(z+x)^{2}(x+y) \\
& (z-y)+(x+y)^{2}(y+z)(x-z) \geqslant 0 \\
\Leftrightarrow & x y^{3}+y z^{3}+z x^{3} \geqslant x^{2} y z+x y^{2} z+x y z^{2} \\
\Leftrightarrow & \f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,457 |
Example 4 Let $a, b, c, d \in \mathbf{R}^{+}$, and
$$\frac{a^{2}}{1+a^{2}}+\frac{b^{2}}{1+b^{2}}+\frac{c^{2}}{1+c^{2}}+\frac{d^{2}}{1+d^{2}}=1$$
Prove: $a b c d \leqslant \frac{1}{9}$. | Let $a=\tan \alpha, b=\tan \beta, c=\tan \gamma, d=\tan \delta . \alpha, \beta, \gamma, \delta \in$ $\left(0, \frac{\pi}{2}\right)$. Then $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma+\sin ^{2} \delta=1$. Therefore,
$3 \cdot \sqrt[3]{\sin ^{2} \alpha \sin ^{2} \beta \sin ^{2} \gamma} \leqslant \sin ^{2} \alpha+\si... | a b c d \leqslant \frac{1}{9} | Inequalities | proof | Yes | Yes | inequalities | false | 738,458 |
Example 5 Let $a, b, c$ be positive real numbers, find the minimum value of
$$\frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c}$$ | Let
$$\left\{\begin{array}{l}
x=a+2 b+c \\
y=a+b+2 c \\
z=a+b+3 c
\end{array}\right.$$
Then we have $x-y=b-c, z-y=c$, which leads to
$$\left\{\begin{array}{l}
a+3 c=2 y-x \\
b=z+x-2 y \\
c=z-y
\end{array}\right.$$
Thus,
$$\begin{aligned}
& \frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c} \\
= & \frac{2 y... | -17+12 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,459 |
Example 8 Let $x, y, z \in \mathbf{R}^{+}$, and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$, prove:
$$\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geqslant \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z}$$ | Let $x=\frac{1}{\alpha}, y=\frac{1}{\beta}, z=\frac{1}{\gamma}$, then $\alpha+\beta+\gamma=1$.
The original inequality is equivalent to
i.e.,
i.e.,
i.e., $\square$
$$\begin{array}{c}
\sum_{o r} \sqrt{\frac{1}{\alpha}+\frac{1}{\beta \gamma}} \geqslant \sqrt{\frac{1}{\alpha \beta \gamma}}+\sum_{o r} \sqrt{\frac{1}{\al... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,462 |
Example 10 Let non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ satisfy the following conditions simultaneously:
(1) $\sum_{i=1}^{n}\left(a_{i}+b_{i}\right)=1$;
(2) $\sum_{i=1}^{n} i\left(a_{i}-b_{i}\right)=0$;
(3) $\sum_{i=1}^{n} i^{2}\left(a_{i}+b_{i}\right)=10$.
Prove that f... | Prove that for any $1 \leqslant k \leqslant n$, we have
$$\begin{aligned}
\left(k a_{k}\right)^{2} & \leqslant\left(\sum_{i=1}^{n} i a_{i}\right)^{2}=\left(\sum_{i=1}^{n} i b_{i}\right)^{2} \leqslant\left(\sum_{i=1}^{n} i^{2} b_{i}\right) \cdot\left(\sum_{i=1}^{n} b_{i}\right) \\
& =\left(10-\sum_{i=1}^{n} i^{2} a_{i}\... | \max \left\{a_{k}, b_{k}\right\} \leqslant \frac{10}{10+k^{2}} | Algebra | proof | Yes | Yes | inequalities | false | 738,463 |
Example 9 Let $x, y, z \in \mathbf{R}^{+}$, prove:
$\frac{y^{2}-x^{2}}{z+x}+\frac{z^{2}-y^{2}}{x+y}+\frac{x^{2}-z^{2}}{y+z} \geqslant 0$. (W. Janous Inequality) | Prove that let $x+y=c, y+z=a, z+x=b$, the original inequality is equivalent to
i.e. $\square$
$$\begin{array}{c}
\frac{c(a-b)}{b}+\frac{a(b-c)}{c}+\frac{b(c-a)}{a} \geqslant 0 \\
\frac{a c^{2}(a-b)+a^{2} b(b-c)+b^{2} c(c-a)}{a b c} \geqslant 0
\end{array}$$
Therefore, it is only necessary to prove
$$a^{2} c^{2}+a^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,464 |
Example 10 Let $x_{1}, x_{2}, x_{3}$ be positive numbers, prove that:
$$x_{1} x_{2} x_{3} \geqslant\left(x_{2}+x_{3}-x_{1}\right)\left(x_{1}+x_{3}-x_{2}\right)\left(x_{1}+x_{2}-x_{3}\right) .$$ | Assume without loss of generality that $x_{1} \geqslant x_{2} \geqslant x_{3}>0$.
Let $x_{1}=x_{3}+\delta_{1}, x_{2}=x_{3}+\delta_{2}$, then $\delta_{1} \geqslant \delta_{2} \geqslant 0$. Therefore,
$$\begin{aligned}
& x_{1} x_{2} x_{3}-\left(x_{2}+x_{3}-x_{1}\right)\left(x_{1}+x_{3}-x_{2}\right)\left(x_{1}+x_{2}-x_{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,465 |
Example 11 Find the largest positive integer $n$, such that there exist $n$ distinct real numbers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying: for any $1 \leqslant i<j \leqslant n$, we have
$$\left(1+x_{i} x_{j}\right)^{2} \leqslant 0.9\left(1+x_{i}^{2}\right)\left(1+x_{j}^{2}\right)$$ | Solve
$$\left(1+x_{i} x_{j}\right)^{2} \leqslant 0.9\left(1+x_{i}^{2}\right)\left(1+x_{j}^{2}\right)$$
This is equivalent to
$$0.1\left(x_{i} x_{j}+1\right)^{2} \leqslant 0.9\left(x_{i}-x_{j}\right)^{2}$$
Which is also
$$\left|x_{i} x_{j}+1\right| \leqslant 3\left|x_{i}-x_{j}\right|$$
Let $x_{i}=\tan \alpha_{i}(1 \l... | 9 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,466 |
Example 12 Given that $x, y, z$ are all positive numbers, prove:
$$\begin{aligned}
& x(y+z-x)^{2}+y(z+x-y)^{2}+z(x+y-z)^{2} \\
\geqslant & 2 x y z\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right),
\end{aligned}$$
equality holds if and only if $x=y=z$. | Prove that if $a=y+z-x, b=x+z-y, c=x+y-z$, then
$$x=\frac{b+c}{2}, y=\frac{a+c}{2}, z=\frac{a+b}{2}$$
Thus, the original inequality is equivalent to
$$\frac{1}{2} \sum a^{2}(b+c) \geqslant 2 \frac{(a+b)(b+c)(c+a)}{8} \cdot \sum \frac{\frac{b+c}{2}}{a+\frac{b+c}{2}}$$
That is,
$$\begin{aligned}
& 2\left[a^{2}(b+c)+b^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,467 |
Example 13 Let $a, b, c \in \mathbf{R}^{+}$, prove that:
$$\frac{b^{3}}{a^{2}+8 b c}+\frac{c^{3}}{b^{2}+8 c a}+\frac{a^{3}}{c^{2}+8 a b} \geqslant \frac{a+b+c}{9}$$ | Prove that if we denote the left side of the inequality as $M$, then
$$\begin{aligned}
S & =\left(a^{2}+8 b c\right)+\left(b^{2}+8 c a\right)+\left(c^{2}+8 a b\right) \\
& =(a+b+c)^{2}+6(a b+b c+c a) \\
& \leqslant 3(a+b+c)^{2} .
\end{aligned}$$
Therefore,
$$\begin{aligned}
3= & \frac{1}{M} \cdot\left(\frac{b^{3}}{a^{... | M \geqslant \frac{1}{9}(a+b+c) | Inequalities | proof | Yes | Yes | inequalities | false | 738,468 |
Example 14 Given non-negative real numbers $a, b, c$ satisfying: $a+b+c=1$, prove that:
$$\begin{aligned}
2 & \leqslant\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2} \\
& \leqslant(1+a)(1+b)(1+c),
\end{aligned}$$
and find the conditions under which equality holds. | Proof: Let $ab + bc + ca = m, abc = n$, then
$$(x-a)(x-b)(x-c) = x^3 - x^2 + mx - n.$$
Let $x = a$, then we have $a^3 = a^2 - ma + n$ (Note: this serves to reduce the degree!). Thus,
$$\begin{array}{l}
\sum_{\mathrm{cc}} a^3 = \sum_{\mathrm{cor}} a^2 - m \cdot \sum_{\mathrm{cr}} a + 3n. \\
\sum_{\mathrm{cc}} a^4 = \su... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,469 |
1 Given $a, b, c \in \mathbf{R}^{+}$. Find the minimum value of $\frac{a}{b+3 c}+\frac{b}{8 c+4 a}+\frac{9 c}{3 a+2 b}$. | Let $b+3c=x, 8c+4a=y, 3a+2b=z$, then the original expression $=\frac{1}{8}\left(\frac{y}{x}+\frac{4x}{y}\right)+\frac{1}{6}\left(\frac{z}{x}+\frac{9x}{z}\right)+\frac{1}{16}\left(\frac{4z}{y}+\frac{9y}{z}\right)-\frac{61}{48} \geqslant \frac{47}{48}$. | \frac{47}{48} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 738,471 |
2 If the ellipse $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1(m, n>0)$ passes through the point $p(a, b)(a b \neq 0,|a| \neq|b|)$, find the minimum value of $m+n$.
The ellipse $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1(m, n>0)$ passes through the point $p(a, b)(a b \neq 0,|a| \neq|b|)$. Determine the minimum value of $m... | 2. Let $m, n \in \mathbf{R}^{+}, a, b \in \mathbf{R}^{+}$, and set $a=m \cos \alpha, b=n \sin \alpha$, where $\alpha \in \left(0, \frac{\pi}{2}\right]$. Then,
$$(m+n)^{2}=\left(\frac{a}{\cos \alpha}+\frac{b}{\sin \alpha}\right)^{2}=\frac{a^{2}}{\cos ^{2} \alpha}+\frac{b^{2}}{\sin ^{2} \alpha}+\frac{2 a b}{\sin \alpha \... | \left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 738,472 |
3 In $\triangle A B C$, prove:
$$\frac{c-a}{b+c-a}+\frac{a-b}{c+a-b}+\frac{b-c}{a+b-c} \leqslant 0 .$$ | 3. Let $b+c-a=2 x, c+a-b=2 y, a+b-c=2 z$, then $x, y, z \in \mathbf{R}^{+}$, and $a=y+z, b=x+z, c=x+y$, so the original inequality is equivalent to $\frac{x-z}{2 x}+\frac{y-x}{2 y}+\frac{z-y}{2 z} \leqslant$ 0, which means $\frac{z}{x}+\frac{x}{y}+\frac{y}{z} \geqslant 3$, obviously true. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 738,473 |
Theorem 1.1. If $a$ and $r$ are real numbers and $r \neq 1$, then
$$\sum_{j=0}^{n} a r^{j}=a+a r+a r^{2}+\cdots+a r^{n}=\frac{a r^{n+1}-a}{r-1}$$ | Proof. To prove that the formula for the sum of terms of a geometric progression is valid, we must first show that it holds for $n=1$. Then, we must show that if the formula is valid for the positive integer $n$, it must also be true for the positive integer $n+1$.
To start things off, let $n=1$. Then, the left side o... | proof | Algebra | proof | Yes | Yes | number_theory | false | 738,474 |
Proposition 1.2. Let $n$ and $k$ be nonnegative integers with $k \leqslant n$. Then
(i ) $\binom{n}{0}=\binom{n}{n}=1$
(ii) $\binom{n}{k}=\binom{n}{n-k}$. | Proof. To see that (i) is true, note that
$$\binom{n}{0}=\frac{n!}{0!n!}=\frac{n!}{n!}=1$$
and
$$\binom{n}{n}=\frac{n!}{n!0!}=\frac{n!}{n!}=1$$
To verify (ii), we see that
$$\binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{n!}{(n-k)!(n-(n-k))!}=\binom{n}{n-k}$$ | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 738,475 |
8. Find a formula for $\prod_{j=1}^{n} 2^{j}$. | 8. $2^{n(n+1) / 2}$ | 2^{n(n+1) / 2} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 738,487 |
23. A dozen equals 12 and a gross equals $12^{2}$. Using base 12 , or duodecimal, arithmetic answer the following questions.
a) If 3 gross, 7 dozen, and 4 eggs are removed from a total of 11 gross and 3 dozen eggs, how many eggs are left?
b) If 5 truckloads of 2 gross, 3 dozen, and 7 eggs each are delivered to the supe... | 23. a) 7 gross, 7 dozen, and 8 eggs
c) 3 gross, 11 dozen, and 6 eggs
b) 11 gross, 5 dozen, and 11 eggs | 3 \text{ gross, 11 dozen, and 6 eggs} | Other | math-word-problem | Yes | Yes | number_theory | false | 738,488 |
Lemma 1.1. Every positive integer greater than one has a prime divisor. | Proof. We prove the lemma by contradiction; we assume that there is a positive integer having no prime divisors. Then, since the set of positive integers with no prime divisors is non-empty, the well-ordering property tells us that there is a least positive integer $n$ with no prime divisors. Since $n$ has no prime div... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,491 |
Theorem 1.8 . There are infinitely many primes. | Proof. Consider the integer
$$Q_{n}=n!+1, \quad n \geqslant 1$$
Lemma 1.1. tells us that $Q_{n}$ has at least one prime divisor, which we denote by $q_{n}$. Thus, $q_{n}$ must be larger than $n$; for if $q_{n} \leqslant n$, it would follow that $q_{n} \mid n!$, and then, by Proposition $1.4, q_{n} \mid\left(Q_{n}-n!\r... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,492 |
Theorem 1.9. If $n$ is a composite integer, then $n$ has a prime factor not exceeding $\sqrt{n}$. | Proof. Since $n$ is composite, we can write $n=a b$, where $a$ and $b$ are integers with $1\sqrt{n}$ and $a b>\sqrt{n} \cdot \sqrt{n}=n$. Now, by Lemma 1.1, $a$ must have a prime divisor, which by Proposition 1.3 is also a divisor of a and which is clearly less than or equal to $\sqrt{n}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,493 |
Proposition 1.8. For any positive integer $n$, there are at least $n$ consecutive composite positive integers. | Proof. Consider the n consecutive positive integers
$$(n+1)!+2,(n+1)!+3, \ldots,(n+1)!+n+1$$
When $2 \leqslant j \leqslant n+1$, we know that $j \mid(n+1)$ !. By Proposition 1.4, it follows that $j \mid(n+1)!+j$. Hence, these $n$ consecutive integers are all composite. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,495 |
7. Let $Q_{n}=p_{1} p_{2} \cdots p_{n}+1$ where $p_{1}, p_{2}, \ldots, p_{n}$ are the $n$ smallest primes. Determine the smallest prime factor of $Q_{n}$ for $n=1,2,3,4,5$, and 6 . Do you think $Q_{n}$ is prime infinitely often? (This is an unresolved question.) | 7. $3,7,31,211,2311,59$ | 3,7,31,211,2311,59 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,503 |
10. a) Find the smallest five consecutive composite integers.
b) Find one million consecutive composite integers. | 10. a) $24,25,26,27,28$
b) 1000001 ! $+2,1000001!+3, \ldots, 1000001!+1000001$ | 24,25,26,27,28 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,506 |
10. What function $f(n)$ is defined recursively by $f(1)=2$ and $f(n+1)=2 f(n)$ for $n \geqslant 1$ ? | $10.2^{n}$ | 2^n | Algebra | math-word-problem | Yes | Yes | number_theory | false | 738,509 |
16. The lucky numbers are generated by the following sieving process. Start with the positive integers. Begin the process by crossing out every second integer in the list, starting your count with the integer 1. Other than 1 the smallest integer left is 3 , so we continue by crossing out every third integer left, start... | 16. a) $1,3,7,9,13,15,21,25,31,33,37,43,49,51,63,67,69,73,75,79,87,93,99$ | 1,3,7,9,13,15,21,25,31,33,37,43,49,51,63,67,69,73,75,79,87,93,99 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,513 |
Proposition 2.1. Let $a, b$, and $c$ be integers with $(a, b)=d$. Then
(i) $(a / d, b / d)=1$
(ii) $(a+c b, b)=(a, b)$. | Proof. (i) Let $a$ and $b$ be integers with $(a, b)=d$. We will show that $a / d$ and $b / d$ have no common positive divisors other than 1. Assume that $e$ is a positive integer such that $e \mid(a / d)$ and $e \mid(b / d)$. Then, there are integers $k$ and $\ell$ with $a / d=k e$ and $b / d=\ell e$, such that $a=d e ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,515 |
Theorem 2.1. The greatest common divisor of the integers $a$ and $b$, that are not both zero, is the least positive integer that is a linear combination of $a$ and b. | Proof. Let $d$ be the least positive integer which is a linear combination of $a$ and $b$. (There is a least such positive integer, using the well-ordering property, since at least one of two linear combinations $1 \cdot a+0 \cdot b$ and $(-1) a+0 \cdot b$, where $a \neq 0$, is positive.) We write
where $m$ and $n$ are... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,516 |
Lemma 2.1. If $a_{1}, a_{2}, \ldots, a_{n}$ are integers, that are not all zero, then $\left(a_{1}, a_{2}, \ldots, a_{n-1}, a_{n}\right)=\left(a_{1}, a_{2}, \ldots,\left(a_{n-1}, a_{n}\right)\right)$ | Proof. Any common divisor of the $n$ integers $a_{1}, a_{2}, \ldots, a_{n-1}, a_{n}$ is, in particular, a divisor of $a_{n-1}$ and $a_{n}$, and therefore, a divisor of $\left(a_{n-1}, a_{n}\right)$. Also, any common divisor of the $n-2$ integers $a_{1}, a_{2}, \ldots, a_{n-2}$, and $\left(a_{n-1}, a_{n}\right)$, must b... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,517 |
5. Periodical cicadas are insects with very long larval periods and brief adult lives. For each species of periodical cicada with larval period of 17 years, there is a similar species with a larval period of 13 years. If both the 17 -year and 13 -year species emerged in a particular location in 1900, when will they nex... | 5. 2121 | 2121 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,523 |
15. Find three mutually relatively prime integers from among the integers $66,105,42,70$, and 165 . | 15. $66,70,105 ; 66,70,165 ;$ or $42,70,165$ | 66,70,105 ; 66,70,165 ; 42,70,165 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,534 |
Lemma 2.2. If $c$ and $d$ are integers and $c=d q+r$ where $c$ and $d$ are integers, then $(c, d)=(d, r)$. | Proof. If an integer $e$ divides both $c$ and $d$, then since $r=c-d q$, Proposition 1.4 shows that $e \mid r$. If $e \mid d$ and $e \mid r$, then since $c=d q+r$, from Proposition 1.4, we see that $e \mid c$. Since the common divisors of $c$ and $d$ are the same as the common divisors of $d$ and $r$, we see that $(c, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,541 |
Theorem 2.2. Let $n$ be a positive integer and let $\alpha=(1+\sqrt{5}) / 2$. Then $u_{n}>\alpha^{n-2}$ for $n \geqslant 3$ | Proof. We use the second principle of mathematical induction to prove the desired inequality. We have $\alpha<2=u_{3}$, so that the theorem is true for $n=3$
Now assume that for all integers $k$ with $k \leqslant n$, the inequality
$$\alpha^{k-2}<u_{k}$$
holds.
Since $\alpha=(1+\sqrt{5}) / 2$ is a solution of $x^{2}-x... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,543 |
Theorem 2.3. Let $u_{n+1}$ and $u_{n+2}$ be successive terms of the Fibonacci sequence. Then the Euclidean algorithm takes exactly $n$ divisions to show that $\left(u_{n+1}, u_{n+2}\right)=1$. | Proof. Applying the Euclidean algorithm, and using the defining relation for the Fibonacci numbers $u_{j}=u_{j-1}+u_{j-2}$ in each step, we see that
$$\begin{aligned}
u_{n+2} & =u_{n+1} \cdot 1+u_{n} \\
u_{n+1} & =u_{n} \cdot 1+u_{n-1} \\
& \cdot \\
& \cdot \\
u_{4} & =u_{3} \cdot 1+u_{2} \\
u_{3} & =u_{2} \cdot 2
\end... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,544 |
Lame's Theorem. The number of divisions needed to find the greatest common divisor of two positive integers using the Euclidean algorithm does not exceed five times the number of digits in the smaller of the two integers. | Proof. When we apply the Euclidean algorithm to find the greatest common divisor of $a=r_{0}$ and $b=r_{1}$ with $a>b$, we obtain the following sequence of equations: $\square$
$$\begin{array}{rlr}
r_{0} & =r_{1} q_{1}+r_{2}, & 0 \leqslant r_{2}\alpha^{n-1}$ for $n>2$ where $\alpha=(1+\sqrt{5}) / 2$. Hence, $b>\alpha^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,545 |
Corollary 2.1. The number of bit operations needed to find the greatest common divisor of two positive integers $a$ and $b$ with $b>a$ is $O\left(\left(\log _{2} a\right)^{3}\right)$. | Proof. We know from Lamé's theorem that $O\left(\log _{2} a\right)$ divisions, each taking $O\left(\left(\log _{2} a\right)^{2}\right)$ bit operations, are needed to find $(a, b)$. Hence, by Proposition $1.7,(a, b)$ may be found using a total of $O\left(\left(\log _{2} a\right)^{3}\right)$ bit operations. | O\left(\left(\log _{2} a\right)^{3}\right) | Number Theory | proof | Yes | Yes | number_theory | false | 738,546 |
Theorem 2.4. Let $a$ and $b$ be positive integers. Then
$$(a, b)=s_{n} a+t_{n} b,$$
for $n=0,1,2, \ldots$, where $s_{n}$ and $t_{n}$ are the $n$th terms of the sequences defined recursively by
$$\begin{array}{l}
s_{0}=1, t_{0}=0 \\
s_{1}=0, t_{1}=1
\end{array}$$
and
$$s_{j}=s_{j-2}-q_{j-1} s_{j-1}, t_{j}=t_{j-2}-q_{j-2... | Proof. We will prove that
$$r_{j}=s_{j} a+t_{j} b$$
for $j=0,1, \ldots, n$. Since $(a, b)=r_{n}$, once we have established $(2.2)$, we will know that
$$(a, b)=s_{n} a+t_{n} b$$
We prove (2.2) using the second principle of mathematical induction. For $j=0$, we have $a=r_{0}=1 \cdot a+0 \cdot b=s_{0} a+t_{0} b$. Hence, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,547 |
3. For each of the following sets of integers, express their greatest common divisor as a linear combination of these integers
a) $6,10,15$
b) $70,98,105$
c) $280,330,405,490$. | 3. a) $1=1 \cdot 6+1 \cdot 10+(-1) 15$
b) $7=0 \cdot 70+(-1) 98+1 \cdot 105$
c) $5=-5 \cdot 280+$
$4 \cdot 330+(-1) 405+1 \cdot 490$ | 7=0 \cdot 70+(-1) \cdot 98+1 \cdot 105 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,550 |
14. a) Let $n$ be a positive integer. By expanding $(1+(-1))^{n}$ with the binomial theorem, show that
$$\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}=0$$
b) Use part (a), and the fact that $\sum_{k=0}^{n}\binom{n}{k}=2^{n}$, to find
$$\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots$$
and
$$\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\c... | None | null | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 738,553 |
Lemma 2.3. If $a, b$, and $c$ are positive integers such that $(a, b)=1$ and $a \mid b c$, then $a \mid c$. | Proof. Since $(a, b)=1$, there are integers $x$ and $y$ such that $a x+b y=1$. Multiplying both sides of this equation by $c$, we have $a c x+b c y=c$. By Proposition 1.4, $a$ divides $a c x+b c y$, since this is a linear combination of $a$ and $b c$, both of which are divisible by $a$. Hence $a \mid c$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,567 |
Corollary 2.2. If $p$ divides $a_{1} a_{2} \cdots a_{n}$ where $p$ is a prime and $a_{1}, a_{2}, \ldots, a_{n}$ are positive integers, then there is an integer $i$ with $1 \leqslant i \leqslant n$ such that $p$ divides $a_{i}$. | Proof. We prove this result by induction. The case where $n=1$ is trivial. Assume that the result is true for $n$. Consider a product of $n+1$, integers, $a_{1} a_{2} \cdots a_{n+1}$ that is divisible by the prime $p$. Since $p \mid a_{1} a_{2} \cdots a_{n+1}=$ $\left(a_{1} a_{2} \cdots a_{n}\right) a_{n+1}$, we know f... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,568 |
Lemma 2.4. If $x$ and $y$ are real numbers, then $\max (x, y)+\min (x, y)$ $=x+y$ | Proof. If $x \geqslant y$, then $\min (x, y)=y$ and $\max (x, y)=x$, so that $\max (x, y)+\min (x, y)=x+y$. If $x<y$, then $\min (x, y)=x$ and $\max (x, y)=y$, and again we find that $\max (x, y)+\min (x, y)=x+y$ | proof | Algebra | proof | Yes | Yes | number_theory | false | 738,569 |
Theorem 2.5. If $a$ and $b$ are positive integers, then $[a, b]=a b /(a, b)$, where $[a, b]$ and $(a, b)$ are the least common multiple and greatest common divisor of $a$ and $b$, respectively. | Proof. Let $a$ and $b$ have prime-power factorizations $a=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{n}^{a_{*}^{*}}$ and $b=p_{1}^{b_{1}} p_{s}^{b_{2}} \cdots p_{n}^{b_{*}}$, where the exponents are nonnegative integers and all primes occurring in either factorization occur in both, perhaps with zero exponents. Now let $M_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,570 |
Proposition 2.2. There are infinitely many primes of the form $4 n+3$, where $n$ is a positive integer. | Proof. Let us assume that there are only a finite number of primes of the form $4 n+3$, say $p_{0}=3, p_{1}, p_{2}, \ldots, p_{r}$. Let
$$Q=4 p_{1} p_{2} \cdots p_{r}+3$$
Then, there is at least one prime in the factorization of $Q$ of the form $4 n+3$. Otherwise, all of these primes would be of the form $4 n+1$, and ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,573 |
Lemma 2.6. If $a$ and $b$ are integers both of the form $4 n+1$, then the product $a b$ is also of this form. | Proof. Since $a$ and $b$ are both of the form $4 n+1$, there exist integers $r$ and $s$ such that $a=4 r+1$ and $b=4 s+1$. Hence,
$$a b=(4 r+1)(4 s+1)=16 r s+4 r+4 s+1=4(4 r s+r+s)+1,$$
which is again of the form $4 n+1$.
We now prove the desired result. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,574 |
8. a) Let $n$ be a positive integer. Show that the power of the prime $p$ occurring in the prime power factorization of $n!$ is
$$[n / p]+\left[n / p^{2}\right]+\left[n / p^{3}\right]+\cdots$$
b) Use part (a) to find the prime-power factorization of 20 !. | 8. b) $2^{18} \cdot 3^{8} \cdot 5^{4} \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19$ | 2^{18} \cdot 3^{8} \cdot 5^{4} \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 | Number Theory | proof | Yes | Yes | number_theory | false | 738,583 |
Theorem 1.2. Let $n$ and $k$ be positive integers with $n \geqslant k$. Then
$$\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}$$ | Proof. We perform the addition
$$\binom{n}{k}+\binom{n}{k-1}=\frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-k+1)!}$$
by using the common denominator $k!(n-k+1)$ !. This gives
$$\begin{aligned}
\binom{n}{k}+\binom{n}{k-1} & =\frac{n!(n-k+1)}{k!(n-k+1)!}+\frac{n!k}{k!(n-k+1)!} \\
& =\frac{n!((n-k+1)+k)}{k!(n-k+1)!} \\
& =\frac{n... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 738,586 |
12. This problem presents an example of a system where unique factorization into primes fails. Let $H$ be the set of all positive integers of the form $4 k+1$, where $k$ is a positive integer.
a) Show that the product of two elements of $H$ is also in $H$.
b) An element $h \neq 1$ in $H$ is called a "Hilbert prime" if ... | 12. b) $5,9,13,17,21,29,33,37,41,49,53,57,61,69,73,77,89,93,97,101$
d) $693=21 \cdot 33=9.77$ | 693=21 \cdot 33=9 \cdot 77 | Number Theory | proof | Yes | Yes | number_theory | false | 738,589 |
14. Find the least common multiple of each of the following pairs of integers
a) 8,12
d) 111,303
b) 14,15
e) 256,5040
c) 28,35
f) 343,999 . | 14. a) 24 b) 210 c) 140 d) 11211 e) 80640 f) 342657 | 140 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,591 |
17. Which pairs of integers $a$ and $b$ have greatest common divisor 18 and least common multiple 540 ? | 17. 18,$540 ; 36,270 ; 54,180 ; 90,108$ | 18,540 ; 36,270 ; 54,180 ; 90,108 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,594 |
Lemma 2.7. If $n$ is an odd positive integer, then there is a one-to-one correspondence between factorizations of $n$ into two positive integers and differences of two squares that equal $n$. | Proof. Let $n$ be an odd positive integer and let $n=a b$ be a factorization of $n$ into two positive integers. Then $n$ can be written as the difference of two squares, since
$$n=a b=\left(\frac{a+b}{2}\right)^{2}-\left(\frac{a-b}{2}\right)^{2}$$
where $(a+b) / 2$ and $(a-b) / 2$ are both integers since $a$ and $b$ ar... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,614 |
Proposition 2.3. The Fermat number $F_{5}=2^{2^{5}}+1$ is divisible by 641 . | Proof. We will prove that $641 \mid F_{5}$ without actually performing the division.
Note that
$$641=5 \cdot 2^{7}+1=2^{4}+5^{4}$$
Hence,
$$\begin{aligned}
2^{2^{5}}+1 & =2^{32}+1=2^{4} \cdot 2^{28}+1=\left(641-5^{4}\right) 2^{28}+1 \\
& =641 \cdot 2^{28}-\left(5 \cdot 2^{7}\right)^{4}+1=641 \cdot 2^{28}-(641-1)^{4}+1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,615 |
Lemma 2.8. Let $F_{k}=2^{2^{k}}+1$ denote the $k$ th Fermat number, where $k$ is a nonnegative integer. Then for all positive integers $n$, we have
$$F_{0} F_{1} F_{2} \cdots F_{n-1}=F_{n}-2$$ | Proof. We will prove the lemma using mathematical induction. For $n=1$, the identity reads
$$F_{0}=F_{1}-2$$
This is obviously true since $F_{0}=3$ and $F_{1}=5$. Now let us assume that the identity holds for the positive integer $n$, so that
$$F_{0} F_{1} F_{2} \cdots F_{n-1}=F_{n}-2$$
With this assumption we can ea... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,617 |
Theorem 2.6. Let $m$ and $n$ be distinct nonnegative integers. Then the Fermat numbers $F_{m}$ and $F_{n}$ are relatively prime. | Proof. Let us assume that $m<n$. From Lemma 2.8, we know that
$$F_{0} F_{1} F_{2} \cdots F_{m} \cdots F_{n-1}=F_{n}-2$$
Assume that $d$ is a common divisor of $F_{m}$ and $F_{n}$. Then, Proposition 1.4 tells us that
$$d \mid\left(F_{n}-F_{0} F_{1} F_{2} \cdots F_{m} \cdots F_{n-1}\right)=2 .$$
Hence, either $d=1$ or ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,618 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.