The product of 3 consecutive positive integers is never a square.

Proof: if we have $a(a^2-1)=b^2$, then$\gcd(a,a^2-1)=\gcd(a^2-1,a^2)=1$, so$a=m^2, a^2-1=n^2$, where$mn=b$, and$\gcd(m,n)=1$. But $a^2-1=n^2$ has no integer solution.

\sqrt[n]{n}

Let $n$ be an positive integer, prove that \[\sqrt{n}\leqslant\sqrt[n]{n!}\leqslant\frac{n+1}{2}\]

Inequality from "Math Bar"

设  $a_{i}>0$

求证：\[\sum_{i=1}^{n}\left(\frac{a_{i}}{a_{i+1}}\right)^{n}\ge\sum_{i=1}^{n}\frac{a_{i+1}}{a_{i}}\]

令$x_{i}=\frac{a_{i}}{a_{i+1}}$

此题可考虑 $(x-1)^2(x^n+x^{n-1}+...+x+1)\ge 0$ 得到$x^{n}\ge x^{n-1}+x-1$再用排序和均值不等式即可。

想复杂了，直接均值也可以，以后做题不能硬套一个思路能简单就尽量简单得好

The last digital of n! is never periodic (Chinese version)

证明：假设是周期数列。设周期长度为$T$（$T$不一定是最小的）。由抽屉原理,在\[10^u, 10^{u+1},...,10^{u+t-1}\]
中必有两个数模$T$同余。 设这两个数是$10^k$ 与 $10^{k+r}$.于是$T|10^{k+r}-10^k$.
考虑到$2\times 10^{k+r}-10^k$的尾数是$9$.
并\[(2\times 10^{k+r}-10^k)!=(2\times 10^{k+r}-10^k-1)!(2\times 10^{k+r}-10^k)\]，
等式左边的尾数是$(10^k)!$的尾数, 等式右边则是$(10^k-1)!\times9$的尾数，
而$(10^k)!$的尾数与$(10^k-1)!$相同，于是$(10^k)!$的尾数只能是$5$.但$(10^k)!$显然是个偶数，矛盾。 证毕。

#数论

Strategy to find the sum

\[\sum_{k=45^{\circ}}^{133^{\circ}}\frac{1}{\sin(k)\sin(k+1^{\circ})}\]

part of solution

let\[\sin(x)=\frac{e^{ix}-\frac{1}{e^{ix}}}{2}\]

then \[\frac{1}{\sin(k)\sin(k+1^{\circ})}=\frac{4e^{(2k+1)i}}{(e^{2ki}-1)(e^{(2k+2)i}-1)}\]\[=\left(\frac{1}{e^{2ki}-1}-\frac{1}{e^{(2k+2)i}-1}\right)\cdot\frac{1}{e^i(e^{2i}-1)}\]

\[\frac{\frac{1}{\sqrt{x+\Delta x}}-\frac{1}{\sqrt{x}}}{\Delta x}=\frac{\sqrt{x}-\sqrt{x+\Delta x}}{\sqrt{x(x+\Delta x)}\Delta x}=\frac{\sqrt{x}-\sqrt{x+\Delta x}}{\sqrt{x(x+\Delta x)}(x-x+\Delta x)}=-\frac{1}{\sqrt{x(x+\Delta x)}(\sqrt{x}+\sqrt{x+\Delta x})}\]

Max and Min

Let $x,y>0$. Show that \[\max\{\min\{x,y,\frac{1}{x}+\frac{1}{y}\}\}=\min\{\max\{x,y,\frac{1}{x}+\frac{1}{y}\}\}=\sqrt{2}\]

Proof:

\[\left(\max\{x,y,\frac{1}{x}+\frac{1}{y}\}\right)^2\geqslant\frac{(x+y)(\frac{1}{x}+\frac{1}{y})}{2}\geqslant2\]

Surpose that $y\geqslant x$, we have \[\left(\min\{x,y,\frac{1}{x}+\frac{1}{y}\}\right)^2\leqslant x\left(\frac{1}{x}+\frac{1}{y}\right)=1+\frac{x}{y}\leqslant2\]

Then it shows that $\max=\sqrt{2}$

Cauchy

若正数$x,y,z$满足$x+y+z=1$,求证:\[\frac{x^4}{z(1-z^2)}+\frac{y^4}{x(1-x^2)}+\frac{z^4}{y(1-y^2)}\geqslant\frac{1}{8}\]
证明：由柯西不等式有\[\frac{x^4}{z(1-z^2)}+\frac{y^4}{x(1-x^2)}+\frac{z^4}{y(1-y^2)}\geqslant\frac{(x^2+y^2+z^2)^2}{x+y+z-x^3-y^3-z^3}=\]
\[\frac{(x^2+y^2+z^2)^2}{1-(x+y+z)(x^3+y^3+z^3)}\geqslant\frac{1/9}{1-1/9}=\frac{1}{8}\]

Polynomial

Let $a,b,c$ be 3 distinct integers, $P(x)$ is a polynomial with integer coefficient, prove that we cannot have $P(a)=b,P(b)=c,$ and $P(c)=a$(USAMO1974)

Prove. we assume that there exsits a $P(x)$ satisfied the conditions.  Notice that $a-b|P(a)-P(b)=b-c$ by the same method, we  have

$a-b|b-c,b-c|c-a,c-a|a-b$ so$|a-b|=|b-c|=|c-a|$, we only need to consider the situation without abs(because other case are obviously contradiction) $a-b\le b-c,b-c\le c-a, c-a\le a-b$, but $a-b+b-c+c-a=0$, therefore we have $a-b=b-c=c-a$

let $b=c+d,a=c+2d$,we have$c-a=-2d=d$ hence $d=0\iff a=b=c $ which is a contradiction.