陈洪葛的博客

问题：设$A$为$m\times n$阶矩阵，$B$为$n\times m$阶，$m>n$,求证
\[ |\lambda I_{m}-AB|=\lambda^{m-n}|\lambda I_{n}-BA| \]
证明
当$\lambda=0$的时候，结论显然成立。下面证明$\lambda\neq 0$的情况。
注意到矩阵
\[ \left(
\begin{array}{cc}
\lambda E_{m} & A \\
B & E_{n} \\
\end{array}
\right)\]
有2种打洞方式。就是
\[ \left(
\begin{array}{cc}
E_{m} & -A \\
O & E_{n} \\
\end{array}
\right)\left(
\begin{array}{cc}
\lambda E_{m} & A \\
B & E_{n} \\
\end{array}
\right)
\left(\begin{array}{cc}
E_{m} & O \\
-B & E_{n} \\
\end{array}
\right)=
\left(
\begin{array}{cc}
\lambda E_{m}-AB & O \\
O & E_{n} \\
\end{array}
\right)\]
和
\[ \left(
\begin{array}{cc}
E_{m} & O \\
-\frac{1}{\lambda} & E_{n} \\
\end{array}
\right)
\left(
\begin{array}{cc}
\lambda E_{m} & A \\
B & E_{n} \\
\end{array}
\right)
\left(
\begin{array}{cc}
E_{m} & -\frac{1}{\lambda}A \\
O & E_{n} \\
\end{array}
\right)=
\left(
\begin{array}{cc}
\lambda E_{m} & O \\
O & E_{n}-\frac{1}{\lambda}BA\\
\end{array}
\right)\]
对上面两个式子取行列式，马上就可以得到
\[ |\lambda I_{m}-AB|=\lambda^{m-n}|\lambda I_{n}-BA| \]
$\square$  

我代数弱，大家不要笑啊。

设$f\in C(\mathbf{R^1})$,令
\[ \delta_{n}(x)=2^{n}\left[f\left(x+\frac{1}{2^n}\right)-f(x)\right] \]
若
\[ |\delta_{n}(x)|\leq M\qquad (x\in \mathbf{R},n\in \mathbf{N}) \]
且
\[  \delta_{n}(x)\rightarrow 0 \qquad (n\rightarrow\infty) \]
试证明$f(x)$是一个常数。
证明对

$ \forall a,b\in \mathbf{R}$,不妨设$a<b$.
由于$f(x)$是$\mathbf{R}$上的连续函数，每个$\delta_{n}(x)$都是$[a,b]$上的Lebesgue可积函数，且满足Lebesgue控制收敛定理的条件。故有
\[ \lim_{n\rightarrow\infty}{\int_{a}^{b}{\delta_{n}(x)dx}}=\int_{a}^{b}{\lim_{n\rightarrow\infty}{\delta_{n}(x)}dx}=0\]
注意到
\begin{align*}
\int_{a}^{b}{\delta_{n}(x)dx}&=2^{n}\int_{a}^{b}{\left(f\left(x+\frac{1}{2^n}\right)-f(x)  \right)dx}\\
&=2^{n}\left(\int_{b}^{b+\frac{1}{2^n}}{f(x)dx}-\int_{a}^{a+\frac{1}{2^n}}{f(x)dx}  \right)
\end{align*}
由于$f(x)$的连续性，我们有
\[ \lim_{n\rightarrow\infty}{2^{n}}{\int_{a}^{a+\frac{1}{2^n}}{f(x)dx}}=f(a) \]
和
\[ \lim_{n\rightarrow\infty}{2^{n}}{\int_{b}^{b+\frac{1}{2^n}}{f(x)dx}}=f(b) \]
这样就有
\[ f(b)-f(a)=\lim_{n\rightarrow\infty}{\int_{a}^{b}{\delta_{n}(x)dx}}=0 \]
命题得证!  $\blacksquare$

设$a,b,c,d>0$ 证明：
\begin{align*}
&\frac{9}{a(b+c+d)}+\frac{9}{b(c+d+a)}+\frac{9}{c(d+b+a)}+\frac{9}{d(a+b+c)}\\
&\geq\frac{16}{(a+b)(c+d)}+\frac{16}{(a+c)(b+d)}+\frac{16}{(a+d)(b+c)}
\end{align*}

（新加坡）
证明
两边同时乘以$a+b+c+d$,不等式变成
\begin{align}
 &\frac{9}{a}+\frac{9}{b}+\frac{9}{c}+\frac{9}{d}+\frac{9}{b+c+d}+\frac{9}{c+d+a}+\frac{9}{d+a+b}+\frac{9}{a+b+c}\\
 &\geq \frac{16}{a+b}+\frac{16}{c+d}+\frac{16}{a+c}+\frac{16}{b+d}+\frac{16}{a+d}+\frac{16}{b+c}
\end{align}
由Popoviciu不等式，我们有
\[ \frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{9}{b+c+d}\geq \frac{4}{b+c}+\frac{4}{c+d}+\frac{4}{b+d}\]
所以有
\[ 3\sum_{sym}{\frac{1}{a}}+9\sum_{sym}{\frac{1}{a+b+c}}\geq 8\sum_{sym}{\frac{1}{a+b}}\]
故只要证明
\[ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\geq 3\sum_{sym}{\frac{1}{a+b+c}} \]
由AM-GM显然。$\square$                                                  

设$x,y,z\in R$,证明

\[ 5(x^4+y^4+z^4)+7(x^3y+y^3z+z^3x)\geq 0 \]
证明：
仿照以前的做法我们可以把不等式加强为有基本取等条件$x=y=z$的那种。
\[ 5(x^4+y^4+z^4)+7(x^3y+y^3z+z^3x)\geq \frac{36}{81}(x+y+z)^4 \]
展开（这个系数有点大，不着急，慢慢展）
\[ 369\sum{x^4}+423\sum{x^3y}-143\sum{xy^3}-216\sum{x^2y^2}-432\sum{x^2yz}\geq 0 \]
到至今为止，这个3元4次不等式已经不是问题了，直接套Vo Quoc Ba Can配方文章中的结论，
只要验证
\[ \left\{
\begin{array}{ll}
 m>0, & \\
 3m(m+n)\geq p^2+pg+g^2, &
 \end{array}
   \right.\]
这里
\[ m=369,n=-216, p=423,g=-143 \]
\[ 3m(m+n)-(p^2+pg+g^2)=169371-(178929+20449-60189)=30182>0 \]
所以不等式得证。                      $\blacksquare$

同样的手段奏效于

\[ 3(x^4+y^4+z^4)+4(x^3y+y^3z+z^3x)\geq 0  \]

事实上，我们可以用pqr来弄出那个最佳系数 :)

\[ (x^4+y^4+z^4)+k(x^3y+y^3z+z^3x)\geq 0 \]

Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$. Prove that
\[ \sqrt[3]{3a^2+2ab+3b^2}+\sqrt[3]{3b^2+2bc+3c^2}+\sqrt[3]{3c^2+2ca+3a^2} \geq 6.\]

Proof：(by gxggs)

By Holder inequality, we have
\[\left(\sum_{cyc}{\sqrt[3]{3a^2+2ab+3b^2}}\right)^3\sum_{cyc}{\left(\frac{(a+b)^4}{3a^2+2ab+3b^2}\right)}\geq 16(a+b+c)^4\]
Therefore it suffices to show that
\[ 16(a+b+c)^4\geq 216\cdot \sum_{cyc}{\left(\frac{(a+b)^4}{3a^2+2ab+3b^2}\right)}\]
Or
\[2(a+b+c)^2\geq 3\cdot \sum_{cyc}{\left(\frac{(a+b)^4}{3a^2+2ab+3b^2}\right)}\]
\[\Longleftrightarrow \sum_{cyc}{\left(a^2+b^2+4ab-\frac{3(a+b)^4}{3a^2+2ab+3b^2}\right)}\geq 0\]
which is equivalent to
\[\frac{2ab(a-b)^2}{3a^2+2ab+3b^2}+\frac{2bc(b-c)^2}{3b^2+2bc+3c^2}+\frac{2ca(c-a)^2}{3c^2+2ca+3a^2}\geq 0\]
which is obvious true. Equal holds when $a=b=c=1$ or $a=3,b=0,c=0$ or their cyclic forms.   $\blacksquare$
另外，越南TaHongQuang 作了如下推广

 1.Let $a, \, b, \, c > 0$ such that $a+b+c =\frac{1}{4}$ .  Prove that
\[  \sqrt [3]{65{a}^{2}+81{b}^{2}-2ab}+\sqrt [3]{65{b}^{2}+81{c}^{2}-2bc}+\sqrt [3]{65{c}^{2}+81{a}^{2}-2ac} \ge 3 \]

2.Let $a, \, b, \, c > 0$ such that $a+b+c =\frac{1}{6}$ .  Prove that
\[  \sqrt [3]{7{a}^{2}+9{b}^{2}-4ba}+\sqrt [3]{7{b}^{2}+9{c}^{2}-4bc}+\sqrt [3]{7{c}^{2}+9{a}^{2}-4ac}\ge 1 \]