陈洪葛的博客

Let$a,b,c>0$ with $a + b+c=3$, Prove that $a\sqrt[3]{a+b}+b\sqrt[3]{b+c}+c\sqrt[3]{c+a} \ge 3 \sqrt[3]2$

(Tran Quoc Anh)

Proof:

by AM-GM inequality,we have
$$ a\sqrt[3]{a+b}=\frac{3\sqrt[3]{2}a(a+b)}{3\sqrt[3]{2(a+b)(a+b)}}\geq 3\sqrt[3]{2}\cdot \frac{a(a+b)}{2+2a+2b} $$
Thus,it's suffice to prove that
$$ \frac{a(a+b)}{a+b+1}+\frac{b(b+c)}{b+c+1}+\frac{c(c+a)}{c+a+1}\geq 2 $$
Or
$$ \frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\leq 1 $$
After homogenous,it's
$$\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\leq \frac{1}{3} $$
Now,multiply $4a+4b+4c$ to each sides.we can rewrite the inequality into
$$ \frac{9ca}{4a+4b+c}+\frac{9ab}{4b+4c+a}+\frac{9bc}{4c+4a+b}\leq a+b+c $$
Using Cauchy-Schwarz inequality,we have
$$ \frac{9}{4a+4b+c}=\frac{(2+1)^2}{2(2a+b)+(2b+c)}\le \frac{2}{2a+b}+\frac{1}{2b+c} $$
Therefore
\begin{align}
  \sum{\frac{9ca}{4a+4b+c}}&\leq \sum{\left(\frac{2ca}{2a+b}+\frac{ca}{2b+c}\right)}\\
&=a+b+c
\end{align}
Hence we are done!

未改进的结果

http://kkkkuingggg.5d6d.net/thread-1119-1-3.html
 

设$f(x)$是$[0,1]$上的非负凹函数，则有

$$ \int_{0}^{1}{f^{p}(x)dx}\leq \frac{2^p}{p+1}\left(\int_{0}^{1}{f(x)dx}\right)^p $$

传说中的Favard不等式(1933年的)，找了好久没找到证明，只谷歌到一篇介绍推广的文章《Some New Results Related to Favard’s Inequality》

好像在Mathematic in Science and Engineering, vol 187,1992年那本上有，可惜下载不到。后来发现《常用不等式》第4版452页的Berwarld不等式就是这个的推广.（首先没看出来，那个范数不太懂)

解答来着风碎便士同学。参看博士家园

http://www.math.org.cn/forum.php?mod=viewthread&tid=26438&extra=page%3D1

With out loss of generally,we consider the case $f(0)=f(1)=0$,and $f(x)$ has order continuous derivative,Therefore
$$f''(x)\leq 0 $$
Thus
$$ f(x)=-\int_{0}^{1}{K(x,t)f''(t)dt} $$
Where $K(x,t)$ is Green function.
$$ K(x,t)=\left\{
  \begin{array}{ll}
   t(1-x)  & \hbox{$0\leq t\le x\le 1$} \\
    x(1-t)  & \hbox{$0\leq x\le t\le 1$}
  \end{array}
\right.
$$
Then by Minkowski inequality,we have
\begin{align}
\left(\int_{0}^{1}{f^{p}(x)dx}\right)^{\frac{1}{p}}&=\left(\int_{0}^{1}{\left(\int_{0}^{1}{K(x,t)(-f''(t))dt}\right)^{p}dx     }\right)^{\frac{1}{p}}\\
&\leq \int_{0}^{1}{\left(\int_{0}^{1}{K^{p}(x,t)(-f''(t))^{p}dx}\right)dt}\\
&=\frac{1}{(p+1)^{\frac{1}{p}}}\int_{0}^{1}{t(1-t)|f''(t)|dt}
\end{align}
On the other hand
\begin{align}
\int_{0}^{1}{f(x)dx}&=-\int_{0}^{1}{\int_{0}^{1}{K(x,t)f''(t)dt}  dx}\\
&=-\int_{0}^{1}{\int_{0}^{1}{K(x,t)f''(t)dx}  dt}\\
&=-\frac{1}{2}\int_{0}^{1}{t(1-t)f''(t)dt} 
\end{align}
Therefore
$$ \int_{0}^{1}{f^{p}(x)dx}\leq \frac{2^p}{p+1}\left(\int_{0}^{1}{f(x)dx}\right)^p $$

设$f(x)$在$[0,1]$上连续，且$\displaystyle \int_{0}^{1}{f(x)dx}=0$,证明:在区间$(0,1)$内存在$c$使得
$$\displaystyle \int_{0}^{c}{xf(x)dx}=0$$,
 

证明
设
\[ F(x)=\int_{0}^{x}{f(t)dt} \]
由条件得
\[ F(0)=F(1)=0 \]
于是，我们要证明区间$(0,1)$内存在$c$使得
\[ cF(c)=\int_{0}^{c}{F(t)dt} \]
构造辅助函数
\[ G(x)=\frac{\int_{0}^{x}{F(t)dt}}{x} \]
并作连续开拓
\[ G(x)=\left\{
     \begin{array}{ll}
       \frac{\int_{0}^{x}{F(t)dt}}{x}, & \hbox{$0<x\leq 1$;} \\
       0, & \hbox{x=0.}
     \end{array}
   \right.\]
显然$G(x)$是$[0,1]$上的连续函数，所以必能取到最大值和最小值。下证明在$x=1$处$G(x)$既不能取到最大值，也不能取到最小值。\\
假设在$x=1$处$G(x)$取到最大值,也即有
\[ G(x)\leq G(1) \]
\[ \Leftrightarrow\int_{0}^{x}{F(t)dt}\leq x\int_{0}^{1}{F(t)dt}=x\int_{0}^{x}{F(t)dt}+x\int_{x}^{1}{F(t)dt} \]
\[\Rightarrow \frac{\int_{0}^{x}{F(t)dt}}{x}\leq \frac{\int_{x}^{1}{F(t)dt}}{1-x} \]
上式令$x\rightarrow1 $
得到
\[ \int_{0}^{1}{F(x)dx}\leq F(1)=0 \]
这说明
\[ G(1)\leq G(0) \]
所以$G(x)$只能为常数函数，否则矛盾。
同样可以说明$G(x)$在$x=1$处不能取到最小值。
故必存在一点$c\in(0,1)$使得$G(x)$取到最值。
显然满足
\[ cF(c)=\int_{0}^{c}{F(t)dt} \]
也即
\[ \int_{0}^{c}{xf(x)dx}=0 \]
由此，命题得证。

设$\{a_{n}\}$是正序列，且$ \displaystyle \varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{n}}}=1 $,证明
\[ \varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{1}+a_{1}+\cdots+a_{n}}}=1 \]
证明 注意到
\[ 1=\varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{n}}}\leq \varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{1}+a_{1}+\cdots+a_{n}}} \]
所以只要估计
\[ \sum_{k=1}^{n}{a_{k}} \]
的一个上界即可。
而由 $ \displaystyle \varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{n}}}=1 $
知，对任意给定的$\varepsilon>0$,存在$N>0$，当$n>N$时
\[ a_{n}\leq (1+\varepsilon)^{n} \]
设\[ A=\sum_{k=1}^{N}{a_{k}} \]
则
\[ \sum_{k=1}^{n}{a_{k}}<A+\frac{(1+\varepsilon)^{n-N+1}}{\varepsilon} \]
而当$n$充分大时$ \frac{(1+\varepsilon)^{n-N+1}}{\varepsilon}>A $
所以
\[ \varlimsup_{n\rightarrow\infty}{\sqrt[n]{ \frac{(1+\varepsilon)^{n-N+1}}{\varepsilon} }}=1 \]
因此
 \[ \varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{1}+a_{1}+\cdots+a_{n}}}=1 \]
 

设$f\in C^{2}[a,\infty)$,若积分
\[ \int_{a}^{+\infty}{f^{2}(x)dx},\qquad \int_{a}^{+\infty}{[f''(x)]^{2}dx} \]
均收敛，则$\displaystyle \int_{a}^{+\infty}{[f'(x)]^{2}dx}$收敛。

证明：由Cauchy-Schwarz不等式知道
\[ \left(\int_{a}^{+\infty}|f(x)f''(x)|dx\right)^{2}\leq \left(\int_{a}^{+\infty}f^{2}(x)dx\right)\left(\int_{0}^{+\infty}(f''(x))^2dx\right)\]
于是反常积分$\displaystyle \int_{a}^{+\infty}f(x)f''(x)dx$收敛。又注意到
\begin{align*}
 \int_{a}^{x}(f'(t))^2dt&=f'(t)f(t)\bigg|_{a}^{x}-\int_{a}^{x}f(t)f''(t)dt\\
 &= f'(x)f(x)-f'(a)f(a)-\int_{a}^{x}f(t)f''(t)dt \tag{1}
\end{align*}
下证明
\[ -\infty<\varliminf_{x\to+\infty}f'(x)f(x)<+\infty \]
反证,若$\displaystyle\varliminf_{x\to+\infty}f'(x)f(x)=-\infty$,则存在数列$\{x_{n}\}$满足$x_{n}\to+\infty$,但$f'(x_{n})f(x_{n})\to-\infty$,这样，
\[ \int_{a}^{x_{n}}(f'(t))^2dt=f'(x_{n})f(x_{n})-f'(a)f(a)-\int_{a}^{x_{n}}f(t)f''(t)dt\to -\infty \qquad (n\to\infty)\]
显然不科学。
若$\displaystyle\varliminf_{x\to+\infty}f'(x)f(x)=+\infty$,则对$\forall L>0$,存在$M>0$,当$x\geq M$时$f'(x)f(x)>L$,这时有
\[ f^{2}(x)-f^{2}(M)=2f'(\theta)f(\theta)(x-M)>2L(x-M)\qquad (\theta\in(M,x))\]
则
\[ f^{2}(x)>f^{2}(M)+2L(x-M) \]
\[ \Rightarrow \int_{M}^{+\infty}f^2(x)dx>\int_{M}^{+\infty}(f^2(M)+2L(x-M))dx=+\infty \]
矛盾。故
\[ -\infty<\varliminf_{x\to+\infty}f'(x)f(x)<+\infty \]
不妨设
\[ A=\varliminf_{x\to+\infty}f'(x)f(x)\]
则存在数列$\{x_{n}\}$满足$x_{n}\to+\infty$,$f'(x_{n})f(x_{n})\to A$,这时
\[ \int_{a}^{x_{n}}(f'(t))^2dt=f'(x_{n})f(x_{n})-f'(a)f(a)-\int_{a}^{x_{n}}f(t)f''(t)dt\]
令$n\to\infty$,则
\[ \int_{a}^{+\infty}(f'(t))^2dt=A-f'(a)f(a)-\int_{a}^{+\infty}f(x)f''(x)dx \]
是收敛的。
真神指出：在讨论$f'(x)f(x)$位于无穷远的情况时，可以考察$f^{2}(x)$的极值情况，若$f^2(x)$在无穷远处恒有极值，则运用Farmat定理知道存在数列$\{x_{n}\}$满足$x_{n}\to+\infty$,$f^{2}(x_{n})$取极值，则$f'(x_{n})f(x_{n})=0$,若$f^{2}(x)$仅在有限区间内有极值，则说明$f^2(x)$在无穷远处单调递减趋于$0$,因此，考虑
\[ f^{2}(n+1)-f^2(n)=2f'(x_{n})f(x_{n})\to 0\qquad (n\to\infty) \]
依然可以找到这样的序列满足$f'(x_{n})f(x_{n})\to 0$.