Jan 30

## some nice inequality books

1.Vo Quoc Ba Can's book

2.Algebraic Inequality- Old and New method

Author: Vasile Cirtoaje

Jan 30

## A New Note On Cauchy-Schwarz

Here is my article about pqr lemma and Cauchy-Schwarz.

Jan 30

## Vo Quoc Ba Can的一个漂亮的引理

For all postive real numbers $a_{1},a_{2},\cdots,a_{n}$,the following inequality holds
$\sum_{k=1}^{n}{a^{\frac{k}{k+1}}_{k}}\leq \sum_{k=1}^{n}{a_{k}}+\sqrt{\frac{2(\pi^2-3)}{9}\sum_{k=1}^{n}{a_{k}}}$
\emph{Proof}:We begin with a preliminary result.\\
\textbf{Lemma}.If
$m_{k}=\left\{ \begin{array}{ll} 1, & \hbox{if \ k=1;} \\ 2\sqrt{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}, & \hbox{if \ k>1 .} \end{array} \right.$
then
$a_{k}^{\frac{k}{k+1}}\leq a_{k}+m_{k}\sqrt{a_{k}}$
Proof,If $k=1$,this is trivially true.Otherwise,if $k>1$,the inequality can be written as
$a_{k}^{\frac{k}{k+1}}\left(1-a^{\frac{1}{k+1}}_{k} \right)\leq m_{k}\sqrt{a_{k}}$
For the nontrivial case $0<a_{k}<1$,this inequality is equivalent to
$a^{\frac{k-1}{k+1}}_{k}\left(1-a^{\frac{1}{k+1}}_{k} \right)^{2}\leq m^{2}_{k}$
To prove it,we use the AM-GM Inequality,as follows\\
\begin{align*}
a^{\frac{k-1}{k+1}}_{k}\left(1-a^{\frac{1}{k+1}}_{k}\right)^{2}
&=4(k-1)^{k-1}\left(\frac{a^{\frac{1}{k+1}}_{k}}{k-1}\right)^{k-1}\left(\frac{1-a^{\frac{1}{k+1}}_{k}}{2}\right)^{2}\\
&\leq4(k-1)^{k-1}\left[\frac{(k-1)\left(\frac{a^{\frac{1}{k+1}}_{k}}{k-1}\right)+2\left(\frac{1-a^{\frac{1}{k+1}}_{k}}{2}\right)}{k+1}\right]^{k+1}\\
&=\frac{4(k-1)^{k-1}}{(k+1)^{k+1}}\\
&=m^{2}_{k}
\end{align*}
and so the lemma is proved.\\
Returning to our problem,we see that it suffices to show that
$\sum_{k=1}^{n}{m_{k}\sqrt{a_{k}}}\leq \sqrt{\frac{2(\pi^2-3)}{9}\sum_{k=1}^{n}{a_{k}}}$
But the \emph{Cauchy-Schwarz} Inequality gives us that
$\sum_{k=1}^{n}{m_{k}\sqrt{a_{k}}}\leq \sqrt{\left(\sum_{k=1}^{n}{m^{2}_{k}}\right)\left(\sum_{k=1}^{n}a_{k}\right)}$
So,it remains to prove that
$\sum_{k=1}^{n}{m^{2}_{k}}\leq \frac{2(\pi^2-3)}{9}$
or
$\frac{1}{4}+\sum_{k=2}^{n}{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}\leq \frac{\pi^2}{18}-\frac{1}{6}$
Now,because
$\frac{(k+1)^{k+1}}{(k-1)^{k-1}}=(k+1)^{2}\left(\frac{k+1}{k-1}\right)^{k-1}=(k+1)^2\left(1+\frac{2}{k-1}\right)^{k-1}$
\emph{Bernoulli's} Inequality yields
$\frac{(k+1)^{k+1}}{(k-1)^{k-1}}\geq 3(k+1)^{2}$
and thus
$\frac{1}{4}+\sum_{k=2}^{n}{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}\leq \frac{1}{4}+\frac{1}{3}\sum_{k=2}^{n}{\frac{1}{(k+1)^{2}}}=\frac{1}{3}\sum_{k=1}^{n+1}{\frac{1}{k^2}}-\frac{1}{6}$
On the other hand
$\lim_{n\rightarrow \infty}{\left(1+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\right)}=\zeta{(2)}=\frac{\pi^2}{6}$
and therefore it follows that
$\frac{1}{4}+\sum_{k=2}^{n}{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}\leq \frac{\pi^2}{18}-\frac{1}{6}$
which finishes our proof.Notice also that equality does not occur.$\square$

Jan 30

## 一个有意思的不等式

Let $a,b,c>0$ with $abc=1$ Prove that:
$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$
Proof:
We have
$\frac{2}{a+2}-\frac{b}{ab+b+1}-\frac{1}{a+b+1}=\frac{a(b-1)^2}{(a+2)(ab+b+1)(a+b+1)}\geq0$
we get
$2\sum{\frac{1}{a+2}}\geq \sum{\frac{b}{ab+b+1}}+\sum{\frac{1}{a+b+1}}$
since
$\sum{\frac{b}{ab+b+1}}=1$
This inequality yields
$\sum{\frac{1}{a+2}}-\sum{\frac{1}{a+b+1}}\geq 1-\frac{1}{a+2}$
Thus,it's suffice to check
$\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\geq 1$
Which is obvious from AM-GM
$\frac{a}{a+2}=\frac{a}{a+2(abc)^{\frac{1}{3}}}\geq \frac{a^{\frac{2}{3}}}{a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}}$
Hence we are done!

Jan 30

## 一道积分不等式的推广

$\int_{0}^{1}{f(x)dx}=\int_{0}^{1}{xf(x)dx}=\cdots=\int_{0}^{1}{x^{n-1}f(x)dx}=1$

(tian_275461)

$\int_{0}^{1}{(f(x))^{2}dx}\geq n^2$

$P(x)=a_{0}+a_{1}x+\cdots+a_{n-1}x^{n-1}$

$\int_{0}^{1}{(P(x))^{2}dx}=a_{0}+a_{1}+\cdots+a_{n-1}$

$\int_{0}^{1}{x^{k}P(x)dx}=1 \qquad k=0,1,\ldots n-1$
$\Rightarrow \frac{a_{0}}{k+1}+\frac{a_{1}}{k+2}+\cdots+\frac{a_{n-1}}{k+n}=1 \qquad k=0,1,\ldots n-1$

$H(x)=\frac{a_{0}}{x+1}+\frac{a_{1}}{x+2}+\cdots+\frac{a_{n-1}}{x+n}-1$

$H(0)=H(1)=\cdots=H(n-1)=0$
$H(x)$应该有
$H(x)=\frac{A\cdot x\cdot(x-1)\cdot(x-2)\cdots(x-n+1)}{(x+1)(x+2)\cdots(x+n)}$

$a_{k}=(-1)^{n-k-1}\frac{(n+k)!}{(k!)^{2}\cdot(n-k-1)!} \qquad k=0,1,\ldots n-1$

$\sum_{k=0}^{n-1}{a_{k}}=n^{2}$

$\int_{0}^{1}{(P(x))^{2}dx}=a_{0}+a_{1}+\cdots+a_{n-1}=n^2$

$\int_{0}^{1}{(P(x))^2dx}\int_{0}^{1}{(f(x))^2dx}\geq \left(\int_{0}^{1}{P(x)f(x)dx} \right)^{2}=n^4$
$\Rightarrow \int_{0}^{1}{(f(x))^{2}dx}\geq n^2$
Done!