早上起来看见kuing粉丝群里的天书同学发了一个有意思的不等式，经过一番思考后弄了出来。
Problem:
Let $a,b,c$ be postive numbers,with $abc=1$,show that
\[ \frac{a}{a^2+3}+\frac{b}{b^2+3}+\frac{c}{c^2+3}\leq \frac{3}{4} \]
证明
先用$ a=\frac{y}{x},b=\frac{z}{y},c=\frac{x}{z} $替换，得到
\[ \frac{xy}{3x^2+y^2}+\frac{yz}{3y^2+z^2}+\frac{xz}{3z^2+x^2}\leq \frac{3}{4} \]
即
\[ \frac{2x^2+(x-y)^2}{3x^2+y^2}+\frac{2y^2+(y-z)^2}{3y^2+z^2}+\frac{2z^2+(z-x)^2}{3z^2+x^2}\geq \frac{3}{2} \]
于是，自然的运用Cauchy-Schwarz，得到
\[ \sum{\frac{2x^2+(x-y)^2}{3x^2+y^2}}\geq \frac{\left(\sum{\sqrt{2x^2+(x-y)^2 }}\right)^{2}}{4\sum{x^2}}\]
故只要证明
\[ \frac{\left(\sum{\sqrt{2x^2+(x-y)^2 }}\right)^{2}}{4\sum{x^2}}\geq \frac{3}{2}\]
经过一番化简，得到
\[ \sum{\sqrt{[2x^2+(x-y)^2][2y^2+(z-y)^2]}}\geq \sum{x^2}+\sum{xy} \]
显然，再次运用Cauchy-Schwarz马上得到答案
\[ \sqrt{[2x^2+(x-y)^2][2y^2+(z-y)^2]}\geq 2xy+(y-x)(y-z) \]
只要把上面类似的3个式子相加就好。
由此，不等式得证。
2011年的圣诞节，越南Le Hai贴出了下面不等式并祝福所有Mathlinker圣诞快乐！
Let $a,\;b,\;c$ be reals such that $a+b+c=0$ and $a^2+b^2+c^2=3.$ Prove that \[{a^5}b + {b^5}c + {c^5}a \le - 3.\]
事实上，用Maple因式分解不难得到它等价于
\[ (a^3+6a^2b+3ab^2-b^3)^2\geq0 \]
只要做点齐次化工作就好。
但过了很久都没看见更理想的证明，直到前不久，tian275461网友给了一个基于三角函数的证明，Can神表示他有Cauchy-Schwarz proof，但他就是不肯说。
Proof (tian 275461)
we use $c=-a-b$put in $a^2+b^2+c^2=3$then $a^2+ab+b^2=\dfrac{3}{2}$
let $a=x-y,b=x+y,\Longrightarrow c=-2x $,and $3x^2+y^2=\dfrac{3}{2}$
so we can let
$x=\dfrac{\sqrt{2}}{2}\cos{t},y=\dfrac{\sqrt{6}}{2}\sin{t},t\in(0,2\pi) $
\[\begin{align*}
a^5+b^5c+c^5a&=(x-y)^5(x+y)-2x(x+y)^5-32x^5(x-y)\\
&=\left(\dfrac{\sqrt{2}}{2}\cos{t}-\dfrac{\sqrt{6}}{2}\sin{t}\right)^5\left(\dfrac{\sqrt{2}}{2}\cos{t}+\dfrac{\sqrt{6}}{2}\sin{t}\right)\\
&-2\dfrac{\sqrt{2}}{2}\cos{t}\left(\dfrac{\sqrt{2}}{2}\cos{t}+\dfrac{\sqrt{6}}{2}\sin{t}\right)^5\\
&-32\left(\dfrac{\sqrt{2}}{2}cost\right)^5\left(\dfrac{\sqrt{2}}{2}\cos{t}-\dfrac{\sqrt{6}}{2}\sin{t}\right)\\
&=-\dfrac{3}{8}\left[10-2\sin{\left(6t-\dfrac{\pi}{6}\right)}\right]\le -3
\end{align*} \]
同时，下面推广可以同样证明：
1 if $a+b+c=0,a^2+b^2+c^2=3$,then
\[-\dfrac{3\sqrt{26}}{4}\le a^4b+b^4c+c^4a\le\dfrac{3\sqrt{26}}{4}\]
2 if $a+b+c=0,a^2+b^2+c^2=3$,then
\[-\dfrac{3\sqrt{2}}{2}\le a^2b+b^2c+c^2a\le\dfrac{3\sqrt{2}}{2}\]
3: if $a+b+c=0,a^2+b^2+c^2=3$,then
\[-\dfrac{9\sqrt{38}}{8}\le a^6b+b^6c+c^6a\le\dfrac{9\sqrt{38}}{8}\]
希望以后能看见Can神的CS proof.
。。。。。
2013年3月1日，Vo Quoc Ba Can 终于发了究极的Cauchy-Schwarz proof
Proof. Since $a+b+c=0$ and $a^2+b^2+c^2=3,$ it is easy to obtain the below results:
\[\left\{
\begin{array}{ll}
ab+bc+ca=-\dfrac{3}{2}, & \\
a^3b+b^3c+c^3a=-(ab+bc+ca)^2=-\dfrac{9}{4}, &\\
ab^2+bc^2+ca^2+3abc=-(a^2b+b^2c+c^2a), & \\
a^3b^3+b^3c^3+c^3a^3=(ab+bc+ca)^3+3a^2b^2c^2=-\dfrac{27}{8}+3a^2b^2c^2, & \\
\displaystyle\sum (4ab+2c^2+6bc+3)^2=54, &
\end{array}
\right.\]
With these results, we have
\begin{align*} {a^5}b + {b^5}c + {c^5}a& = \sum {{a^5}b}\\
&= \sum {{a^3}b(3 - {b^2} - {c^2})} \\
&= 3\sum {{a^3}b} - \sum {{a^3}{b^3}} - abc\sum {a{b^2}} \\
& = - \frac{{27}}{4} + \frac{{27}}{8} - 3{a^2}{b^2}{c^2} - abc\sum {a{b^2}}\\
&= - \frac{{27}}{8} + abc\sum {{a^2}b} .
\end{align*}
Therefore, it suffices to prove that
\begin{equation}
abc(a^2b+b^2c+c^2a) \le \frac{3}{8}.
\end{equation}
On the other hand, using the Cauchy-Schwarz inequality, we have
\[{\left[ {\sum {a(4ab + 2{c^2} + 6bc + 3)} } \right]^2} \le \left( {\sum {{a^2}} } \right)\left[ {\sum {{{(4ab + 2{c^2} + 6bc + 3)}^2}} } \right] = 162.\]
From this, it follows that
\[-9\sqrt{2} \le \sum a(4ab+2c^2+6bc+3) \le 9\sqrt{2},\]
or
\[-\frac{3}{\sqrt{2}} \le a^2b+b^2c+c^2a+3abc \le \frac{3}{\sqrt{2}}.\]
The last inequality yields:
\begin{equation}
(a^2b+b^2c+c^2a+3abc)^2 \le \frac{9}{2}
\end{equation}
Using (2) and the AM-GM inequality, we have
\begin{align*}
abc(a^2b+b^2c+c^2a) &=\frac{1}{3}\cdot 3abc\cdot (a^2b+b^2c+c^2a)\\ & \le \frac{1}{3} \left(\frac{3abc+a^2b+b^2c+c^2a}{2}\right)^2\le \frac{3}{8},
\end{align*}
which is (1). So, we are done.
设$x,y,z$为非负数
\[ \sum{\frac{xy}{\sqrt{xy+yz}}}\leq \frac{3\sqrt{3}}{4}\sqrt{\frac{(x+y)(y+z)(x+z)}{x+y+z}} \]
证明：
首先，用$x=bc,y=ac,z=ab$替换，我们可以把不等式写成
\[ \frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leq \frac{3\sqrt{3}}{4}\sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+xz}} \]
由于不等式是cyclic型，故可以设$(y-x)(y-z)\leq 0 $\\
\[ \Leftrightarrow \sum_{cyc}{\frac{x}{\sqrt{(x+y)(x+z)}}\cdot\frac{1}{\sqrt{(x+y)(y+z)}}}\leq \frac{3\sqrt{3}}{4\sqrt{xy+yz+xz}} \]
现在，使用排序不等式
$
LHS\leq \frac{x}{\sqrt{(x+y)(x+z)}}\cdot\frac{1}{\sqrt{(y+z)(z+x)}}+
\frac{y}{\sqrt{(y+z)(y+x)}}\cdot\frac{1}{\sqrt{(y+z)(y+x)}}+\frac{z}{\sqrt{(z+x)(z+y)}}\cdot\frac{1}{\sqrt{(x+y)(x+z)}} $
\[ =\frac{y}{(y+z)(y+x)}+\frac{1}{\sqrt{(y+z)(y+x)}}=\frac{1}{\sqrt{xy+yz+xz}}\cdot\sqrt{\frac{xy+yz+xz}{(y+z)(y+x)}}\cdot\left( 1+\frac{y}{\sqrt{(y+z)(y+x)}} \right ) \]
记$t=\frac{y}{\sqrt{(y+x)(y+z)}}\leq 1 $我们有
\[ \sqrt{\frac{xy+yz+xz}{(y+x)(y+z)}}=\sqrt{1-t^2} \]
使用AM-GM
\[ \frac{1}{\sqrt{xy+yz+xz}}\cdot\sqrt{\frac{xy+yz+xz}{(y+z)(y+x)}}\cdot\left( 1+\frac{y}{\sqrt{(y+z)(y+x)}} \right )=(1+t)\sqrt{1-t^2}\]
\[
=\sqrt{\frac{(1+t)(1+t)(1+t)\cdot 3(1-t)}{3}}\leq \frac{3\sqrt{3}}{4} \]
Done!
$\square$
______________________________________________________________
我想
利用上面那个证明
\[ \sum{\frac{xy}{\sqrt{xy+yz}}}\leq \frac{3\sqrt{3}}{4}\sqrt{\frac{(x+y)(y+z)(x+z)}{x+y+z}}\leq \frac{\sqrt{2}}{2}(x+y+z) \]
应该没有技术上的困难。 
1.Let $a,b,c \in R$ .Show that:
\[ 4(a^6+b^6+c^6)+5(a^5b+b^5c+c^5a)\geq \frac{1}{27}(a+b+c)^6 \]
Ps:下面更弱不等式已经由严文兰老师解决。
\[ 4(a^6+b^6+c^6)+5(a^5b+b^5c+c^5a)\geq 0 \]
2.Let.$ a,b,c \in R^{+}$ .Prove that
\[ \frac{bc}{b^2+c^2+3a^2}+\frac{ca}{a^2+c^2+3b^2}+\frac{ab}{a^2+b^2+3c^2}\leq \frac{3}{5} \]
Ps:虽然这个不等式已经可以由SOS或者SOS-Schur解决。但是是否能用Cauchy-Schwarz呢?
Let $x,y,z$ be postive real numbers,with $ (x-2)(y-2)(z-2)\geq xyz-2 $.show that
\[ \frac{x}{\sqrt{x^5+y^3+z}}+\frac{y}{\sqrt{y^5+z^3+x}}+\frac{z}{\sqrt{z^5+x^3+y}}\leq \frac{3}{\sqrt{x+y+z}} \]
Proof:
by Holder inequality
\[ (x^5+y^3+z)\left(\frac{1}{x}+1+z\right)^{2}\geq (x+y+z)^{3} \]
Hence
\[ \sum_{cyc}{\frac{x}{\sqrt{x^5+y^3+z}}}\leq \sum_{cyc}{\frac{1+x+xz}{(x+y+z)\sqrt{x+y+z}}}=\frac{3+x+y+z+xy+yz+xz}{(x+y+z)\sqrt{x+y+z}} \]
it's suffice to check
\[ \frac{3+x+y+z+xy+yz+xz}{(x+y+z)\sqrt{x+y+z}}\leq \frac{3}{\sqrt{x+y+z}} \]
Or
\[ 2(x+y+z)\geq 3+xy+yz+xz \]
Which is obvious from the condition.
Done!
$ \square$
