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!

$\square$

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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。