一道不等式

longzaifei

$x,y,z>0,x+y+z=3 $,请证明：\[  \dfrac{xy}{z+xy}+\dfrac{yz}{x+yz}+\dfrac{zx}{y+zx}\le \dfrac32  \]

kuing

SOS就可以了
\begin{align*}
\sum\frac{xy}{z+xy}\leqslant \frac32&\iff\sum\frac z{z+xy}\geqslant \frac32 \\
&\iff \sum\frac{z(x+y+z)}{z(x+y+z)+3xy}\geqslant \frac32 \\
&\iff \sum\left( \frac{z(x+y+z)}{z(x+y+z)+3xy}-\frac12 \right)\geqslant 0 \\
&\iff \sum\frac{(z-x)(z+3y)+(z-y)(z+3x)}{z(x+y+z)+3xy}\geqslant 0 \\
&\iff \sum\left( \frac{(z-x)(z+3y)}{z(x+y+z)+3xy}+\frac{(x-z)(x+3y)}{x(x+y+z)+3yz} \right)\geqslant 0 \\
&\iff \sum\frac{6y^2(x-z)^2}{\bigl(z(x+y+z)+3xy\bigr)\bigl(x(x+y+z)+3yz\bigr)}\geqslant 0.
\end{align*}

写完感觉以前可能写过同样的东西……

longzaifei

谢谢kuing!!!!!!!!!!!!!