设 $x,y,z>0$，且 $x+y+z=1$，求证：$$\dfrac{xy}{\sqrt{xy+yz}}+\dfrac{yz}{\sqrt{yz+zx}}+\dfrac{zx}{\sqrt{zx+xy}}\leqslant \dfrac{\sqrt 2}2.$$
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【答案】
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【解析】
原不等式等价于$$\sum_{cyc}\left(xy\sqrt{\dfrac{2}{xy+yz}}\right)\leqslant 1,$$而\begin{eqnarray*}\begin{split} xy\sqrt{\dfrac{2}{xy+yz}}&=\dfrac{x\sqrt{2(xy+yz)}}{z+x}
=\dfrac{x\sqrt{2(xy+yz)\cdot (y+z)^2}}{(y+z)(z+x)}\\
&\leqslant \dfrac{x\left(2xy+4yz+y^2+z^2\right)}{2(y+z)(z+x)},\end{split} \end{eqnarray*}而$$(x+y+z)-\sum_{cyc}\dfrac{x\left(2xy+4yz+y^2+z^2\right)}{2(y+z)(z+x)}=\dfrac{\sum_{cyc}\left[zx(x^2-y^2)\right]}{2(x+y)(y+z)(z+x)},$$不妨设 $z$ 最小，考虑分子有\begin{eqnarray*}\begin{split} \sum_{cyc}\left[zx(x^2-y^2)\right]=&xz(x^2-y^2)+xy(y^2-z^2)+yz(z^2-y^2+y^2-x^2)\\=&z(x-y)(x^2-y^2)+y(x-z)(y^2-z^2)\geqslant 0.\end{split} \end{eqnarray*}于是原不等式得证．
=\dfrac{x\sqrt{2(xy+yz)\cdot (y+z)^2}}{(y+z)(z+x)}\\
&\leqslant \dfrac{x\left(2xy+4yz+y^2+z^2\right)}{2(y+z)(z+x)},\end{split} \end{eqnarray*}而$$(x+y+z)-\sum_{cyc}\dfrac{x\left(2xy+4yz+y^2+z^2\right)}{2(y+z)(z+x)}=\dfrac{\sum_{cyc}\left[zx(x^2-y^2)\right]}{2(x+y)(y+z)(z+x)},$$不妨设 $z$ 最小，考虑分子有\begin{eqnarray*}\begin{split} \sum_{cyc}\left[zx(x^2-y^2)\right]=&xz(x^2-y^2)+xy(y^2-z^2)+yz(z^2-y^2+y^2-x^2)\\=&z(x-y)(x^2-y^2)+y(x-z)(y^2-z^2)\geqslant 0.\end{split} \end{eqnarray*}于是原不等式得证．
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