已知$x,y,z>0$，则$\min\left\{2x,\dfrac 1y,y+\dfrac 1x\right\}$的最大值为______．
正确答案是$\sqrt 3$．
分析与解 注意到当$2x=\dfrac 1y=y+\dfrac 1x$时，有$\left(x,y\right)=\left(\dfrac{\sqrt 3}2,\dfrac{\sqrt 3}3\right)$，于是考虑使用加权平均．
根据题意，有\[\forall x,y\in\mathbb R^+,\min\left\{2x,\dfrac 1y,y+\dfrac 1x\right\}\leqslant \dfrac{2x\cdot \dfrac 23+\dfrac 1y\cdot \dfrac 13+y+\dfrac 1x}{\dfrac 23+\dfrac 13+1},\]即\[\forall x,y\in\mathbb R^+,\min\left\{2x,\dfrac 1y,y+\dfrac 1x\right\}\leqslant \dfrac 12\cdot\left(\dfrac {4x}3+\dfrac 1x+y+\dfrac 1{3y}\right),\]而\[\dfrac 12\cdot\left(\dfrac {4x}3+\dfrac 1x+y+\dfrac 1{3y}\right)\geqslant \dfrac 12\cdot \left(\dfrac{4}{\sqrt 3}+\dfrac{2}{\sqrt 3}\right)=\sqrt 3,\]等号当且仅当$x=\dfrac{\sqrt 3}2$，$y=\dfrac{\sqrt 3}3$时取得．因此所求的最大值为$\sqrt 3$．