965

已知 ${\displaystyle x,y\ge 0}$, 且 ${\displaystyle x+y=1}$, 则 ${\displaystyle (x+1)\sqrt{y}+(y+1)\sqrt{x}}$ 的最小值为_____, 最大值为_____.

设 ${\displaystyle x={\sin }^2\theta ,y={\cos }^2\theta }$, 不妨设 ${\displaystyle \theta \in {\displaystyle [0,\frac{\pi }{2}]}}$, 再记 ${\displaystyle t=\sin \theta +\cos \theta \in [1,\sqrt{2}]}$, 则$$\begin{array}{r}\text{原式}=(\sin \theta +\cos \theta )(1+\sin \theta \cos \theta )=\frac{1}{2}(t^3+t)\in [1,\frac{3\sqrt{2}}{2}].\end{array}$$
因此最小值为${\displaystyle 1}$, 最大值为${\displaystyle {\displaystyle \frac{3\sqrt{2}}{2}}}$.