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已知 ${\displaystyle a,b,c}$ 是正实数，且 ${\displaystyle b+c=\sqrt{6}}$, 求 ${\displaystyle {\displaystyle \frac{a{c}^2+2a}{bc}+\frac{16}{a+2}}}$ 的最小值.

首先注意到当 ${\displaystyle 0<c<\sqrt{6}}$,$$\begin{array}{r}(\sqrt{3}c-\sqrt{2}{)}^2=3c^2-2\sqrt{6}c+2\ge 0\Rightarrow \frac{c^2}{c(\sqrt{6}-c)}\ge 2,\end{array}$$
从而$$\begin{array}{r}\text{原式}\ge 2a+\frac{16}{a+2}=2(a+2)+\frac{16}{a+2}-4\ge 8\sqrt{2}-4.\end{array}$$