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

解钱莘芪

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