940

对任意的正实数 ${\displaystyle a,b,c}$, 满足 ${\displaystyle b+c=1}$, 求 ${\displaystyle {\displaystyle \frac{a+8a{b}^2}{bc}+\frac{16}{a+1}}}$ 的最小值.

由题意$$\begin{array}{r}\frac{1+8b^2}{bc}=\frac{(b+c{)}^2+8b^2}{bc}=\frac{c^2+9b^2}{bc}+2\ge 8,\end{array}$$
从而$$\begin{array}{r}\text{原式}\ge 8a+\frac{16}{a+1}=8(a+1)+\frac{16}{a+1}-8\ge 16\sqrt{2}-8,\end{array}$$
等号成立当且仅当 ${\displaystyle {\displaystyle (a,b,c)=(\sqrt{2}-1,\frac{1}{4},\frac{3}{4})}}$.