894

若 $a,b,c>0$，满足 $a+b+c=1$，求 $a+\sqrt{b}+\sqrt[4]{c}$ 的最大值.

由均值不等式得$$\begin{array}{r}B+\frac{1}{4}\ge \sqrt{b},c+{\left(\frac{1}{4}\right)}^{4/3}+{\left(\frac{1}{4}\right)}^{4/3}+{\left(\frac{1}{4}\right)}^{4/3}\ge \sqrt[4]{c},\end{array}$$故消去 $a$ 得$$\begin{array}{r}A+\sqrt{b}+\sqrt[4]{c}=1+(\sqrt{b}-b)+(\sqrt[4]{c}-c)\le \frac{5}{4}+\frac{3\sqrt[3]{2}}{8},\end{array}$$
等号成立当且仅当 ${\displaystyle (a,b,c)=(\frac{3}{4}-\frac{\sqrt[3]{2}}{8},\frac{1}{4},\frac{\sqrt[3]{2}}{8})}$.