Let $a;b;c\in R+$ such that $a+\frac{b}{16}+\frac{c}{81}\le \:3;\frac{b}{16}+\frac{c}{81}\le 2;c\le 81$. Find maxima of $A$

Question

Let $a;b;c\in R+$ such that $a+\frac{b}{16}+\frac{c}{81}\le \:3;\frac{b}{16}+\frac{c}{81}\le 2;c\le 81$. Find the maxima of function $$A=\sqrt[4]{a}+\sqrt[4]{b}+\sqrt[4]{c}$$

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Wlog $a\le b\le c$

$f''(a)=-3/16a^{7/4}$ then $f(a)=a^{\frac{1}{4}}$ is concave function ,$(a;b;c)\succ (1;16;81)$

By Karamata :$A=\sqrt[4]{a}+\sqrt[4]{b}+\sqrt[4]{c}\le 1+2+3=6$

It is one of the most difficult problem on my exam. Then this is my try, check it for me

Answer 1

Let $a=x$, $\frac{b}{16}=y$ and $\frac{c}{81}=z.$

Thus, $$x+y+z\leq3,$$ $$y+z\leq2$$ and $$z\leq1$$ and since $f(x)=\sqrt[4]{x}$ is a concave function, by Jensen we obtain: $$\sqrt[4]a+\sqrt[4]b+\sqrt[4]c=\sqrt[4]x+2\sqrt[4]y+3\sqrt[4]z=$$ $$=\sqrt[4]x+\sqrt[4]y+\sqrt[4]z+\sqrt[4]y+\sqrt[4]z+\sqrt[4]z\leq$$ $$\leq3\sqrt[4]{\frac{x+y+z}{3}}+2\sqrt[4]{\frac{y+z}{2}}+\sqrt[4]z\leq3+2+1=6.$$ The equality occurs for $x=y=z=1,$ which says that we got a maximal value.