TITLE: $a,b,c>0,a+b+c=21$ prove that $a+\sqrt{ab} +\sqrt[3]{abc} \leq 28$
QUESTION [9 upvotes]: $a,b,c>0,a+b+c=21$ prove that $a+\sqrt{ab} +\sqrt[3]{abc} \leq 28$
I have tried to use AM-GM inequality, but get no result as follows:
$$a+\sqrt{ab}+\sqrt[3]{abc}\leq a+\frac{a+b}{2}+\frac{a+b+c}{3}$$

REPLY [17 votes]: $a+\sqrt{ab}+\sqrt[3]{abc}=a+\frac{\sqrt{4ab}}{2}+\frac{\sqrt[3]{64abc}}{4}\le a+\frac{a+4b}{4}+\frac{a+4b+16c}{12}=\frac{4(a+b+c)}{3}=28$