<s>[INST] Jensen's inequality application I tried finding the maximum value of the expression
$$\frac{2a}{3a+b}{\sqrt{3a+b}}+\frac{2b}{3b+c}{\sqrt{3b+c}}+\frac{2c}{3c+a}{\sqrt{3c+a}}$$
where a,b,c are positive real numbers.
I used Jensen's inequality for concave function: $$f(x)=\sqrt{x}$$
The expression is homogeneous, so WLOG I assume:
$$\frac{2a}{3a+b}+\frac{2b}{3b+c}+\frac{2c}{3c+a}=1$$
Therefore: $$\frac{2a}{3a+b}{\sqrt{3a+b}}+\frac{2b}{3b+c}{\sqrt{3b+c}}+\frac{2c}{3c+a}{\sqrt{3c+a}}\le\sqrt{\frac{2a}{3a+b}(3a+b)+\frac{2b}{3b+c}(3b+c)+\frac{2c}{3c+a}(3c+a)}=\sqrt{2a+2b+2c}$$
And that is not true. Is there someone nice who could tell me why is this solution wrong?
 [/INST] The maximum does not exist. Try $a=b=c\rightarrow+\infty$.
Your reasoning is true and gives
$\sum\limits_{cyc}\frac{2a}{3a+b}\sqrt{3a+b}\leq\sqrt{2a+2b+2c}=\sqrt{(2a+2b+2c)\sum\limits_{cyc}\frac{2a}{3a+b}}$, 
which gives again that the maximum does not exist.
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