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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?

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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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  • $\begingroup$ Thank you, I meant a maximum for given a,b,c (in a form of another expression). For example for a=1, b=2, c=3 it would give $$4,125\le3,464$$ which shows that the inequality is wrong $\endgroup$
    – Norbert
    Nov 6, 2016 at 18:47
  • $\begingroup$ @Norbert If $a$, $b$ and $c$ are given so the value of $\sum\limits_{cyc}\frac{2a}{\sqrt{3a+b}}$ is a maximum is a minimum or it's what which you want, but it's just a value of the expression. By the way, for your values $\sum\limits_{cyc}\frac{2a}{3a+b}=1$ does not hold. $\endgroup$ Nov 6, 2016 at 18:56
  • $\begingroup$ I see that I have expressed myself quite incorectly. I wanted to use the above inequality as a step to prove another inequality, but I am so confused by why does the inequality not work for all real positive a, b, c. $\endgroup$
    – Norbert
    Nov 6, 2016 at 19:23
  • $\begingroup$ @Norbert Your inequality is true for all positives $a$, $b$ and $c$ such that $\sum\limits_{cyc}\frac{2a}{3a+b}=1$. $\endgroup$ Nov 6, 2016 at 19:32
  • $\begingroup$ @Norbert If so, use the arrow ;) $\endgroup$ Nov 6, 2016 at 20:43

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