<s>[INST] Prove that $\sum_{cyc}\frac{2a}{\sqrt{3a+b}}\le \sqrt{3(a+b+c)}$ Prove that if $a,b,c\in \mathbb{R}^+$ then $\sum_{cyc}\frac{2a}{\sqrt{3a+b}}\le \sqrt{3(a+b+c)}$
  The equality case seems to hold if $a=b=c$ so I was thinking of using AM-GM but cant operate in any creative ways. Please help or provide alternate solutions.
 [/INST] By C-S $$\left(\sum_{cyc}\frac{a}{\sqrt{3a+b}}\right)^2\leq(a+b+c)\sum_{cyc}\frac{a}{3a+b}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{a}{3a+b}\leq\frac{3}{4}$$ or
$$\sum_{cyc}\left(\frac{a}{3a+b}-\frac{1}{3}\right)\leq\frac{3}{4}-1$$ or
$$\sum_{cyc}\frac{b}{3a+b}\geq\frac{3}{4},$$
which is true by C-S:
$$\sum_{cyc}\frac{b}{3a+b}=\sum_{cyc}\frac{b^2}{3ab+b^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(3ab+b^2)}\geq\frac{3}{4}$$
because the last inequality it's just
$$\sum_{cyc}(a-b)^2\geq0.$$
Done!
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