Prove that $\sum_{cyc}\frac{2a}{\sqrt{3a+b}}\le \sqrt{3(a+b+c)}$

Question

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.

Answer 1

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!

Answer 2

we write the left-hand side as $$\left(\frac{2a}{3a+b}\sqrt{3a+b}+\frac{2b}{3b+c}\sqrt{3b+c}+\frac{2c}{3c+a}\sqrt{3c+a}\right)^2$$ and now we use CS and we get $$\le \left(\left(\frac{2a}{3a+b}\right)^2+\left(\frac{2b}{3b+c}\right)^2+\left(\frac{2c}{3c+a}\right)^2\right)(4(a+b+c))\le 3(a+b+c)$$ this is equivalent to $$\left(\frac{2a}{3a+b}\right)^2+\left(\frac{2b}{3b+c}\right)^2+\left(\frac{2c}{3c+a}\right)^2\le \frac{3}{4}$$ we moving all to the right and multiplying out and we have: $$-45\,{a}^{4}{b}^{2}+66\,{a}^{4}bc+11\,{a}^{4}{c}^{2}+66\,{a}^{3}{b}^{3 }-162\,{a}^{3}{b}^{2}c+414\,{a}^{3}b{c}^{2}+66\,{a}^{3}{c}^{3}+11\,{a} ^{2}{b}^{4}+414\,{a}^{2}{b}^{3}c-1050\,{a}^{2}{b}^{2}{c}^{2}-162\,{a}^ {2}b{c}^{3}-45\,{a}^{2}{c}^{4}+66\,a{b}^{4}c-162\,a{b}^{3}{c}^{2}+414 \,a{b}^{2}{c}^{3}+66\,ab{c}^{4}-45\,{b}^{4}{c}^{2}+66\,{b}^{3}{c}^{3}+ 11\,{b}^{2}{c}^{4} \geq 0$$ setting $$b=a+u,c=a+u+v$$ and simplifying we get $$\left( 512\,{u}^{2}+512\,uv+512\,{v}^{2} \right) {a}^{4}+ \left( 1536 \,{u}^{3}+2816\,{u}^{2}v+2304\,u{v}^{2}+512\,{v}^{3} \right) {a}^{3}+ \left( 1568\,{u}^{4}+4160\,{u}^{3}v+3936\,{u}^{2}{v}^{2}+1344\,u{v}^{ 3}+32\,{v}^{4} \right) {a}^{2}+ \left( 576\,{u}^{5}+2008\,{u}^{4}v+ 2352\,{u}^{3}{v}^{2}+1008\,{u}^{2}{v}^{3}+88\,u{v}^{4} \right) a+32\,{ u}^{6}+152\,{u}^{5}v+219\,{u}^{4}{v}^{2}+110\,{u}^{3}{v}^{3}+11\,{u}^{ 2}{v}^{4} \geq 0$$ which is true.