Prove inequality $\sum _{cyc}\frac{2ab}{\sqrt{c+3}}\le 3$

Question

For $a>0$, $b>0$, $c>0$ and $a^3+b^3+c^3=3$ Prove that $$\frac{2ab}{\sqrt{c+3}}+\frac{2bc}{\sqrt{a+3}}+\frac{2ca}{\sqrt{b+3}}\le 3$$

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We have: $abc\le \frac{a^3+b^3+c^3}{3}=1$

$\Leftrightarrow 2abc\left(\frac{1}{c\sqrt{c+3}}+\frac{1}{a\sqrt{a+3}}+\frac{1}{b\sqrt{b+3}}\right)\le 3$

$\Leftrightarrow \frac{1}{c\sqrt{c+3}}+\frac{1}{a\sqrt{a+3}}+\frac{1}{b\sqrt{b+3}}\le \frac{3}{2}$

I can't countinue. Help

Answer 1

By C-S $$\left(\sum_{cyc}\frac{ab}{\sqrt{c+3}}\right)^2\leq\sum_{cyc}ab\sum_{cyc}\frac{ab}{c+3}.$$ Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, it remains to prove that $$\sum_{cyc}ab\sum_{cyc}\frac{ab}{c+3}\leq\frac{9}{4}$$ or $$3v^2\cdot\frac{\sum\limits_{cyc}ab(a+3)(b+c)}{\prod\limits_{cyc}(a+3)}\leq\frac{9}{4}$$ or $$4v^2\sum_{cyc}(a^2b^2+3a^2b+3a^2c+9ab)\leq3\left(abc+27+\sum_{cyc}(3ab+9a)\right)$$ or $$4v^2(9v^4-6uw^3+27uv^2-9w^3+27v^2)\leq3(w^3+27+9v^2+27u)$$ or $$w^3(1+12v^2+8uv^2)+27+27u+9v^2\geq12v^6+36uv^4+36v^4.$$ Now, by Schur $w^3\geq4uv^2-3u^3.$.

Thus, it's enough to prove that $$(4uv^2-3u^3)(1+12v^2+8uv^2)+27+27u+9v^2\geq12v^6+36uv^4+36v^4$$ or $$27+27u+9v^2+4uv^2+32u^2v^4+12uv^4\geq12v^6+36v^4+3u^3+24u^4v^2+36u^3v^2,$$ which is obvious by Power Mean inequality and AM-GM.

Indeed, this inequality we can get after summing of the following inequalities.

1. $$12uv^4\geq12v^6;$$
2. $$12+12u^2v^4\geq24u^4v^2;$$
3. $$18u^2v^4+18u\geq36u^3v^2;$$
4. $$2+2u^2v^4\geq4v^4$$ and 5. $$13+9u+9v^2+4uv^2\geq32v^4+3u^3.$$ Done!

Answer 2

Another leaner way.

Some preparation: From the Power mean inequalities we obtain $$1 \:=\: \left(\frac{a^3+b^3+c^3}3\right)^{\tfrac13} \:\geqslant\:\left(\frac{a^2+b^2+c^2}3\right)^{\tfrac12} \:\geqslant\:\frac{a+b+c}3\,,$$ hence $$3\:\geqslant\: a+b+c\,,\text{ and by AGM,}\quad ab+bc+ca\:\leqslant\: a^2+b^2+c^2\:\leqslant\: 3\,.$$

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$$\begin{align} \frac{2ab}{\sqrt{c+3}}+ \frac{2bc}{\sqrt{a+3}}+\frac{2ca}{\sqrt{b+3}} \: & =\: 2\sum_\text{cyc}\sqrt{ab}\:\sqrt{\frac{ab}{c+3}}\\[2ex] & \leqslant\: 2\sqrt{\sum_\text{cyc}ab}\:\sqrt{\sum_\text{cyc}\frac{ab}{c+3}}\quad\text{by Cauchy-Bunyakovsky-Schwarz} \\[2ex] & \leqslant\: 2\sqrt{\,3\sum_\text{cyc}\frac{ab}{c+a+b+c}}\quad\text{using the preparations} \\[2ex] & \leqslant\: 2\sqrt{\,3\sum_\text{cyc}\frac{ab}{4}\left(\frac{1}{c+a}+\frac{1}{b+c}\right)}\quad\text{by }\:\frac1{\text{AM}}\leqslant\frac1{\text{HM}} \\[2ex] & =\:\sqrt{\,3\sum_\text{cyc}\left(\frac{ab}{c+a}+\frac{ab}{b+c}\right)} \;=\;\sqrt{\,3\,(a+b+c)}\:\leqslant\:3\,. \end{align}$$