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

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

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

• i used to U.C.T (Un Coefficient Technique) but it's wrong – Word Shallow Jul 11 '17 at 11:52
• The last line isn't equivalent to the one before it; it only implies the one before it. (So it is sufficient to prove the last line that you have written, but it is not necessary.) Unfortunately, the last line which you have written is not true, as can be seen by letting $a$ become very small. – Dylan Jul 11 '17 at 12:24
• I have a proof, but it's very ugly. – Michael Rozenberg Jul 11 '17 at 12:31
• @MichaelRozenberg: No problem – Word Shallow Jul 11 '17 at 12:49

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.
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!