Prove $\frac{ab}{\sqrt{2c+a+b}}+\frac{bc}{\sqrt{2a+b+c}}+\frac{ca}{\sqrt{2b+c+a}}\le\sqrt\frac{2}{3}$ if $a+b+c=2$

Question

Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=2$. Prove: $$\frac{ab}{\sqrt{2c+a+b}}+\frac{bc}{\sqrt{2a+b+c}}+\frac{ca}{\sqrt{2b+c+a}}\le\sqrt\frac{2}{3}$$

Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz, Hölder and AM-GM because I have background in them.

Things I have tried: I was thinking to making the denominator smaller using AM-GM.but I was not successful. My other idea was to re write LHS into this form.something like my idea on this question $$A-\frac{ab}{\sqrt{2c+a+b}}+B-\frac{bc}{\sqrt{2a+b+c}}+C-\frac{ca}{\sqrt{2b+c+a}}$$

But I was not able to observe something good.

I don't know this will lead to something useful but,here is my other idea. let $x=2c+a+b,y=2a+b+c,z=2b+c+a$. rewriting LHS:$$\sum\limits_{cyc}\frac{(3y-(x+z))(3z-(y+x))}{16\sqrt x} \le \sqrt\frac{2}{3}$$ $\sum\limits_{cyc}$denotes sums over cyclic permutations of the symbols $x,y,z$ . another thing I observed that $(x-y-z)^2-4(y-z)^2 = (3y-(x+z))(3z-(y+x))$

I looked at related problems and I think this and (Prove $\frac{a}{ab+2c}+\frac{b}{bc+2a}+\frac{c}{ca+2b} \ge \frac 98$) may have some common idea in proof with my question inequality.

Well,it seems like someone posted it a little after on AoPS.right now there is a solution there by $uvw$ and Cauchy-Schwarz. I post the starting part of solution that is with Cauchy (credits to arqady of AoPS). By Cauchy-Schwarz:$$\left(\sum_{cyc}\frac{ab}{\sqrt{2c+a+b}}\right)^2\leq(ab+ac+bc)\sum_{cyc}\frac{ab}{2c+a+b}$$

Hence, it remains to prove that: $$(ab+ac+bc)\sum_{cyc}\frac{ab}{2c+a+b}\leq\frac{(a+b+c)^3}{12}$$

I stuck here.

Answer 1

Well I think figured out . I will write it down from first.

By Cauchy-Schwarz: $$\left(\sum_{cyc}\frac{ab}{\sqrt{2c+a+b}}\right)^2\leq(ab+ac+bc)\sum_{cyc}\frac{ab}{2c+a+b}$$

Hence, it remains to prove that: $$(ab+ac+bc)\sum_{cyc}\frac{ab}{2c+a+b}\leq\frac{(a+b+c)^3}{12}$$

by the well known inequality:$$3(ab+bc+ca)\le(a+b+c)^2$$

So $$(ab+bc+ca)\le\frac{(a+b+c)^2}{3}$$

is true.it remain to show that $$\sum_{cyc}\frac{ab}{2c+a+b}\le \frac{(a+b+c)}{4}=\frac{1}{2}$$

by well known inequality $$\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}$$

So $$\frac{ab}{c+a}+\frac{ab}{c+b}\ge\frac{4ab}{2c+a+b} \rightarrow \frac{1}{4}(\frac{ab}{c+a}+\frac{ab}{c+b}) \ge\frac{ab}{2c+a+b} $$

so we can conclude that $$\frac{1}{4}\sum\limits_{cyc}(\frac{ab}{c+a}+\frac{ab}{c+b})=\frac{1}{4}(a+b+c)=\frac{1}{2}\ge\sum_{cyc}\frac{ab}{2c+a+b}$$

Done.

Answer 2

By AM-GM and C-S we obtain: $$\sum_{cyc}\frac{ab}{\sqrt{2c+a+b}}=\sum_{cyc}\frac{ab\sqrt{\frac{3}{8}}\cdot2\sqrt{\frac{8}{3}(2c+a+b)}}{2(2c+a+b)}\leq\sqrt{\frac{3}{32}}\sum_{cyc}\frac{ab\left(\frac{8}{3}+2c+a+b\right)}{2c+a+b}=$$ $$=\sqrt{\frac{1}{96}}\sum_{cyc}\frac{ab(4(a+b+c)+6c+3a+3b)}{2c+a+b}=\sqrt{\frac{1}{96}}\sum_{cyc}\frac{ab(10c+7a+7b)}{2c+a+b}=$$ $$=\sqrt{\frac{1}{96}}\sum_{cyc}\left(\frac{-4abc}{2c+a+b}+7ab\right)\leq\sqrt{\frac{1}{96}}\left(\frac{-36abc}{\sum\limits_{cyc}(2c+a+b)}+7(ab+ac+bc)\right)=$$ $$=\sqrt{\frac{1}{96}}\left(\frac{-9abc}{a+b+c}+7(ab+ac+bc)\right).$$ Thus, it remains to prove that $$\sqrt{\frac{1}{96}}\left(\frac{-9abc}{a+b+c}+7(ab+ac+bc)\right)\leq\sqrt{\frac{2}{3}}$$ or $$\frac{-9abc}{a+b+c}+7(ab+ac+bc)\leq8$$ or $$\frac{-9abc}{a+b+c}+7(ab+ac+bc)\leq2(a+b+c)^2$$ or $$\sum_{cyc}(2a^3-a^2b-a^2c)\geq0$$ or $$\sum_{cyc}(a-b)^2(a+b)\geq0.$$ Done!