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

Question

Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3$.Prove that $$\frac{a}{\sqrt{2a^2+3bc}}+\frac{b}{\sqrt{2b^2+3ca}}+\frac{c}{\sqrt{2c^2+3ab}} \le \sqrt{ab+bc+ca}.$$

By C-S we need proof $$4(a+b+c)^2[\frac{a}{(2a^2+3bc)(4a+b+7c)}+\frac{b}{(2b^2+3ca)(4b+c+7a)}+\frac{c}{(2c^2+3ab)(4c+a+7b)}]-ab-bc-ca \le 0$$ Please try to give an analytical solution rather than a graphical computer oriented one. How to proceed with the working?

Answer 1

Unfortunately, I think this problem is wrong.

Put $a=2.5$, $b=0.5$, $c=0$.