Inequality Challenge with $a+b+c=3$

Question

I tried to prove this inequality, but I failed. It seemed that if one uses Cauchy-Schwarz inequality on the right hand side, it may expand RHS's value larger than LHS's. Problem is presented as follows:

If $a,b,c\ge0$ and $a+b+c=3$, prove that $$2(ab+bc+ca)-3abc\ge a\cdot \sqrt{\frac{b^2+c^2}{2}}+b\cdot\sqrt{\frac{c^2+a^2}{2}}+c\cdot\sqrt{\frac{a^2+b^2}{2}}$$

Answer 1

After using $\sqrt{\frac{a^2+b^2}{2}}\leq\frac{3a^2+2ab+3b^2}{4(a+b)}$,

which is just $(a-b)^4\geq0$, you'll get something obvious.

Indeed, it remains to prove that $$2(ab+ac+bc)-\frac{9abc}{a+b+c}\geq\sum\limits_{cyc}\frac{c(3a^2+2ab+3b^2)}{4(a+b)}$$ or

$$2(ab+ac+bc)-\frac{9abc}{a+b+c}\geq\frac{3}{2}(ab+ac+bc)-abc\sum\limits_{cyc}\frac{1}{a+b}$$ or $$\frac{1}{2}(ab+ac+bc)-\frac{9abc}{2(a+b+c)}-\frac{9abc}{2(a+b+c)}+abc\sum\limits_{cyc}\frac{1}{a+b}\geq0$$ or $$\sum\limits_{cyc}c(a-b)^2+abc\left(\sum\limits_{cyc}(a+b)\sum\limits_{cyc}\frac{1}{a+b}-9\right)\geq0$$

Done!