Proving $\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}\geq \sum_{cyc} a\sqrt{2a^2+bc}$ for non-negative $a$, $b$, $c$ I was trying this question with factorization and other similar methods,

Let $a, b, c \geq 0$. Prove that
$$\begin{array}{c}
\sqrt{a^4+a^2b^2+b^4}+\sqrt{b^4+b^2c^2+c^4}+\sqrt{c^4+c^2a^2+a^4} \\[4pt]
\geq a\sqrt{2a^2+bc}+b\sqrt{2b^2+ca}+c\sqrt{2c^2+ab}
\end{array}$$

This is one of Hoojoo-Lee's Inequality. This seems very intuitive at first as if we square each term, $$ 2\sum_{cyc}{a^4}+\sum_{cyc}{a^2b^2} \geq 2\sum_{cyc}{a^4}+\sum_{cyc}{a^2bc} \Rightarrow \sum_{cyc}{a^2b^2} \geq \sum_{cyc}{a^2bc} $$ which is quite clear. I noticed it but can not exploit it. May be taking the square on each side could help? But I couldn't find a solution.
Please help!

By C-S $$\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}=$$
$$=\sqrt{\sum_{cyc}\left(a^4+a^2b^2+b^4+2\sqrt{(a^4+a^2b^2+b^4)(a^4+a^2c^2+c^4)}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}(a^4+a^2b^2+b^4+2(a^4+a^2bc+b^2c^2))}=\sqrt{\sum_{cyc}(4a^4+3a^2b^2+2a^2bc)}.$$
In another hand, by AM-GM
$$\sum_{cyc}a\sqrt{2a^2+bc}=\sqrt{\sum_{cyc}(2a^4+a^2bc+2ab\sqrt{(2a^2+bc)(2b^2+ac)})}\leq$$
$$\leq\sqrt{\sum_{cyc}(2a^4+a^2bc+ab(2a^2+bc+2b^2+ac))}=\sqrt{\sum_{cyc}(2a^4+2a^3b+2a^3c+3a^2bc)}.$$
Id est, it's enough to prove that:
$$\sum_{cyc}(2a^4-2a^3b-2a^3c+3a^2b^2-a^2bc)\geq0$$ or
$$2\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+3\sum_{cyc}(a^2b^2-a^2bc)\geq0,$$ which is true by Schur and Muirhead.
