<s>[INST] 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!
 [/INST] I think the key tools here are Cauchy-Schwarz and AM-GM. I just bound one term on the LHS. The other two are similar:
$$\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}=\sum_{cyc}\sqrt{(a^4+\frac{a^2b^2}{2})+(b^4+\frac{a^2b^2}{2})}\ge$$$$\frac{1}{\sqrt{2}}\sum_{cyc}\left(\sqrt{a^4+\frac{a^2b^2}{2}}+\sqrt{b^4+\frac{a^2b^2}{2}}\right)$$$$=\frac{1}{\sqrt{2}}\sum_{cyc}\left(\sqrt{a^4+\frac{a^2b^2}{2}}+\sqrt{a^4+\frac{a^2c^2}{2}}\right)\ge\sqrt{2}\sum_{cyc}\left(\sqrt{a^4+\frac{a^2b^2}{2}}\sqrt{a^4+\frac{a^2c^2}{2}}\right)^{1/4}$$$$\ge\sqrt{2}\sum_{cyc}\sqrt{a^4+\frac{a^2bc}{2}}=\sum_{cyc}\sqrt{2a^4+a^2bc}$$
I use Cauchy-Schwarz at the first and third inequality, and AM-GM at the second.
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