<s>[INST] $(a^2+b^2)\sqrt{a^2b^2+c^2}+(b^2+c^2)\sqrt{b^2c^2+a^2}+(c^2+a^2)\sqrt{c^2a^2+b^2}+(a-b)(b-c)(c-a)\ge2(a^2b^2+b^2c^2+c^2a^2)$ Problem: Let $a,b,c\ge0.$ Prove that: $$(a^2+b^2)\sqrt{a^2b^2+c^2}+(b^2+c^2)\sqrt{b^2c^2+a^2}+(c^2+a^2)\sqrt{c^2a^2+b^2}+(a-b)(b-c)(c-a)\ge2(a^2b^2+b^2c^2+c^2a^2)$$
This is strange form for me.
My approach: By AM-GM inequality:$(ab)^2+c^2\ge\frac{1}{2}(ab+c)^2$ and $a^2+b^2\ge2ab$
It is enough to show: $\sqrt{2}\sum_{cyc}{ab(ab+c)}+(a-b)(b-c)(c-a)\ge2(a^2b^2+b^2c^2+c^2a^2)$
But it seems not true by calculation example. I think Equality holds at (0,0,0). Any one help me? Thanks!
 [/INST] We will prove the stronger inequality
$$
\sum (a^2 + b^2)\sqrt{a^2b^2 + c^2} \ge 2\sum(ab)^2 + |(a-b)(b-c)(c-a)|
$$
The inequality is equivalent to
$$
\sum \left[(a-b)^2\sqrt{a^2b^2+c^2} + 2(\sqrt{a^2b^2 + c^2} - ab)ab\right] \ge |(a-b)(b-c)(c-a)|
$$
However, this is true as
$$
\sum \left[(a-b)^2\sqrt{a^2b^2+c^2}\right] \ge \sum \left[(a-b)^2c\right] \ge |(a-b)(b-c)(c-a)|
$$
and
$$
\sum 2(\sqrt{a^2b^2 + c^2} - ab)ab \ge 0
$$
This completes our proof. Hopefully it is okay this time round :)
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