An inequality on three constrained positive numbers Assume $a,b,c$ are all positive numbers, and $2a^3b+2b^3c+2c^3a=a^2b^2+b^2c^2+c^2a^2$.
Prove that: 
$$2ab(a-b)^2+2bc(b-c)^2+2ca(c-a)^2\ge(ab+bc+ca)^2$$

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition gives $\sum\limits_{cyc}(a^3b+a^3c-a^2b^2)=\sum\limits_{cyc}(a^3c-a^3b)$ 
or $9u^2v^2-9v^4+uw^3=u\prod\limits_{cyc}(a-b)$
or $(9u^2v^2-9v^4+uw^3)^2=27u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$
or $28u^2w^6+18u(6u^4-8u^2v^2-v^4)w^3-54u^2v^6+81v^8=0$, which gives
$3(6u^4-8u^2v^2-v^4)^2-28(-2u^2v^6+3v^8)\geq0$ 
or $108u^8-288u^6v^2+156u^4v^4+104u^2v^6-81v^8\geq0$, which gives
$u^2\geq1.49098...v^2$.
By the way, we get $28u^2w^6+18u(6u^4-8u^2v^2-v^4)w^3=9v^4(6u^2v^2-9v^4)$.
In another hand, we need to prove that $2\sum\limits_{cyc}(a^3b+a^3c-2a^2b^2)\geq9v^4$ or 
$6u^2v^2-9v^4+2uw^3\geq0$ or $9v^4(6u^2v^2-9v^4)+18uv^4w^3\geq0$
or $28u^2w^6+18u(6u^4-8u^2v^2-v^4)w^3+18uv^4w^3\geq0$ 
or $27u^3-36uv^2+7w^3\geq0$.
By Schur $w^3\geq4uv^2-3u^3$.
Hence, it remains to prove that $u^2\geq\frac{4}{3}v^2$, which is obvious.
