prove that $ a^2 + b^2 + c^2 \ge 2\left( {a^3 b^3 + a^3 c^3 + c^3 b^3 + 4a^2 b^2 c^2 } \right)$ I have:


  let $a$, $b$ and $c$ be non-negative real numbers with sum $2$. Prove that 
  $$a^2  + b^2  + c^2  \ge 2\left( {a^3 b^3  + a^3 c^3  + c^3 b^3  + 4a^2 b^2 c^2 } \right)$$


I should determine whether this is a convergent or divergent integral. The problem is that I don't know how to start.
This inequality is come from?  This inequality can use this 

$$\Longleftrightarrow \frac{(a^2+b^2+c^2)(a+b+c)^4}{16}-2(\sum{a^3b^3}+4a^2b^2c^2)=\frac{3abc(a+b+c)}{4}\sum{(a-b)^2}+\sum\frac{c(2a+2b+c)(a^2+b^2-c^2)^2}{16}+\frac{(a-b)^2(b-c)^2(c-a)^2}{2}+\frac{a^2b^2c^2}{4}$$
You can use the KKT condition: http://en.wikipedia.org/wiki/Karush-Kuhn-Tucker_conditions

As a hint, here's a physicist's argument: given the symmetry among the 3 variables, the optimal value is likely to occur at the high-symmetry scenario when a=b=c. And you'll see that the inequality holds.