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.
 A: This inequality is come from?  This inequality can use this 
$$\Longleftrightarrow \frac{(a^2+b^2+c^2)(a+b+c)^4}{16}-2(\sum{a^3b^3}+4a^2b^2c^2)=\frac{3abc(a+b+c)}{4}\sum{(a-b)^2}+\sum\frac{c(2a+2b+c)(a^2+b^2-c^2)^2}{16}+\frac{(a-b)^2(b-c)^2(c-a)^2}{2}+\frac{a^2b^2c^2}{4}$$
A: Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Since $\sum\limits_{cyc}a^3b^3=27v^6-27uv^2w^3+3w^6$, we see that our inequality is equivalent to $f(w^3)\geq0$,
where $f$ is a concave function, which says that it's enough to prove our inequality 
for an extremal value of $w^3$, which happence in the following cases.


*

*$w^3=0$. 


Let $c=0$. Thus, we need to prove that $$a^2+b^2\geq2a^3b^3,$$
which is AM-GM: $a^2+b^2\geq2ab$ and it's enough to prove that $ab\leq1$,
which follows from our condition: $2=a+b\geq2\sqrt{ab}$;


*$b=a$ and $c=2-2a$, where $0\leq a\leq1$.


In this case we need to prove that
$$2a^2+4(1-a)^2\geq2(a^6+2a^3(2-2a)^3+4a^4(2-2a)^2)$$ or
$$(1-a)(a^5+17a^4-15a^3+a^2-2a+2)\geq0,$$
which is true.
Done!
A: 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.
