Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$ 
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$.

I expanded the brackets and applied AM-GM on all of the eight terms to get :
$$\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big) \geq 3\sqrt{3}abc$$
, which is horribly weak !
I can not use equality constraint whichever method I use. Thanks to Wolfram|Alpha, I know the answer is $125$ when $(a,b,c) \equiv (1,1,1).$
Any help would be appreciated. :)

More way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$(3u^2+2a^2)(3+2b^2)(3+2c^2)\geq125u^6$$ or
$$27u^6+18u^4(9u^2-6v^2)+12u^2(9v^4-6uw^3)+8w^6\geq125u^6$$ or $f(w^3)\geq0$, where
$$f(w^3)=2w^6-18u^3w^3+16u^6-27u^4v^2+27u^2v^4.$$
But $f'(w^3)=2w^3-18u^3<0$, which says that $f$ is a decreasing function.
Thus, it remains to prove our inequality for a maximal value of $w^3$, 
which happens for equality case of two variables.
Let $b=a$. Hence, $c=3-2a$ and we get
$$(a-1)^2(8a^4-8a^3+21a^2-22a+16)\geq0$$
again. Done!
