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. :)

Another way.
Let $c=\max\{a,b,c\}$. 
Hence, $c\geq1$, $a+b\leq2$ and
 $$(2a^2+3)(2b^2+3)\geq\left(\frac{(a+b)^2}{2}+3\right)^2$$
because it's just $$(a-b)^2(12-a^2-6ab-b^2)\geq0$$ and
$$12-a^2-6ab-b^2=12-(a+b)^2-4ab\geq12-2(a+b)^2\geq12-8>0.$$
Thus, it remains to prove that $$\left(\frac{(a+b)^2}{2}+3\right)^2(2c^2+3)\geq125$$ or
$$\left(\frac{(3-c)^2}{2}+3\right)^2(2c^2+3)\geq125$$ or
$$(c-1)^2(2c^4-20c^3+93x^2-190c+175)\geq0,$$
which is obvious.
Done!
I have else proofs, but I think the last is the best.
