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. :)

For $a=b=c=1$ we get the value $125$.
We'll prove that it's the minimal value.
Indeed, let $f(x)=-\ln(3+2x^2)$.
Hence, $f''(x)=\frac{4(2x^2-3)}{(2x^2+3)^2}<0$ for all $0<x<1$.
Thus, by Vasc's LCF Theorem it's enough to prove our inequality for $b=a$ and $c=3-2a.$
Hence, it's enough to prove that
$$(3+2a^2)^2(3+2(3-2a)^2)\geq125$$ or
$$(a-1)^2(8a^4-8a^3+21a^2-22a+16)\geq0$$
which is obvious.
Done!
