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.
We need to prove that $$27\prod_{cyc}((a+b+c)^2+6a^2)\geq125(a+b+c)^6$$ or
$$\sum_{sym}\left(8a^6+15a^5b+159a^4b^2+160a^3b^3-\frac{249}{2}a^4bc-336a^3b^2c+\frac{237}{2}a^2b^2c^2\right)\geq0$$ or
$$168\sum_{sym}(a^3b^2-2a^3b^2c+a^2b^2c^2)+$$
$$+\sum_{sym}\left(8a^6+15a^5b+159a^4b^2-8a^3b^3-\frac{249}{2}a^4bc-\frac{99}{2}a^2b^2c^2\right)\geq0,$$
which is obviously true by Schur and Muirhead.
