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. :)

Set $g(a,b,c)=a+b+c-3$ and $f(a,b,c)=(3+2a^2)(3+2b^2)(3+2c^2)$. We shall use the Lagrange multipliers method.
We have to minimize $h(a,b,c)=f(a,b,c)-\lambda\cdot g(a,b,c)$.
$\frac{d\ h(a,b,c)}{d\ a}=0$
$\frac{d\ h(a,b,c)}{d\ b}=0$
$\frac{d\ h(a,b,c)}{d\ c}=0$
$\frac{d\ h(a,b,c)}{d\ \lambda}=0$
You obtain
$4a(3+2b^2)(3+2c^2)-\lambda=0$
$4b(3+2a^2)(3+2c^2)-\lambda=0$
$4c(3+2a^2)(3+2b^2)-\lambda=0$
$a+b+c=3$
Solve the above system and you have the solution. Indeed, you'll find that the only real solutions are given when $a=b=c$ and hence from last equation you have $a=b=c=1$. Hence the min is $125$.
