<s>[INST] 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. :)
 [/INST] By Lagrange multipliers, $(a,b,c)=(1,1,1)$ is the only stationary point inside the closed region $(a,b,c)\geq 0, a+b+c=3$. To prove it is a minimum is is enough to study the given function under the assumptions $c=0$ and $b=3-a$. The one variable function
$$ f(a) = 3(3+2a^2)(3+2(3-a)^2) $$
has absolute minima over the interval $[0,3]$ at $a=\frac{3\pm\sqrt{3}}{2}$. It follows that the minimum value of the original function over $(a,b,c)\geq 0, a+b+c=3$ is $162>125$.

An algebraic alternative: by looking at the LHS as a norm in $\mathbb{R}[i]$, we get that is equals
$$ 3(3-2ab-2ac-2bc)^2 +2 (3a+3b+3c-2abc)^2\\ = 3(a^2+b^2+c^2-6)^2+2(9-2abc)^2$$
and by GM-AM-QM is is trivial that the minimum is achieved only at $a=b=c=1$.
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