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. :)
 A: 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 $a=b=c=1$.
A: 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!
A: 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 $125$.
A: More way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$(3u^2+2a^2)(3+2b^2)(3+2c^2)\geq125u^6$$ or
$$27u^6+18u^4(9u^2-6v^2)+12u^2(9v^4-6uw^3)+8w^6\geq125u^6$$ or $f(w^3)\geq0$, where
$$f(w^3)=2w^6-18u^3w^3+16u^6-27u^4v^2+27u^2v^4.$$
But $f'(w^3)=2w^3-18u^3<0$, which says that $f$ is a decreasing function.
Thus, it remains to prove our inequality for a maximal value of $w^3$, 
which happens for equality case of two variables.
Let $b=a$. Hence, $c=3-2a$ and we get
$$(a-1)^2(8a^4-8a^3+21a^2-22a+16)\geq0$$
again. Done!
A: 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.
A: 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.
