Prove that if $a+b+c+d=4$, then $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq256$ 
Given $a,b,c,d$ such that $a + b + c + d = 4$ show that $$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 256$$

What I have tried so far is using CBS:
$(a^2 + 3)(b^2 + 3) \geq (a\sqrt{3} + b\sqrt{3})^2 = 3(a + b)^2$
$(c^2 + 3)(d^2 + 3) \geq 3(c + d)^2$
$(a^2 + b^2)(c^2 + d^2) \geq (ac + bd)^2$
Then, we have:
$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 9(a + b)^2(c + d)^2$. 
Thus, we have to prove that $9(a + b)^2(c + d)^2 \geq 256$.
Then, I used the following substitution:
$c + d = t$ and $a + b = 4 - t$.
We assume wlog that $a \leq b \leq c \leq d$.
Then, $4 = a + b + c + d \leq 2(c + d) = 2t$. Thus, $t \geq 2$.
Then, what we have to prove is:
$9t^2(4 - t)^2 \geq 256$.
We can rewrite this as:
$(3t(4 - t) - 16)(3t(4 - t) + 16) \geq 0$, or
$(3t^2 + 2t + 16)(3t^2 - 12t - 16) \geq 0$,
at which point I got stuck.

Alternatively, you could solve
$min_{(a, b, c, d) |a+b+c+d=4} f(a, b, c, d),$
$f$ being your term with the squares. 
The convexity guarantees a global minimum. And the symmetry of the problem will yield the answer. 
Edit
According to the comments, $f$ is not convex but it is subharmonic (thanks to Jack D'Aurizio), i.e. for the given $f$ we have $\Delta f \geq 0$ on $\mathbb{R}^4$. $f$ is obviously not constant. So by the maximum principle it cannot have a maximum on the interior of its domain. Hence, it has a unique (unconstrained) minimum in $\mathbb{R}^4$. The constraint defining function $g(a, b, c, d) =a+b+c+d$ is harmonic. I did not go further but maybe it then can be shown that the constrained min problem has a unique solution and that it must be symmetric because otherwise we could move towards the origin (0,0,0,0) and further minimize the function (using a closed ball argument based in the definition of an harmonic function according to Wikipedia). 
