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.

We need to prove that
$$\sum_{cyc}\ln(a^2+3)\geq4\ln4$$ or 
$$\sum_{cyc}\left(\ln(a^2+3)-\ln4-\frac{1}{2}(a-1)\right)\geq0.$$
Now, let $f(x)=\ln(x^2+3)-\ln4-\frac{1}{2}(x-1).$
Thus, $$f'(x)=\frac{(x-1)(3-x)}{2(x^2+3)},$$ which says that $f$ decreases on $[3,+\infty)$ and since $f(3)>0$, there is unique $x_0>3$, for which $f(x_0)=0$.
And indeed, $x_0=4.586...$ and since $f(1)=0$, our inequality is proven for $\max\{a,b,c,d\}\leq x_0$.
Let $a>x_0$.
Thus, $$\prod_{cyc}(a^2+3)\geq(x_0^2+3)\cdot3^3>256$$ and we are done!
