<s>[INST] 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.
 [/INST] We have
\begin{align}
(a^2+3)(b^2+3)
& =(a^2+1+2)(b^2+1+2) \\
& =(a^2+1)(b^2+1)+2(a^2+1+b^2+1)+4 \\
& = (a^2+1)(b^2+1)+2(a^2+b^2+2)+4 \\
& =(a^2+1)(b^2+1)+2(a^2+b^2)+8
\end{align}
By Cauchy Schwartz we obtain
$$(a^2+1)(b^2+1)\geq(a+b)^2\tag{$\star$}$$
$$2(a^2+b^2)=(1^2+1^2)(a^2+b^2)\geq(a+b)^2\tag{$\star \star$}$$
Puting together ($\star$) and ($\star$) we get
$$(a^2+3)(b^2+3)=(a^2+1)(b^2+1)+2(a^2+b^2)+8\geq 2(a+b)^2+8=2[(a+b)^2+4]\tag{1}$$
In the same way we get 
$$(c^2+3)(d^2+3)=(c^2+1)(d^2+1)+2(c^2+d^2)+8\geq 2(c+d)^2+8=2[(c+d)^2+4]\tag{2}$$
Multiplying (1) and (2) we get the conclusion
\begin{align}
(a^2+3)(b^2+3)(c^2+3)(d^2+3)
& \ge 4[(a+b)^2+4][(c+d)^2+4] \\
& \ge 4[2(a+b)+2(c+d)]^2 \tag{Cauchy Schwartz}\\
& =16(a+b+c+d)^2
=256
\end{align}
</s>