<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] 
Minimize $f(a,b,c,d)=(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3)$ subject to $a+b+c+d=4$.

The Lagrange function: $L(a,b,c,d,t)=f(a,b,c,d)+t(4-a-b-c-d)$. 
FOC:
$$\begin{cases}L_a=2a(b^2 + 3)(c^2 + 3)(d^2 + 3)-t=0\\
L_b=2b(a^2 + 3)(c^2 + 3)(d^2 + 3)-t=0\\
L_c=2c(a^2 + 3)(b^2 + 3)(d^2 + 3)-t=0\\
L_d=2d(a^2 + 3)(b^2 + 3)(c^2 + 3)-t=0\\
L_t=4-a-b-c-d=0\\
\end{cases}$$
Consider the difference:
$$L_a-L_b=(c^2+3)(d^2+3)(2ab^2+6a-2ba^2-6b)=0 \iff \\
(2ab-6)(b-a)=0 \Rightarrow 1) \ a=b; \ \ 2) \ ab=3.$$
Similarly, other differences are found:
$$1)\ a=c; \ \ 2) \ ac=3\\
1)\ a=d; \ \ 2) \ ad=3\\
1)\ b=c; \ \ 2) \ bc=3\\
1)\ b=d; \ \ 2) \ bd=3\\
1)\ c=d; \ \ 2) \ cd=3\\$$
Cases:
$$\begin{align} \ &1) \ a=b=c\ne d \Rightarrow ad=bd=cd=3 \Rightarrow 4-\frac 3d-\frac 3d-\frac 3d-d=0 \Rightarrow \emptyset;\\
&2) \ a=b\ne c=d \Rightarrow ac=ad=bc=bd=3 \Rightarrow 4-\frac 3d-\frac3d-d-d=0 \Rightarrow \emptyset \\
&3) \ a=b\ne c\ne d\ne a \Rightarrow ac=bc=cd=ad=bd=3 \Rightarrow 4-\frac 3d-\frac3d-\frac3d-d=0 \Rightarrow \emptyset; \\
&4) \ a=b=c=d \Rightarrow a+b+c+d=4 \Rightarrow a=b=c=d=1.\end{align}$$
The only solution is: $(a,b,c,d)=(1,1,1,1)$.
Now we will check bordered Hessian (let $g(x)=a+b+c+d$):
$$\bar{H}=\begin{vmatrix}
0&g_a&g_b&g_c&g_d\\ 
g_a&L_{aa}&L_{ab}&L_{ac}&L_{ad}\\
g_b&L_{ba}&L_{bb}&L_{bc}&L_{bd}\\
g_c&L_{ca}&L_{cb}&L_{cc}&L_{cd}\\
g_d&L_{da}&L_{db}&L_{dc}&L_{dd}\\
\end{vmatrix}=\begin{vmatrix}
0&1&1&1&1\\ 
1&128&64&64&64\\
1&64&128&64&64\\
1&64&64&128&64\\
1&64&64&64&128\\
\end{vmatrix} \Rightarrow \\
\bar{H}_1=-1<0; \ \bar{H}_2=-128<0; \ \bar{H}_3=-12288<0; \ \bar{H}_4=-1048576<0,$$
which implies $f(1,1,1,1)=256$ is minimum.
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