The solution for $a+b+c+d = 4$ and $\left( \frac{1}{a^{12}} + ... + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right) = 16$ $a,b,c,d > 0$, then find the solution for 
$$a+b+c+d = 4$$
and
$$ \left( \frac{1}{a^{12}} + \frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right) = 16$$

Attempt:
We know that $a=b=c=d=1$ is a solution. Are there any other solutions? By AM-GM, 
$$ \frac{4}{4} \ge (abcd)^{1/4} \implies abcd \le 1 $$
Also by HM-GM we have
$$ \frac{4}{ \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}}  } \le \left( a^{12} b^{12} c^{12} d^{12} \right)^{1/4} $$
$$ \frac{4}{(abcd)^{3}} \le \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}}  $$
then
$$ 4 \le \frac{4}{(abcd)^{3}} \le \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}}  $$
and of course $1 + 3abcd \le 4$.
Next, 
$$ \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} = 
\frac{ (bcd)^{12} + (acd)^{12} + (abd)^{12} + (abc)^{12}  }{(abcd)^{12}} $$
so we have 
$$\left( \frac{1}{a^{12}} + \frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right)
= \frac{ (bcd)^{12} + (acd)^{12} + (abd)^{12} + (abc)^{12}  }{(abcd)^{12}} 
 + \frac{ 3(bcd)^{12} + 3(acd)^{12} + 3(abd)^{12} + 3(abc)^{12}  }{(abcd)^{11}} 
= 16 $$

Note that
\begin{align*}
\left(\frac{1}{a^{12}} + \frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}}\right)(1 + 3abcd) 
&\geq \frac{4}{(abcd)^3}(1 + 3abcd) \\
&= \frac{4}{(abcd)^3} + \frac{12}{(abcd)^2} \\
&\geq 4 + 12 \\
&= 16
\end{align*}
where the first inequality holds by AM-GM, and the last inequality holds since $abcd \leq 1$ (again established by AM-GM). But by the equality conditions of AM-GM, equality is attained in each of these inequalities only when $a, b, c, d$ are equal.
