Classical Inequalities addicts can do this. Part 4

Using A.M.-G.M. for reals $x,y$ we have $x^2+y^2 \geq 2xy$

Here coefficient of $ab$ and $cd$ is unity unlike the coefficient of $bc$

$a^2 +m b^2 \geq 2 \sqrt{m} ab \\ d^2+n c^2 \geq 2 \sqrt{n} cd \\ (1-m) b^2 + (1-n) c^2 \geq 2 \sqrt{(1-m)(1-n)} bc$

Let the minimum value is $k$ then $a^2+b^2+c^2+d^2 \geq k ab +4k bc+ kcd$

Adding the above three equations we get $a^1+b^2+c^2+d^2 \geq 2 \sqrt{m} ab+ 2 \sqrt{(1-m)(1-n)} bc + + 2 \sqrt{n} cd$

Comparing above two equations $k=2 \sqrt{n}=2 \sqrt{m} \implies m=n$

Putting this in third part $4k= 2 \sqrt{(1-m)(1-n)} \\ 2k= \sqrt{(1-m)(1-n)} \\ 2k=1-m=4\sqrt{m}$

On solving we get $m=9-4 \sqrt{5}$ (As m,n are less than one)

So, $k=2 \sqrt{m}=2 \sqrt{9-4\sqrt{5}}=2(\sqrt{5}-2)$ which is the required minimum value.

By AM-GM ,

$\begin{aligned} a^2 + (9 - 4\sqrt{5})b^2 &\ge 2(\sqrt{5} - 2)ab \\ (4\sqrt{5} - 8)b^2 + (4\sqrt{5} - 8)c^2 &\ge 2(\sqrt{5} - 2)4bc \\ (9 - 4\sqrt{5})c^2 + d^2 &\ge 2(\sqrt{5} - 2)cd \end{aligned}$

Adding the 3 inequalities yields $a^2 + b^2 + c^2 + d^2 \ge 2(\sqrt{5} - 2)(ab + 4bc + cd)$

Dividing both sides by $ab + 4bc + cd$

$\implies \dfrac{a^2 + b^2 + c^2 + d^2}{ab + 4bc + cd} \ge 2\sqrt{5} - 4 \\ \implies \alpha = 2\sqrt{5} - 4 \\ \therefore \dfrac{(\alpha + 4)^6}{64} = \boxed{125}$