From Europe with love

Applying AM $\geq$ HM $\frac{(a^4+1)+(b^4+1)+(c^4+1)+(d^4+1)}{4}\geq\frac{4}{\frac{1}{a^4+1}+\frac{1}{b^4+1}+\frac{1}{c^4+1}+\frac{1}{d^4+1}}=4$ $\Rightarrow \displaystyle \sum_{cyc} a^4 \geq 12$ Now applying AM $\geq$ GM $\Rightarrow \displaystyle \sum_{cyc} a^4 \geq 12 \geq 4(\prod_{cyc} a^4)^{\frac{1}{4}}$ $\Rightarrow \prod_{cyc} a\leq 3$ Hence $\Large 20\times 3=60$ $\small{\color{#D61F06}{\text{Equality occurs when }a=b=c=d=(3)^{\frac{1}{4}}}}$

From the condition we have $\displaystyle\sum_{cyc}^{a,b,c,d} \frac{\frac{1}{a^4}}{1+\frac{1}{a^4}}=1$ Now by Cauchy-Schwarz Inequality we have $1=\displaystyle\sum_{cyc}^{a,b,c,d} \frac{\frac{1}{a^4}}{1+\frac{1}{a^4}}\geq\frac{\big(\displaystyle\sum_{cyc}^{a,b,c,d} \frac{1}{a^2}\big)^2}{4+\displaystyle\sum_{cyc}^{a,b,c,d} \frac{1}{a^4}}$ Therefore $\displaystyle\sum_{cyc}^{a,b,c,d} \frac{1}{a^4}+4\geq(\displaystyle\sum_{cyc}^{a,b,c,d} \frac{1}{a^2})^2$ $\Leftrightarrow\displaystyle\sum_{cyc}^{a,b,c,d} \frac{2}{ab}\leq 4$ By AM-GM we have $LHS\geq\frac{12}{abcd}$ , so finally $abcd\geq 3 \Rightarrow 20abcd\geq 60$ The equality holds when $a=b=c=d=\sqrt[4]{3}$

$\sum(b^4+1)(c^4+1)(d^4+1)=(a^4+1)(b^4+1)(c^4+1)(d^4+1)$ $Left= \sum(b^4c^4d^4+\sum b^4c^4+\sum b^4+1)$ $=\sum a^4b^4c^4+2\sum a^4b^4+3\sum a^4+4=a^4b^4c^4d^4+\sum a^4b^4c^4 +\sum a^4b^4+\sum a^4 +1$ $\Rightarrow a^4b^4c^4d^4=\sum a^4b^4+2\sum a^4+3$ $\geq 6a^2b^2c^2d^2+8abcd+3$ In fact,we can conclude that $abcd\geq 3$ .But I 'd like to point out why only 3: $$ Let $t=abcd$ then $t^4-6t^2-8t-3\geq 0$ Let $f(t)=t^4-6t^2-8t-3$ $f(3)=0,f(2)<0,f(0)<0$

$f^{'} (t)=4t^3-12t-8=4(t^3-3t-2)$ Find that $f^{'}(2)=0,f^{'}(1)<0,f^{'}(0)<0$ and $f^{''}(t)=12(t^2-1)$ hence $t=3$ is the only positive solution to $f(t)=0$