If $\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$, prove $abcd \geq 3$. Given four positive real numbers $a,b,c,d$. It is given that  $\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$. Prove $abcd \geq 3$.
I've applied AM-HM inequality and got the result, $a^4+b^4+c^4+d^4 \geq 12$. Then by AM-GM, $a^4+b^4+c^4+d^4\geq 4abcd; 12\geq 4abcd \implies 3\geq abcd$.
But this is contradictory. Then I realized I was doing all wrong. Does anyone have method for this one?

By AM-GM, $\frac{a^4}{1+a^4}=\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}\ge \frac{3}{\sqrt[3]{(1+b^4)(1+c^4)(1+d^4)}}$, etc. Multiplying all these together we have $\frac{(abcd)^4}{\prod (1+a^4)}\ge \frac{3^4}{\prod (1+a^4)}$, i.e. $abcd\ge 3$.
