<s>[INST] 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?
 [/INST] Clearing all denominators, we have that
$$
\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1
$$
is equivalent  to
$$
a^4b^4c^4d^4 = (a^4b^4 + a^4c^4 + a^4d^4 + b^4c^4 + b^4d^4 +  c^4d^4) + 2(a^4  + b^4  + c^4 + d^4) + 3
$$
Applying arithmetic mean $\ge$ geometric mean to  the brackets on the right gives
$$
a^4b^4c^4d^4 \ge 6  a^2b^2c^2d^2 + 8 a bcd   + 3
$$
This is again equivalent to
$$
 (a bcd   - 3) (a bcd   +1)^3 \ge 0
$$
Hence $ a bcd   \ge 3$.
$\qquad \Box$
</s>