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?
 A: 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$.
$\qquad \Box$
A: Let
$$w=\frac{1}{1+a^4}, x=\frac{1}{1+b^4}$$$$ y=\frac{1}{1+c^4}, z=\frac{1}{1+b^4}$$
So, $$a^4=\frac{1-w}{w}=\frac{x+y+z}{w}$$$$ b^4=\frac{1-x}{x}=\frac{w+y+z}{x}$$$$ c^4=\frac{1-y}{y}=\frac{w+x+z}{y}$$$$ d^4=\frac{1-z}{z}=\frac{w+x+y}{z}$$
$$a^4b^4c^4d^4=(\frac{x+y+z}{w}) (\frac{w+y+z}{x})( \frac{w+x+z}{y} )(\frac{w+x+y}{z})-(1)$$
Apply AM-GM for these four terms on RHS individually in eqation $(1)$,
RHS$\geq$$(\frac{3(xyz)^{\frac{1}{3}}}{w})$$(\frac{3(wyz)^{\frac{1}{3}}}{x})$$(\frac{3(wxz)^{\frac{1}{3}}}{y})$$(\frac{3(wxy)^{\frac{1}{3}}}{z})$
$$⇒RHS\geq 81$$
RHS is also equal to $a^4b^4c^4d^4$.
$$a^4b^4c^4d^4\geq81 ⇒ abcd\geq3$$
A: 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. $abcd\ge 3$.
