MHB Proving $abcd\ge 3$ with $a, b, c, d>0$

[anemone]

Let $a,\,b,\,c$ and $d>0$ and $\dfrac{1}{1+a^4}+\dfrac{1}{1+b^4}+\dfrac{1}{1+c^4}+\dfrac{1}{1+d^4}=1$. Prove that $abcd\ge 3$.

 

[anemone]

Spoiler: Solution of other

Let $a^2=\tan \alpha,\,b^2=\tan \beta,\,c^2=\tan \gamma,\,d^2=\tan \delta$. Then $\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma+\cos^2 \delta=1$. By the AM-GM inequality,

$\sin^2 \alpha=\cos^2 \beta+\cos^2 \gamma+\cos^2 \delta \ge 3(\cos \beta \cos \gamma \cos \delta)^{\small \dfrac{2}{3}}$

Multiplying this and three other similar inequalities, we have

$\sin^2 \alpha \sin^2 \beta \sin^2 \gamma \sin^2 \delta=81\cos^2 \alpha \cos^2 \beta \cos^2 \gamma \cos^2 \delta$

Consequently we get $abcd=\sqrt{\tan \alpha \tan \beta \tan \gamma \tan \delta}\ge 3$