Maximum of sum of fractions with tans Let $ABC$ be an acute triangle. What is the maximum value of
$$\frac{\tan^2A+\tan^2B}{\tan^4A+\tan^4B}+\frac{\tan^2B+\tan^2C}{\tan^4B+\tan^4C}+\frac{\tan^2C+\tan^2A}{\tan^4C+\tan^4A}?$$
For the equilateral triangle, $\tan A=\tan B=\tan C=\sqrt{3}$, and the sum is $1$. For the isosceles triangle with $\angle A=\angle B\rightarrow\pi/2$ and $\angle C\rightarrow 0$, the sum approaches zero.

Say $x= \tan A$, $y= \tan B$ and $z = \tan C$. It is
well-known that $x+y+z=xyz$. Then we have $$E = {x^2+y^2\over
x^4+y^4} +{y^2+z^2\over y^4+z^4} + {z^2+x^2\over
z^4+x^4}$$
By Cauchy inequality and then by $AM-GM$ we have
$${x^2+y^2\over x^4+y^4} \leq {2\over x^2+y^2}\leq {1\over xy}$$
So $$ E \leq {1\over xy} +{1\over yz}+{1\over zx} = 1$$
We get equality exactly when $x=y=z$, thus $3x=x^3 \Longrightarrow  x=\sqrt{3}$.  
