Simplifying $\sum_{cyc}\tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)$. I get $0$, but the answer is $\pi$. So the question is
$$  \tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)+\tan^{-1}\left(\sqrt{\frac{y(x+y+z)}{xz}}\right)+\tan^{-1}\left(\sqrt{\frac{z(x+y+z)}{yx}}\right) =\ ? $$
So my take on the question is to rewrite it as 
$$  \tan^{-1}\left(x\sqrt{\frac{(x+y+z)}{xyz}}\right)+\tan^{-1}\left(y\sqrt{\frac{(x+y+z)}{xyz}}\right)+\tan^{-1}\left(z\sqrt{\frac{(x+y+z)}{yzx}}\right) $$
Then say $$\frac{x+y+z}{yzx}= a^2.$$
We get
$$ \tan^{-1}\left( \frac{a((x+y+z)-a^2xyz)}{1-a^2(xy+yz+zx)}\right)$$
And since $ (x+y+z) = a^2xyz $ , this is just equal to $\tan^{-1}(0)= 0 $ but the answer given is $\pi.$

Use https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values,
$-\dfrac\pi2\le \tan^{-1}a\le\dfrac\pi2$
Now $\sqrt b\ge0$ for real $\sqrt b$
$\implies0\le\tan^{-1}\sqrt b\le\dfrac\pi2$
So, here the sum will lie in $\in[0,3\pi/2]$
Now the sum will be $=0$ only if each term under is individually $=0$
i.e. if $x+y+z=0$
Otherwise the sum will be $\ne0$
Also, the general value of the sum is $n\pi$ where $n$ is an integer 
So, for $x+y+z\ne0,n=1$
