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 $\pi.$

Let $x$, $y$ and $z$ be positive numbers. We consider
 a triangle $ABC$ with side lengths $a=BC=y+z$, $b=CA=x+z$ and $c=AB=x+y$. The semi-perimeter $s=x+y+z$ inradius $r$. Now, by Heron’s formula we have
$$\eqalign{\cot(A/2)&=\frac{s-a}{r}=\frac{s(s-a)}{{\rm Area}(ABC)}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}\cr
&=\sqrt{\frac{x(x+y+z)}{yz}}}$$
So,
$$\eqalign{\tan^{-1}\sqrt{\frac{x(x+y+z)}{yz}}&=\frac{\pi-A}{2}\cr
\tan^{-1}\sqrt{\frac{y(x+y+z)}{zx}}&=\frac{\pi-B}{2}\cr
\tan^{-1}\sqrt{\frac{z(x+y+z)}{xy}}&=\frac{\pi-C}{2}}$$
Adding we get $\pi$ as a sum.
