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 $\pi.$
 A: 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.
A: 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 $x+y+z\ne0,n=1$
A: Hint:-
$$ \tan^{-1}a + \tan^{-1}b + \tan^{-1}c= \pi$$
Only and only if 

$$a+b+c=abc$$

A: Like  Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$$\tan^{-1}\sqrt{\dfrac{x(x+y+z)}{yz}}+\tan^{-1}\sqrt{\dfrac{y(x+y+z)}{zx}}$$ $$=\begin{cases} \tan^{-1}\left(\dfrac{\sqrt{\dfrac{x(x+y+z)}{yz}}+\sqrt{\dfrac{y(x+y+z)}{zx}}}{1-\sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}}\right) &\mbox{if } \sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}<1 \ \ \ \  (1) \\ 
\pi+\tan^{-1}\left(\dfrac{\sqrt{\dfrac{x(x+y+z)}{yz}}+\sqrt{\dfrac{y(x+y+z)}{zx}}}{1-\sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}}\right) & \mbox{if } \sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}>1 \ \ \ \  (2) \end{cases} $$
Now $R=\dfrac{\sqrt{\dfrac{x(x+y+z)}{yz}}+\sqrt{\dfrac{y(x+y+z)}{zx}}}{1-\sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}}=\sqrt{\dfrac{x+y+z}{xyz}}\cdot\dfrac{|z|(|x|+|y|)}{|z|-|x+y+z|}$
If $|x+y|,|z|,|x+y+z|\ge0,$ $$R=-\sqrt{\dfrac{x+y+z}{xyz}}\cdot z=-\sqrt{\dfrac{z(x+y+z)}{xy}}$$
Again  $(2)$ will hold true if $\sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}>1 \iff (x+y+z)^2>z^2$
which is true if $x,y,z>0$
Finally $\tan^{-1}(-u)=-\tan^{-1}u$
