<s>[INST] 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 $\pi.$
 [/INST] Hint:-
$$ \tan^{-1}a + \tan^{-1}b + \tan^{-1}c= \pi$$
Only and only if 

$$a+b+c=abc$$

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