If $\;$ $\tan^{-1} \left(\sqrt{\dfrac{a \; r}{b \; c}}\right) + \tan^{-1} \left(\sqrt{\dfrac{b \; r}{c \; a}}\right) + \tan^{-1} \left(\sqrt{\dfrac{c \; r}{a \; b}}\right) = \pi$, $\;$ then prove that $\;$ $a + b + c = r$, $\;\;\;$ $a, \; b, \; c, \; r > 0$
Let $\;$ $\tan^{-1} \left(\sqrt{\dfrac{a \; r}{b \; c}}\right) = \alpha$, $\;$ $\tan^{-1} \left(\sqrt{\dfrac{b \; r}{c \; a}}\right) = \beta$, $\;$ $\tan^{-1} \left(\sqrt{\dfrac{c \; r}{a \; b}}\right) = \gamma$, $\;$ $\alpha, \; \beta, \; \gamma \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
Then, $\;$ $\tan \alpha = \sqrt{\dfrac{a \; r}{b \; c}}$, $\;$ $\tan \beta = \sqrt{\dfrac{b \; r}{c \; a}}$, $\;$ $\tan \gamma = \sqrt{\dfrac{c \; r}{a \; b}}$
$\therefore \;$ $\tan^{-1} \left(\sqrt{\dfrac{a \; r}{b \; c}}\right) + \tan^{-1} \left(\sqrt{\dfrac{b \; r}{c \; a}}\right) + \tan^{-1} \left(\sqrt{\dfrac{c \; r}{a \; b}}\right) = \pi$
$\implies$ $\alpha + \beta + \gamma = \pi$
i.e. $\;$ $\alpha + \beta = \pi - \gamma$
$\therefore \;$ $\tan \left(\alpha + \beta\right) = \tan \left(\pi - \gamma\right)$
i.e. $\;$ $\dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta} = - \tan \gamma$
i.e. $\;$ $\dfrac{\sqrt{\dfrac{a \; r}{b \; c}} + \sqrt{\dfrac{b \; r}{c \; a}} }{1 - \sqrt{\dfrac{a \; r}{b \; c}} \times \sqrt{\dfrac{b \; r}{c \; a}}} = - \sqrt{\dfrac{c \; r}{a \; b}} $
i.e. $\;$ $\dfrac{\dfrac{a \sqrt{c \; r} + b \sqrt{c \; r}}{c \sqrt{a \; b}}}{\dfrac{c \sqrt{a \; b} - r \sqrt{a \; b}}{c \sqrt{a \; b}}} = - \sqrt{\dfrac{c \; r}{a \; b}} $
i.e. $\;$ $\dfrac{\left(a + b\right) \sqrt{c \; r}}{\left(c - r\right) \sqrt{a \; b}} = - \sqrt{\dfrac{c \; r}{a \; b}} $
i.e. $\;$ $\dfrac{a + b}{c - r} = - 1$
i.e. $\;$ $a + b = r - c$
i.e. $\;$ $a + b + c = r$
Hence proved.