Well, it is an very interesting problem... Let $BC=a$,$CA=b$,$AB=c$. You get $a=2(\sqrt{rr_{2}}+\sqrt{rr_{3}}+\frac{r_{2}\sqrt{rr_{2}}}{r-r_{2}}+\frac{r_{3}\sqrt{rr_{3}}}{r-r_{3}})$, like this , you can also get $b$ and$ c$ in$ r$,$r_{1}$,$r_{2}$,$r_{3}$. From these elements, we can calculate $S$ (square) of$ ABC$ in 2 ways. From Helon's folmula, $S=\sqrt{t(t-a)(t-b)(t-c)}(t=\frac{a+b+c}{2})$　$(p1)$ And, $S=\frac{1}{2}r(a+b+c)$ $(p2)$ By substituting above mentioned $a$,$b$,$c$(in $r$,$r_{1}$,$r_{2}$,$r_{3}$) for $(p1)$and $(p2)$, we get the following $\frac{4\sqrt{rr_{1}}\sqrt{rr_{2}}\sqrt{rr_{3}}}{(r-r_{1})(r-r_{2})(r-r_{3})}=\sqrt{\frac{rr_{1}}{r-r_{1}}}+\sqrt{\frac{rr_{2}}{r-r_{2}}}+\sqrt{\frac{rr_{3}}{r-r_{3}}}$. From this, we can get the conclusion.

IZHO 2021 P4