For a triangle, prove that $\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}=\frac{1}{r^2} $

Question

Context: I made this question and am contemplating about submitting it as a contest question. The contest is not a large one; its scope does not comprise even our whole class. Its just a friendly, small contest within a small group.
This originally occurred to me when we were asked to prove $\tan \frac A2\tan\frac B2 +\tan \frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1$ in class. I tried a geometrical proof but failed, but I got the following for my efforts.

Derivation: Given $\triangle ABC$ and its incircle $\Im$ with inradius $r$, we have the known (and easily provable) formula $$\tan \frac A2\tan\frac B2 +\tan \frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1.\tag{1}\label{1}$$ We know, $IE=IF=IG=r$. Now, $$\tan \frac A2=\frac{r}{AE}=\frac{r}{AF}$$$$\tan \frac B2=\frac{r}{BE}=\frac{r}{BG}$$$$\tan \frac C2=\frac{r}{CG}=\frac{r}{CF}$$ Putting appropriate values in the formula $(1)$, we get $$\frac{r^2}{AE\cdot BE}+ \frac{r^2}{CG\cdot BG}+ \frac{r^2}{AF\cdot CF}=1$$ or $$\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}=\frac{1}{r^2}$$ Finally, using the identities $\Delta=rs, s=\dfrac{a+b+c}{2}, \Delta=\dfrac{abc}{4R}$ (where R is the circumradius) and the sine law we get $$\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}=\dfrac{s^2}{\Delta^2}=\dfrac{(a+b+c)^2}{4\Delta^2}=\dfrac{(a+b+c)^2\cdot 16R^2}{4a^2b^2c^2} $$ Thus, $$\sqrt{\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}}=2R\frac{a+b+c}{abc}$$ $$=2R\frac{2R(\sin A+\sin B+\sin C)}{8R^3\sin A\sin B\sin C}$$$$=\frac{1}{2R}\left(\frac{1}{\sin B\sin C}+ \frac{1}{\sin A\sin C}+ \frac{1}{\sin B\sin A}\right)$$ so that $$2R\sqrt{\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}}=\left(\frac{1}{\sin B\sin C}+ \frac{1}{\sin A\sin C}+ \frac{1}{\sin B\sin A}\right)$$

Question: Prove that for a triangle ABC, $$2R\sqrt{\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}}=\left(\frac{1}{\sin B\sin C}+ \frac{1}{\sin A\sin C}+ \frac{1}{\sin B\sin A}\right)$$

The result looks pretty symmetrical and nice, and also reminds one of Ceva’s theorem because of the segments taken consecutively.

If you were given this question and did not know the above derivation, how would you approach it? Basically, I am looking for alternative solutions to this problem just for reference.

If the question has any flaws, please tell me. Also, I would really like to know how you would rate this question for a high school level informal contest. If any of you can come up with variants of this, especially as inequalities, I would be grateful. Thanks in advance.

Answer 1

Note that $AE=s-a, BE=s-b, BG=s-b, GC=s-c, CF=s-c, FA=s-a$. The expression under the square root becomes $$\sqrt{\frac{1}{(s-a)(s-b)}+\frac{1}{(s-b)(s-c)}+\frac{1}{(s-a)(s-c)}}$$ By Heron's formula, $K=\sqrt{s(s-a)(s-b)(s-c)}\implies \frac{1}{(s-b)(s-c)}=\frac{s(s-a)}{K^2}$, etc. and so the expression under square root becomes $$\frac{1}{K}\sqrt{s(s-a)+s(s-b)+s(s-c)}=\frac{s}{K}$$ So we want to show that $$\frac{2Rs}{K}=\sum_{cyc}\frac{1}{\sin A\sin B}$$ Now, by the extended law of sines, $\frac{a}{\sin A}=2R\implies \frac{1}{\sin A}=\frac{2R}{a}$, so the RHS is $$(2R)^2\sum_{cyc}\frac{1}{ab}=\frac{(2R)^2(a+b+c)}{abc}$$ Thus we want to show $$\frac{s}{K}=\frac{2R(a+b+c)}{abc}$$ Now this is easy because it rearranges to $\frac{1}{K}=\frac{4R}{abc}$, which is $K=\frac{abc}{4R}$ which is true.

Answer 2

By the result $1$ found in my post, $$\displaystyle x yz=r^{2}(x+y+z),\tag*{(*)} $$

where $r$ is the in-radius of the triangle ABC of sides $a,b$ and $c$ and $a=y+z, b=z+x $ and $c=x+y$.

Dividing the equation (*) by $xyzr^2$ yields the required equation $$ \frac{1}{y z}+\frac{1}{x z}+\frac{1}{x y}=\frac{1}{r^{2}} $$