Show that $\frac{r_1}{bc}+\frac{r_2}{ca}+\frac{r_3}{ab}=\frac{1}{r}-\frac{1}{2R}$

$LHS=\frac{r_1}{bc}+\frac{r_2}{ca}+\frac{r_3}{ab}=\frac{a{r}_1+b{r}_2+c{r}_3}{abc}$

$=\sum \frac{a{r}_1}{abc}=\frac{1}{abc}\sum as\cdot \tan \frac{A}{2}$

$=\frac{S}{abc}\sum 2R\sin A\tan \frac{A}{2}$

$=\frac{S}{abc}\sum 2R2\sin \frac{A}{2}\cos \frac{A}{2}\cdot \frac{\sin A/2}{\cos A/2}$

$=\frac{4RS}{abc}\sum \frac{1-\cos A}{2}=\frac{S}{\mathrm{\Delta }}\cdot \frac{1}{2}\sum (1-\cos A)$

$=\frac{S}{2\mathrm{\Delta }}[1-\cos A+1-\cos B+1-\cos C]$

$=\frac{S}{2\mathrm{\Delta }}[3-(\cos A+\cos B+\cos C\left)\right]$

Consider $\cos A+\cos B+\cos C$ 

$=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+\cos C$

$=2\cos \left(\frac{\pi }{2}-\frac{c}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}$

$=2\sin \frac{C}{2}\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}$

$=1+2\sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\sin \frac{C}{2}\right]$

$=1+2\sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{\pi }{2}-\left(\frac{A+B}{2}\right)\right)\right]$

$=1+2\sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)\right]$

$=1+2\sin \frac{C}{2}\left[2\sin \frac{A}{2}\sin \frac{B}{2}\right]$

$\cos A+\cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

$=\frac{1}{2r}\left(2-4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\right)$

$=\frac{1}{2r}\left(2-\frac{4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}{R}\right)$

$=\frac{1}{2r}\left(2-\frac{r}{R}\right)=\frac{1}{r}-\frac{1}{2R}=\text{ R.H.S. }$