Mehmet Sahin's Identity in Triangle

Solution

Below, we are going to use several known identities, $S=rs,$ $S=s_a(s-a),$ $abc=4RS,$ $\displaystyle [\Delta I_aI_bI_c]=\frac{abc}{2r^2s}{S}.$ Thus,

\displaystyle \begin{align} \sum_{cycl}\frac{a^3}{r_b+r_c} &=\sum_{cycl}\frac{a^3}{\displaystyle \frac{S}{s-b}+\frac{S}{s-c}}\\ &=\frac{1}{S}\sum_{cycl}\frac{a^3(s-b)(s-c)}{2s-a-b}\\ &=\frac{1}{S}\sum_{cycl}a^2(s-b)(s-c)\\ &=\frac{1}{rs}\cdot 4rs^2(R-r)=4s(R-r)=4sR-4sr\\ &=\frac{4Rrs}{r}-4S=\frac{4RS}{r}-4S=2\cdot\frac{abcS}{2r^2s}-4S\\ &=2[\Delta I_aI_bI_c]-4S. \end{align}

Acknowledgment

The problem from the Romanian Mathematical Magazine has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page, along with the above solution of his. The problem is by Mehmet Sahin.