$\displaystyle \tan C=\frac{BA'}{HA'},\,\tan B=\frac{A'C}{HA'}.$

$\displaystyle \begin{align} \tan B+\tan C&=\frac{BA'+A'C}{HA'}=\frac{a}{HA'}\Rightarrow HA'=\frac{a}{\tan B+\tan C}\\ S[BHC]&=\frac{HA'\cdot BC}{2}=\frac{1}{2}\cdot \frac{a^2}{\tan B+\tan C}\\ S[AHC]&=\frac{b^2}{2(\tan A+\tan C)};\\ S[AHB]&=\frac{c^2}{2(\tan A+\tan B)}\\ S[ABC]&=S[AHB]+S[AHC]+S[AHB]\\ S&=\frac{1}{2}\sum_{cycl}\frac{a^2}{\tan B+\tan C}\\ &\sum_{cycl}\frac{a^2}{\tan B+\tan C}=2S=2rs\overbrace{\leq}^{Euler}sR. \end{align}$

The problem (from the Romanian Mathematical Magazine) has been posted by Dan Sitaru at the CutTheKnotMath facebook page, Dan later has kindly communicated his solution in a LaTeX file.