Since $h_a = \frac {bc}{2R}$ and so on, it follows that $$\sum h_bh_c = \sum \frac {a^2bc}{4R^2} = \frac {abc}{4R^2}\cdot\sum a = \frac {4Rrp}{4R^2}\cdot 2p = \frac {2r}{R}\cdot p^2.$$ According to Euler's Inequality we have $2r\le R,$ so $$\boxed{\sum h_bh_c \le p^2}\, .$$