### Solution

Let $p_a,\,p_b,\,p_c\,$ be the distances from the circumcenter to the corresponding side lines. Then, obviously,

$m_a\le R+p_a,\, m_b\le R+p_b,\,m_c\le R+p_c.$

By adding these inequalities, we get

(1)

$m_a+m_b+m_c\le 3R+(p_a+p_b+p_c).$

Looking at the sides of the triangle as chords in the circumcircle,

$ap_a=R^2\sin 2\alpha,\,bp_b=R^2\sin 2\beta,\, cp_c=R^2\sin 2\gamma.$

Thus we have $p_a+p_b+p_c)=R(\cos\alpha+\cos\beta+\cos\gamma)\,$ and since $\cos\alpha+\cos\beta+\cos\gamma=\displaystyle 1+\frac{r}{R},\,$ (1) implies Leuenberger's inequality.

### Acknowledgment

The proof came from O. Bottema et al Geometric Inequalities (1968, 8.2, 73-74) where they credit F. Leuenberger (1958).