# Geometric Inequality Related To Median, Altitude

For a triangle $ABC$, let $m_{a}$, $h_{a}$ be $A$-median, $A$-altitude.
Define $m_{b}$,$h_{b}$ and $m_{c}$,$h_{c}$ likewise.

Prove that $\dfrac{h_{a}}{m_{b}}+\dfrac{h_{b}}{m_{c}}+\dfrac{h_{c}}{m_{a}}\leq 3$

I have no solution.

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The only reference I can find is: Walther Janous, "Further Inequalities of Erdos-Mordell Type", Forum Geometricorum 4 (2004), pp. 203–206 (PDF), citing p. 315 of D. S. Mitrinovic et al., Recent Advances in Geometric Inequalities, (Kluwer 1989; there's an expensive Springer reprint), which attributes to Klamkin and Meir the result that $\frac{\overline{h_1}}{m_1} + \frac{\overline{h_2}}{m_2} + \frac{\overline{h_3}}{m_3} \leqslant 3$, where $(\overline{h_1}, \overline{h_2}, \overline{h_3})$ is any permutation of $(h_1, h_2, h_3)$. – Calum Gilhooley Mar 25 at 0:08

Denote the side of the triangle opposite vertex $A$ as $a$. Now consider the points on side $a$ where $m_a,h_a$ intersect side $a$. Label the intersection of $a$ and $h_a$ as the point $X$, and label the intersection of $a$ and $m_a$ as the point $Y$. Then the triangle $XYA$ is a right triangle. The ratio $\dfrac{h_a}{m_a}$ is actually the sine of the angle that $m_a$ makes with side $a$. Since the sine is always less than or equal to 1, we see that $\dfrac{h_a}{m_a} \leq 1$. The same argument shows that $\dfrac{h_b}{m_b} \leq 1$ and $\dfrac{h_c}{m_c} \leq 1$. Taking the sum of all three inequalities, you have your result.
Not $\dfrac{h_{a}}{m_{a}}$ but $\dfrac{h_{a}}{m_{b}}$ – chloe_shi Jun 11 '13 at 16:11