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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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1 Answer 1

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.

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Not $\dfrac{h_{a}}{m_{a}}$ but $\dfrac{h_{a}}{m_{b}}$ –  chloe_shi Jun 11 '13 at 16:11
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