Let $h_{1},h_{2},h_{3}$ be the altitudes and $m_{1},m_{2},m_{3}$ be the medians of the triangle ABC.

Question

Show that:$$\frac{h_1}{m_1}+\frac{h_2}{m_2}+\frac{h_3}{m_3}\leq3$$ So, I was wondering if we could prevent all the hefty geometry and solve this using Chebyshev's or the Rearrangement inequality. Assume WLOG that $a\leq b \leq c.$ We must have $h_1\geq h_2 \geq h_3$ because the altitude inversely depends on the opposite side length. For similar reasons we have $m_1 \geq m_2 \geq m_3$. But I am not sure how to proceed any further than this, as the application of Chebyshev's inequality ceases to provide any result. $$3(h_1m_1+h_2m_2+h_3m_3)\geq (h_1+h_2+h_3)(m_1+m_2+m_3)$$ Any hints to this problem are much appreciated. Thanks!

Answer 1

Let $h_1,h_2,h_3$ and $m_1,m_2,m_3$ are the altitudes and the medians to the same side of a triangle, respectively.

Altitude is the shortest distance from a vertex to opposite side.

$h_1\leq m_1$ then $\frac{h_1}{m_1}\leq 1$

$h_2\leq m_2$ then $\frac{h_2}{m_2}\leq 1$

$h_3\leq m_3$ then $\frac{h_3}{m_3}\leq 1$

By adding inequalities side by side

$$\frac{h_1}{m_1}+\frac{h_2}{m_2}+\frac{h_3}{m_3}\leq 3$$

Equality holds, when $h_1=m_1$, $h_2=m_2$ and $h_3=m_3$