Find the maximum value in a triangle

Let $r$ , $R$ and $S$ represent the inradius , circumradius and area of the triangle respectively.

Claim: $\frac{1}{h_A}+\frac{1}{h_B}+\frac{1}{h_C}=\frac{1}{r}$ Proof: $\frac{1}{h_A}+\frac{1}{h_B}+\frac{1}{h_C}=\frac{a}{2S}+\frac{b}{2S}+\frac{c}{2S}=\frac{a+b+c}{2S}=\frac{1}{r}$

Let $O$ the center of the circumcircle of $\triangle ABC$ and $D, E, F$ be the midpoints of side $BC, CA, AB$ respectively.

Claim: $\frac{OD}{h_A}+\frac{OE}{h_B}+\frac{OF}{h_C}=1$ Proof: $OD\cdot a + OE\cdot b + OF\cdot c = a\cdot h_A = b\cdot h_B = c\cdot h_C=2S$ $\implies \frac{OD}{h_A}+\frac{OE}{h_A}+\frac{OF}{h_C}=1$

Also, $m_A =AD \le OA+OD \implies m_A\le R+OD$ . Writing similar inequalities and adding up we get, $\frac{m_A}{h_A}+\frac{m_B}{h_B}+\frac{m_C}{h_C} \le R\left (\frac{1}{h_A}+\frac{1}{h_B}+\frac{1}{h_C} \right ) + \left(\frac{OD}{h_A}+\frac{OE}{h_A}+\frac{OF}{h_C} \right )$

Using the above claims we get the well known inequality, $\boxed{\frac{m_A}{h_A}+\frac{m_B}{h_B}+\frac{m_C}{h_C} \le 1+\frac{R}{r}}$

Therefore the maximum value of $\frac{m_A}{h_A}+\frac{m_B}{h_B}+\frac{m_C}{h_C}$ is $\boxed{4}$