$\\$

Gọi $O$ là tâm $(ABC)$ và $O'$ là tâm nội tiếp $\Delta ABC$

Gọi $X,Y,Z$ là trung điểm $BC, AC,AB$

Ta có: $AX\le OA + OX=R+OX$

$\Rightarrow m_a\le R+OX$

Tương tự: $m_b\le R+OY, m_c\le R+OZ$

$\Rightarrow \dfrac{m_a}{h_a}+\dfrac{m_b}{h_b}+\dfrac{m_c}{h_a}\le \dfrac{R+OX}{h_a}+\dfrac{R+OY}{h_b}+\dfrac{R+OZ}{h_c}\\\le R\left(\dfrac{1}{h_a}+\dfrac{1}{h_b }+\dfrac{1}{h_c}\right)+\left(\dfrac{OX}{h_a}+\dfrac{OY}{h_b}+\dfrac{OZ}{h_c}\right)$

Đặt: $(BC;CA;AB)=(a;b;c)(a;b;c>0)$

$\bullet$ $S_{ABC}=\dfrac{1}{2}(OX.a + OY.b +OZ.c)$

$\Rightarrow 2S_{ABC}=OX.a+OY.b+OZ.c$

Mà $S_{ABC}=\dfrac{1}{2}.h_a .a \Rightarrow a=\dfrac{2S_{ABC}}{h_a}$

Tương tự: $b=\dfrac{2S_{ABC}}{h_b},c=\dfrac{2S_{ABC}}{h_c}$

$\Rightarrow 2S_{ABC}=2S\left(\dfrac{OX}{h_a}+\dfrac{OY}{h_b}+\dfrac{OZ}{h_c}\right)\\\Rightarrow \dfrac{OX}{h_a}+\dfrac{OY}{h_b}+\dfrac{OZ}{h_c}=1$

$\bullet$ $S_{ABC}=p.r(p$ là nửa chu vi $\Delta ABC)$

$\Rightarrow S_{ABC}=\dfrac{a+b+c}{2}.r\\\Rightarrow a+b+c=\dfrac{2S_{ABC}}{r}$

Theo chứng minh trên: $a+b+c=2S_{ABC}\left(\dfrac{1}{h_a}+\dfrac{1}{h_b}+\dfrac{1}{h_c}\right)$

$\Rightarrow \dfrac{1}{h_a}+\dfrac{1}{h_b}+\dfrac{1}{h_c}=\dfrac{1}{r}$

$\Rightarrow \dfrac{m_a}{h_a}+\dfrac{m_b}{h_b}+\dfrac{m_c}{h_c}\le \dfrac{R}{r}+1=\dfrac{R+r}{r}$

Dấu "=" xảy ra khi: $AB=BC=AC$ hay $\Delta ABC$ đều.