Ta có:
$\quad ah_a = bh_b = ch_c =2S$
$\to \begin{cases}h_a =\dfrac{2S}{a}\\h_b=\dfrac{2S}{b}\\h_c=\dfrac{2S}{c}\end{cases}$
Ta được:
$h_a + h_b + h_c = \dfrac{2S}{a}+\dfrac{2S}{b}+\dfrac{2S}{c} = 2S\left(\dfrac1a +\dfrac1b +\dfrac1c\right)$
$\to h_a + h_b + h_c \geq 2S\cdot\dfrac{9}{(a+b+c)}$
$\to h_a + h_b + h_c \geq 2pr\cdot\dfrac{9}{(a+b+c)}$
$\to h_a + h_b + h_c \geq r\cdot(a+b+c)\cdot\dfrac{9}{(a+b+c)}$
$\to h_a + h_b + h_c \geq 9r$
Dấu $=$ xảy ra $\Leftrightarrow a = b = c\Leftrightarrow$ tam giác đều
Vậy $h_a + h_b + h_c = 9r \Leftrightarrow$ tam giác đã cho là tam giác đều
