Question

If $h_1,h_2,h_3$ are the altitudes of a triangle, then the minimum value of $\frac{1}{h_1}+\frac{1}{h_2}+\frac{1}{h_3}$ =$\frac{a}{R}$ then a is

Answer 1

$△=\frac{a{h}_1}{2}\Rightarrow \frac{1}{h_1}=\frac{a}{2△}$
Similarly, others
$\Rightarrow \frac{1}{h_1}+\frac{1}{h_2}+\frac{1}{h_3}=\frac{a+b+c}{2△}$
$=\frac{2s}{2△}$
$=\frac{s}{△}$
$=\frac{△}{r.△}=\frac{1}{r}$
But $R\ge 2r\Rightarrow 2r\le R$
$\Rightarrow \frac{1}{r}\ge \frac{2}{R}$
Hence, minimum value is $\frac{2}{R}=\frac{a}{R}$

$a=2$