# Inequality between heights and inscribed circle radius.

Prove that for every triangle the following inequality is true $$\dfrac{1}{2r} < \dfrac{1}{h_1} + \dfrac{1}{h_2} < \dfrac{1}{r}$$

My attempt was trying to get somthing from connection between area, heights and radius, but without any effect.

• can you please explain what is $h1,h2$? – Sujit Bhattacharyya Mar 24 '18 at 12:01
• $h_1a=h_2b=r s$ where $a,b$ are two sides and $s$ is the perimeter – Lozenges Mar 24 '18 at 12:13
• They are any two heights – user128409235 Mar 24 '18 at 12:36
• @Lozenges shouldn't it be $h_1 a = h_2 b = 2rs$? – kayush Mar 24 '18 at 12:55

$h_1a=h_2b=2rs$ Equating area

$h_1 = \cfrac{2rs}a$

$h_2 = \cfrac{2rs}b$

$\cfrac1{h_1} + \cfrac1{h_2} = \cfrac{a+b}{2rs} = \cfrac{2(a+b)}{2r(a+b+c)}$

Using traingular inequality, $a+b > c$

$\cfrac1{h_1} + \cfrac1{h_2} > \cfrac{1}{2r}$

and since c > 0, $a+b+c > a+b \implies (a+b)/(a+b+c) < 1$

$\cfrac1{h_1} + \cfrac1{h_2} < \cfrac{1}{r}$