2015/011) prove that if $a_1$,$a_2$ and $a_3$ are altitudes of a triangle and r is

  
radius of inscribed triangle then

$\dfrac{1}{a_1} + \dfrac{1}{a_2} + \dfrac{1}{a_3}= \dfrac{1}{r}$

let x,y,z be the sides of triangle with corresponding altitudes $a_1$.$a_2$ and $a_3$ and area be A

so $xa_1 = ya_2 = za_3= 2A$

so
$\dfrac{1}{a_1} + \dfrac{1}{a_2} + \dfrac{1}{a_3}$

= $\dfrac{x}{2A} + \dfrac{y}{2A} + \dfrac{z}{2A}$

=  $\dfrac{x+y+z}{2A}$

= $\dfrac{2s}{2A}$

= $\dfrac{s}{A}\cdots(1)$

now because

$r^2= \dfrac{(s - x)*(s - y)*(s - z)}{s}$

= $\dfrac{s(s - x)*(s - y)*(s - z)}{s^2}$
=  $\dfrac{A^2}{s^2}$
so $r = \dfrac{A}{s}$

Putting the above in (1) we get the result

Note:

I  have  taken the problem from the http://mathhelpboards.com/potw-secondary-school-high-school-students-35/problem-week-148-january-26th-2015-a-14133.html where this appeared as problem of the week. Other solution provided there are more elegant