2015/008) show that if $x,y,z$ are 3 sides of a triangle so are $\dfrac{1}{x+y},\dfrac{1}{x+z},\dfrac{1}{y+z}$

we need to show that for arbitrary sides of length x,y z

$\dfrac{1}{x+z} + \dfrac{1}{y+z}\gt \dfrac{1}{x+y}$

or $(y+z)(x+y) + (x+z)(x+y) \ge (x+z)(x+z)$
or simlifying
$x^2+y^2 + xy + yz + xz \ge z^2$


now
$x^2+y^2 + xy + yz + xz = z(x+y) + x^2+y^2 + xy$
$ \gt z(x+y) $
$\gt z^2$ as $x+y \gt z$

hence proved

I have taken the problem from http://mathhelpboards.com/challenge-questions-puzzles-28/lengths-three-segments-triangle-14063.html#post66760 where I provided the above solution