MHB Lengths of three segments of a triangle

[anemone]

If $x,\,y,\,z$ are lengths of three segments which can form a triangle, show that the same is true for $\dfrac{1}{x+z},\,\dfrac{1}{y+z},\,\dfrac{1}{x+y}$.

 

[kaliprasad]

If $x,\,y,\,z$ are lengths of three segments which can form a triangle, show that the same is true for $\dfrac{1}{x+z},\,\dfrac{1}{y+z},\,\dfrac{1}{x+y}$.

Spoiler

we need to show that for arbritary 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

 

[anemone]

Spoiler

we need to show that for arbritary 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

Very well done, kaliprasad! And thanks for participating!

 

[anemone]

Here is another solution of other that I want to share with MHB:

Spoiler

WLOG we can assume that $z\ge y \ge x$ so that $\dfrac{1}{x+y}\ge \dfrac{1}{z+x}\ge\dfrac{1}{y+z}$ and now, it remains to prove that $\dfrac{1}{y+z}+\dfrac{1}{z+x}>\dfrac{1}{x+y}$.

We have

$\dfrac{1}{y+z}+\dfrac{1}{z+x}-\dfrac{1}{x+y}=\dfrac{x^2+xy+y^2+z(x+y-z)}{(y+z)(z+x)(x+y)}>0$ since $x+y>z$ and the result follows.

Note that this is an identical approach to kaliprasad's, but written in a slightly different way.

 

[kaliprasad]

Here is another solution of other that I want to share with MHB:

Spoiler

WLOG we can assume that $z\ge y \ge z$ so that $\dfrac{1}{x+y}\ge \dfrac{1}{z+x}\ge\dfrac{1}{y+z}$ and now, it remains to prove that $\dfrac{1}{y+z}+\dfrac{1}{z+x}>\dfrac{1}{x+y}$.

We have

$\dfrac{1}{y+z}+\dfrac{1}{z+x}-\dfrac{1}{x+y}=\dfrac{x^2+xy+y^2+z(x+y-z)}{(y+z)(z+x)(x+y)}>0$ since $x+y>z$ and the result follows.

Note that this is an identical approach to kaliprasad's, but written in a slightly different way.

Spoiler

There is a typo error in WLOG iline,
it should be $z\ge y \ge x $
Secondly we do not need WLOG line as
$x+y-z \gt 0$ for any x,y,z sides of the triangle