MHB Lengths of three segments of a triangle

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The discussion focuses on proving that if lengths \(x, y, z\) can form a triangle, then the lengths \(\frac{1}{x+z}, \frac{1}{y+z}, \frac{1}{x+y}\) can also form a triangle. Participants share different approaches to the proof, with one method highlighted as particularly effective. The conversation emphasizes the importance of understanding the properties of triangle inequalities in both cases. The solutions presented are noted for their clarity and effectiveness in demonstrating the concept. Overall, the thread contributes valuable insights into triangle segment relationships.
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If $x,\,y,\,z$ are lengths of three segments which can form a triangle, show that the same is true for $\dfrac{1}{x+z},\,\dfrac{1}{y+z},\,\dfrac{1}{x+y}$.
 
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anemone said:
If $x,\,y,\,z$ are lengths of three segments which can form a triangle, show that the same is true for $\dfrac{1}{x+z},\,\dfrac{1}{y+z},\,\dfrac{1}{x+y}$.

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
 
kaliprasad said:
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! :cool: And thanks for participating!
 
Here is another solution of other that I want to share with MHB:

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

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.
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
 
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