Help with Ptolemy metric space

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Ptolemy metric space. Help!!

The problem is :
"Let x,y,z,t belongs to R^n where d(x,y)=||x-y||.
Show that(Ptolemy's inequality):
d(x,y)d(z,t)<=d(x,z)d(y,t)+d(x,t)d(y,z)"



I have found this related to the topic paper but I cannot show that the Euclidean space R^n is Ptolemy.
The paper in the second page "2.Preliminaries" says that "To show that the Euclidean space R^n is Ptolemy, consider again four points x,y,z,w. Applying a suitable Mobius transformation we can assume that z is a midpoint of y and w, i.e. |yz|=|zw|=1/2 |yw|. For this configuration...."
But how can we extract from the above paragraph that the Euclidean space R^n is Ptolemy and which is the "suitable Mobius transformation"??


Thanks anyone in advance.
 

fresh_42

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In a tendon quadrangle
\ square ABCD
look at the triangle
 \ triangle ABC
with the separate point D on its circumference with radius r and the associated base triangle
\ triangle LMN
, The formula for calculating the side lengths of a base triangle then returns for
\ triangle LMN
:

{\ begin {aligned} & | MN | & = {\ frac {| AD | \ cdot | BC |} {2r}} \\ & | LN | & = {\ frac {| BD | \ cdot | AC |} {2r}} \\ & | LM | & = {\ frac {| CD | \ cdot | AB |} {2r}} \\\ end {aligned}}

220px-Ptolemy_inequality_proof.svg.png


But now that D is on the perimeter of
\ triangle ABC
is, is
\ triangle LMN
degenerate and its sides lie on the corresponding Simson line , so that the two sides LM and NM complement each other to the third side LN . It therefore applies:

| LM | + | NM | = | LN |

With the above equations, this provides:

| AB | \ cdot | CD | + | BC | \ cdot | AD | = | AC | \ cdot | BD |

If D is not on the circumference, then due to the triangle inequality for
\ triangle LMN
:

|> | | LM | + | NM LN |

The above equations then provide the inequality of Ptolemy:

| AB | \ cdot | CD | + | BC | \ cdot | AD |> | AC | \ cdot | BD |



Source: https://de.wikipedia.org/wiki/Satz_von_Ptolemäus
Translation: Google chrome
 

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