Thread: I can't prove the triangle inequality holds for this metric space

In this real analysis book I have there is a proposition that if (X,d) is a metric space then so is (X,p) where p=d(x,y)/(1+d(x,y))

The proof that this is a metric space is left for the reader.

I have shown that it satisfies all requirements for a metric space apart from the triangle inequality i.e. p(x,y)<=p(x,z)+p(z,y)

Could someone give assistance please?

to simplify the algebra, let:

a = d(x,y)
b = d(x,z)
c = d(z,y), and we are given that a ≤ b+c, and also that a,b,c are all non-negative.

then:

p(x,y) = a/(1+a)
p(x,z) = b/(1+b)
p(z,y) = c/(1+c)

what we want to prove is:

,

that is:



or, equivalently, that:



which is to say:



so:



since

, it follows that:

, from which the desired result follows.

Originally Posted by Hefotos2004

In this real analysis book I have there is a proposition that if (X,d) is a metric space then so is (X,p) where p=d(x,y)/(1+d(x,y))

Here is a second way.
Lemma: If


We know that .
So

Thank you both very much. Very helpful.