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

1. ## 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)

2. ## Re: I can't prove the triangle inequality holds for this metric space

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:

$\displaystyle \frac{a}{1+a} \leq \frac{b}{1+b} + \frac{c}{1+c}$,

that is:

$\displaystyle \frac{a}{1+a} \leq \frac{b+c+2bc}{1+b+c+bc}$

or, equivalently, that:

$\displaystyle a(1 + b + c + bc) \leq (1+a)(b+c+2bc)$

which is to say:

$\displaystyle a + ab + ac + abc \leq b + c + 2bc + ab + ac + 2abc$

so:

$\displaystyle a \leq b + c + 2bc + abc$

since

$\displaystyle a \leq b+c$, it follows that:

$\displaystyle a \leq b+c = b+c+0 \leq b + c + 2bc + abc$, from which the desired result follows.

3. ## Re: I can't prove the triangle inequality holds for this metric space

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 $\displaystyle 0\le a\le b$
\displaystyle \begin{align*}a+ab &\le b+ab \\ a(1+b)&\le b(1+a) \\ \frac{a}{1+a}&\le\frac{b}{1 +b}\end{align*}

We know that $\displaystyle d(x,y)\le d(x,z)+d(z,y)$.
So
$\displaystyle \frac{d(x,y)}{1+ d(x,y)} &\le \frac{d(x,z)+d(z,y)}{1+ d(x,z)+d(z,y)}$$\displaystyle \le \frac{d(x,z)}{1+d(x,z)}+\frac{d(z,y)}{1+d(z,y)}$

4. ## Re: I can't prove the triangle inequality holds for this metric space

Thank you both very much. Very helpful.