Proof of Metric Space Triangle Inequality

[mynameisfunk]

ok i am stumped on a proof i am trying to construct of a metric:
d(x,y)=$$\frac{|x-y|}{1+|x-y|}$$

so, out of the 3 requirements to be a metric, the first 2 are trivial and I am just working on proving the triangle inequality...

i need $$\frac{|x-y|}{1+|x-y|}$$ $$\leq$$ $$\frac{|x-z|}{1+|x-z|}$$ + $$\frac{|z-y|}{1+|z-y|}$$

p²(1+q+r+qr) $$\leq$$ q²(1+p+r+pr)+r²(1+p+q+pq)

can i now go to:
p(1+q+r+qr) $$\leq$$ q(1+p+r+pr)+r(1+p+q+pq) ?

 

[csopi]

If you simplify the inequality you get:

p<=q+r+(some other positive expressions)

This is true, because p<=q+r by the properties of the absolute value.

 

[fluxions]

hint: first show that
$$\frac{u}{1+u} \leq \frac{v}{1+v}$$
whenever
$$0 \leq u \leq v$$.

 

[snipez90]

The method you're suggesting is clever, but the OP almost had a correct solution. He seems to have written down an unjustified inequality and then questioned the correct inequality (where the solution then immediate falls out by multiplying everything out and canceling).

 

[mynameisfunk]

i don't know If i posted enough info but, I had set |x-y|= p , |x-z| = q , |z-r| = r , and since the triangle equality holds, proving p <= q + r will suffice

 

[snipez90]

Defining your variables would have probably gotten you more responses, but it was pretty easy to figure out what p,q,r was. The only confusing part was that the last inequality you wrote in the original post is not a consequence of the inequality before it. The last inequality is what you should get upon clearing denominators. Then multiply out and you should get what csopi wrote. It's also clear that the steps are reversible.