Show $\frac{d(a,c)}{1+d(a,c)}\leq \frac{d(a, b)+d(b,c)}{1+d(a, b)+d(b,c)}$

Question

If $0\leq d(a,c)\leq d(a, b)+d(b,c) $ then $$\frac{d(a,c)}{1+d(a,c)}\leq \frac{d(a, b)+d(b,c)}{1+d(a, b)+d(b,c)}$$ The problem is when I say $$1+d(a,c)\leq 1+d(a, b)+d(b,c)$$ $$\implies \frac{1}{1+d(a, b)+d(b,c)}\leq \frac{1}{1+d(a,c)}$$ But if I multiply the first and the last inequiality gives $$\frac{d(a,c)}{1+d(a, b)+d(b,c)}\leq \frac{d(a, b)+d(b,c)}{1+d(a,c)}$$ Is that ok or how to proceed?

Answer 1

We have $$\frac A {1+A} = 1-\frac{1}{1+A}.$$ Therefore, if $0\le A\le B$, $$\frac1{1+A} \ge \frac 1{1+B}$$ so that $$\frac A{1+A}=1-\frac1{1+A} \le 1-\frac 1{1+B} = \frac B{1+B}.$$ Put $A = d(a,c)$ and $B=d(a,b)+d(b,c)$ to complete the proof.

Answer 2

I think there is a better way. Suppose you know that $0 \leq A \leq B.$ Then clearly we have $$A+AB \leq B+AB.$$ Now this gives $$A(1+B) \leq B(1+A)$$ and dividing we get $$\frac{A}{1+A} \leq \frac{B}{1+A}.$$

Taking $A = d(a, c)$, $B = d(a,b) + d(b,c)$ and using the triangle inequality gets you the result.

In all likelihood you are trying to prove something like "if $d(x, y)$ is a metric, then so is $\rho(x, y) = \frac{d(x,y)}{1 + d(x,y)}$" or that they are equivalent, which is an important enough result to make sure you understand how to get this inequality. At some point, you will likely have to prove the result that given a sequence of seminorms defining a topology, a series with similar-looking terms gives a metric with equivalent topology (see the fifth example here).