Proof that a function is a b-metric

Question

I'm trying to prove that this function is a b-metric. I can't prove the triangle inequality. Here's the definition of a b-metric space.bmetric definition. Here's the example. I wanna prove the last inequality.example Any idea would be helpful. My teacher told me that this is a basic and easy prove but I spent more than 2 hours and still can't understand how they got s=5/2. (Sorry for my english). Thank you!

Answer 1

HINT.-What you need is that for all $m,n,p$ in $\mathbb N\cup\{\infty\}$ $$\frac{d(m,p)}{d(m,n)+d(n,p)}\le\frac52\tag1$$ You don't need to prove that in $(1)$ the maximum of the $LHS$ is equal to $\dfrac52$ (in fact, it seems that never $LHS=RHS$).

The maximum of the numerator is $5$ so $(m,p)=(2r+1,2s+1)$ or $(m,p)=(2r+1,\infty)$ (or $(\infty,2s+1)$ of course) and the minimum of the denominator is easily seen to be $2+2=4$ because of the taken values of $(m,p)$ (note that taking $m=n$ we would have $0+5\gt4$).

On the other hand one has

$$\left|\frac1m-\frac1p\right|=\left|\left(\frac1m-\frac1n\right)+\left(\frac1n-\frac1p\right)\right|$$ and the triangular inequality would work to discard this possibility.

Finally one has $$\frac{d(m,p)}{d(m,n)+d(n,p)}\le\frac54\le\frac52$$

Answer 2

Hint: if $d(m,n) \geq 2$ or $d(n,p) \geq 2$ the RHS is $\geq 5$ and it is obvious from the definition that LHS $\leq 5$. Now consider the case where $d(m,n)=|\frac 1 m -\frac 1 n|$ and $d(n,p)=|\frac 1 n -\frac 1 p|$. Verify that in this case we must have $d(m,p)=|\frac 1 m -\frac 1 p|$.