Prove that $d(x,y) \ge d(a,b)-d(x,a)-d(y,b)$

Question

So far I've attempted algebraic manipulation with absolute values.

I know that by symmetry, $d(x,y) = d(y,x)$,
I also know that the Triangle Inequality holds for distances; i.e. $d(x,z)\le d(x,y)+d(y,z)$.

But I just can't figure out how to go about this proof.

Answer 1

$$\underbrace{d(x,y)+d(x,a)}_{\geq d(a,y)}+d(b,y)\geq d(a,y)+d(b,y)\geq d(a,b)$$

Answer 2

We have, by using the triangle inequality twice, that $$ d(a,b)\leq d(a,x)+d(x,b)\leq d(a,x)+d(x,y)+d(y,b) $$ Remove the middle party, and reorganize it a bit, and you get your inequality.