You have gotten to . Great job. From this we have . We need this denominator to be at its maximum so that the fraction on the right will be at its minimum, to give us a value we can set for . To maximise the denominator, we need an upper bound, so we can use the triangle inequality...

Unfortunately, this denominator is unconstrained, but no matter, we only care about what's happening NEAR , so we restrict the distance between and to within some small value, say 1. Then .

The first thing we can determine from is that

So that means

Therefore we can say is minimised at .

So finally we can let and reverse each step to complete your proof.