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.