(In reply to re(3): Solution
I started by looking at PT of the form (2k+1, 2k^2+2k, 2k^2+2k+1). They have the property that the longer leg and the hypotenuse are consecutive, which makes the ratio between the legs large as k increases. At k=3 the ratio was already greater than 3,which gave PT(7,24,25).
Proving this was the smallest with a ratio greater than 3 is as simple as listing all the smaller PT and showing their ratios are less than 3. (3,4,5)-4/3=1.333; (6,8,10)-8/6=1.333; (5,12,13)-12/5=2.4; (8,15,17)-15/8=1.875.
BTW, I was also able to prove your expression of a,b, and r is satisfied by my right triangle by direct substitution.