Variation on Pythagoras: If a^2 + b^2 = c^2, then a + b = c*sqrt(2)?

Question

This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with derivative of the Pythagorean Theorem using calculus, trigonometry, geometry, or plain old algebra, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes: Prove that for a, b, c > 0, if a^2 + b^2 = c^2, then a + b = c*sqrt(2)

Answer 1

Here is a counter example:

Right triangle a=5, b=12, C=13

a^2+b^2 = 5^2 + 12^2 = 25+144 = 169 = 13^2 = c^2.

BUT....

a+b = 17 while c*sqrt(2) = 13*sqrt(2) > 17

Please re-check the statement of your theorem and repost.