Hi,

If you're ever stuck trying to prove a logical implication (a statement of the form "if A, then B"), it's always a good idea to try proof by contradiction. Proof by contradiction starts by assuming statement A holds, but then supposing statement B does not; if you're able to derive a contradiction (a mathematical statement that is necessarily false), you may conclude statement B must hold. (This is a widely used technique in mathematics). If you want to try this technique for yourself, without seeing my solution, stop here and give it a shot (hint: it should not involve anything more complicated than some basic algebra after starting the proof this way).

For this particular problem, a proof by contradiction would go as follows:

Let a, b, and c be positive numbers. Suppose a^{2} + b^{2} = c^{2}.

Assume for sake of contradiction that a + b > c(√2). We also know c(√2) > 0 because c > 0. Thus we have that 0 < c(√2) · c(√2) < (a + b) · (a + b), so we have that 2c^{2} < (a + b)^{2} = a^{2} + 2ab + b^{2} = c^{2} + 2ab. Subtracting c^{2} from both sides, we then have c^{2} < 2ab, so a^{2} + b^{2} < 2ab, and therefore a^{2} - 2ab + b^{2} < 0. Factoring the left hand side, we get (a - b)^{2} < 0. This is a contradiction because any real number squared is necessarily non-negative.

Therefore, our assumption that a + b > c(√2) must have been false (as we were able to derive a contradiction), and so we conclude that we must have a + b ≤ c(√2).

You could just as easily make a constructive argument, completely avoiding contradiction, essentially just doing these steps in reverse order with the inequalities flipped. This is not a natural argument though, and it should seem to come out of thin air; what's done above is really the motivation for what I'm about to write. Here's how a constructive argument would go:

Let a, b, and c be positive numbers. Suppose a^{2} + b^{2} = c^{2}.

We know that (a - b)^{2} ≥ 0 because any real number squared is non-negative. Then a^{2 }- 2ab + b^{2} ≥ 0 by expanding the left hand side, and so a^{2} + b^{2} ≥ 2ab. From here, we have that c^{2} ≥ 2ab, making use of our assumption. Therefore 2c^{2} ≥ c^{2} + 2ab = a^{2} + 2ab + b^{2} = (a + b)^{2} by adding c^{2} to both sides and then factoring the right side. Now, since both (a + b) and c(√2) are positive numbers, we may take the square root of both sides of the inequality to obtain a + b ≤ c(√2), as desired.

It is likely that there are other more geometric-flavored proofs, but this is what came to mind most immediately for me. (If you're seeking that kind of argument, I'd start by assuming there is a right triangle with side lengths a, b, and c, and then try to work with that geometrically). Let me know if you have any follow-up questions.

Thanks,

Aaron