Variation on Pythagoras: If $a^2 + b^2 = c^2$, then $a + b \\leq c\\sqrt{2}$?

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 $a + b \\leq c\\sqrt{2}$

Answer 1

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 a² + b² = c².

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 2c² < (a + b)² = a² + 2ab + b² = c² + 2ab. Subtracting c² from both sides, we then have c² < 2ab, so a² + b² < 2ab, and therefore a² - 2ab + b² < 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 a² + b² = c².

We know that (a - b)² ≥ 0 because any real number squared is non-negative. Then a² - 2ab + b² ≥ 0 by expanding the left hand side, and so a² + b² ≥ 2ab. From here, we have that c² ≥ 2ab, making use of our assumption. Therefore 2c² ≥ c² + 2ab = a² + 2ab + b² = (a + b)² by adding c² 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