Question

Problem with my proof [calculus]

Answer 1

I don't see any any errors.

But as soon as you've proved xy >= 0, you know that 4x\^2 + 6xy + 4y\^2 is a sum of nonnegative terms, which immediately leads to a proof.

Another approach is to write 4x\^2 + 6xy + 4y\^2 = (7/2)(x + y)\^2 + (1/2)(x - y)\^2.

Answer 2

Isn't 4x^2 + 6xy + 4y^2 = x^2 + y^2 + 3(x+y)^(2)?

It seems like that would be easier

Answer 3

I haven't been able to spot any logical flaws (doesn't mean there aren't any) but the hint might just be trying to tell you that your proof is unnecessarily complex. If it were just that there are easier methods then I don't think that would be good enough reason to criticise it, but you did quite a lot of work only to create a problem harder than the original question. Near the end, starting from this line:

4x^(2) \+ 8xy + 4y^(2) <= 4xy

you prove indirectly that 4x^(2) \+ 4xy + 4y^(2) <= 0 is contradictory, using the fact proved early on that xy >= 0. But as soon as you had that, you could have proved that 4x^(2) \+ 6xy + 4y^(2) can't be negative using the same method, by considering 4x^(2) \+ 4y^(2) <= -6xy <= 0.

Answer 4

There is a simpler method which utilizes the AMGM inequality which is not hard to prove for two inputs. Then you wont have to use a proof by contradiction.

Answer 5

This proof is correct (someone tell me if I’m wrong). If the instructor is saying that there’s a logical flaw, then I’d ask what it is, because as far as I can tell you didn’t make any mistakes.

The way you go about proving this is unnecessarily convoluted, but I don’t think it’s wrong.