Question

how do you prove that the sum of the squares of two numbers is always greater than or equal to twice their product?

Answer 1

Could you kindly help me lokisan

Answer 2

? on your question?

Answer 3

My problem is posted at http://openstudy.com/groups/mathematics#/updates/4d92ba770b9d8b0bd98b08a9

Answer 4

lol

Answer 5

no help?

Answer 6

Yes I understand taht

Answer 7

Something strange is happening

Answer 8

Hello vern, I thought someone was helping you. Consider expanding,\[(a+b)^2\]

Answer 9

The answers are getting distributed

Answer 10

Oh..

Answer 11

Vern, I can help you in a minute.

Answer 12

@lokisan, you can continue with vern, as my problem is solved

Answer 13

Thanks a lot, and I must say you are a genius

Answer 14

Okay, vern, this is what you should do: assume what you have to prove is FALSE. Note, you're trying to eventually show that\[a^2+b^2\ge 2ab\]Let's assume that's false. Then the following would be true:\[a^2+b^2<2ab\]but then, \[a^2+b^2-2ab<0\]that is,\[(a-b)^2<0\] understanding that (a-b)^2 = a^2 + b^2 -2ab. But this conclusion is false, since this is saying that the square of a real number is negative (you can't have such a thing). Therefore, our assumption that \[a^2+b^2<2ab\]leads to a contradiction. We must therefore conclude that the assumption that lead us to this contradiction is false, which means\[a^2+b^2 \ge 2ab\] is true.

Answer 15

great! thanks