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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Could you kindly help me lokisan

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0My problem is posted at http://openstudy.com/groups/mathematics#/updates/4d92ba770b9d8b0bd98b08a9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes I understand taht

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Something strange is happening

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hello vern, I thought someone was helping you. Consider expanding,\[(a+b)^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The answers are getting distributed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Vern, I can help you in a minute.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@lokisan, you can continue with vern, as my problem is solved

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks a lot, and I must say you are a genius

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, 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, \[a^2+b^22ab<0\]that is,\[(ab)^2<0\] understanding that (ab)^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.
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