anonymous
  • anonymous
how do you prove that the sum of the squares of two numbers is always greater than or equal to twice their product?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Could you kindly help me lokisan
anonymous
  • anonymous
? on your question?

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anonymous
  • anonymous
lol
anonymous
  • anonymous
no help?
anonymous
  • anonymous
Yes I understand taht
anonymous
  • anonymous
Something strange is happening
anonymous
  • anonymous
Hello vern, I thought someone was helping you. Consider expanding,\[(a+b)^2\]
anonymous
  • anonymous
The answers are getting distributed
anonymous
  • anonymous
Oh..
anonymous
  • anonymous
Vern, I can help you in a minute.
anonymous
  • anonymous
@lokisan, you can continue with vern, as my problem is solved
anonymous
  • anonymous
Thanks a lot, and I must say you are a genius
anonymous
  • anonymous
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.
anonymous
  • anonymous
great! thanks

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