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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?

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  1. anonymous
    • 5 years ago
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    Could you kindly help me lokisan

  2. anonymous
    • 5 years ago
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    ? on your question?

  3. anonymous
    • 5 years ago
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    My problem is posted at http://openstudy.com/groups/mathematics#/updates/4d92ba770b9d8b0bd98b08a9

  4. anonymous
    • 5 years ago
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    lol

  5. anonymous
    • 5 years ago
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    no help?

  6. anonymous
    • 5 years ago
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    Yes I understand taht

  7. anonymous
    • 5 years ago
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    Something strange is happening

  8. anonymous
    • 5 years ago
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    Hello vern, I thought someone was helping you. Consider expanding,\[(a+b)^2\]

  9. anonymous
    • 5 years ago
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    The answers are getting distributed

  10. anonymous
    • 5 years ago
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    Oh..

  11. anonymous
    • 5 years ago
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    Vern, I can help you in a minute.

  12. anonymous
    • 5 years ago
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    @lokisan, you can continue with vern, as my problem is solved

  13. anonymous
    • 5 years ago
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    Thanks a lot, and I must say you are a genius

  14. anonymous
    • 5 years ago
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    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.

  15. anonymous
    • 5 years ago
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    great! thanks

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