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Prove inequality 2ab <= a^2 + b^2 from 0 <= (a - b)^2

  1. Sep 30, 2005 #1
    Show that: 2ab <= a^2 + b^2

    It follows that 0 <= (a - b)^2 is going to be always positive, then inequelity holds. But I think I need to prove that 2 times a times b has to be less than equal to the sum of the squares of a and b.
    Any suggestions?
  2. jcsd
  3. Sep 30, 2005 #2
    I believe that's all you need to do (expand (a-b)2).

    [tex]\left(a-b\right)^2\geq 0\implies a^2+b^2\geq 2ab[/tex]

    ...so I think your initial argument is proof enough.
  4. Sep 30, 2005 #3


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    Homework Helper

    Well, you practically solved it yourself, no?

    [tex]\left( {a - b} \right)^2 \ge 0 \Leftrightarrow a^2 - 2ab + b^2 \ge 0 \Leftrightarrow a^2 + b^2 \ge 2ab[/tex]
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