1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    TD

    User Avatar
    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?