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[tex](ax+by)^2 \leq ax^2+by^2[/tex] for every a,b which satisfy

  1. Jul 8, 2011 #1
    [tex](ax+by)^2 \leq ax^2+by^2[/tex] for every a,b which satisfy [tex]0\leq a,b\leq1[/tex] and a+b=1.

    My book consider this as trivial however I have hard time to prove this, will appreciate your help.
     
  2. jcsd
  3. Jul 8, 2011 #2

    Mentallic

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    Re: inequality

    Expand the left side, let b=1-a and simplify. See if you can convert all factors involving a into a2-a :wink:
     
  4. Jul 8, 2011 #3
    Re: inequality

    I have tried many things including this one, how should I approach these kind of problems?
    What is the intuition? Is the solution process of these kind of problems really involves trial and error method or I miss something?
     
  5. Jul 8, 2011 #4
    Re: inequality

    Eventually i came to this nasty prove:

    [tex]ax^2+by^2-(ax+by)^2=ax^2+by^2-a^2x^2-2abxy-b^2y^2=(a-a^2)x^2-2abxy+(b-b^2)y^2=(a-a^2)x^2-2(a-a^2)xy+(b-b^2)y^2=(a-a^2)[(x-y)^2-y^2]+(b-b^2)y^2=
    (a-a^2)(x-y)^2+y^2[b-b^2+a^2-a]=(a-a^2)(x-y)^2+y^2[2ab] \geq 0[/tex]

    Is there a more efficient approaches to such problems?
     
  6. Jul 9, 2011 #5

    Mentallic

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    Re: inequality

    Yes, use the fact that a+b=1, thus b=1-a. Use this conversion early on.
     
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