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Help with algebra

  1. Mar 19, 2005 #1
    How can you proof that

    [tex]a^2 (1 + b^4) + b^2(1 + a^4) \leq (1 + a^4)(1 + b^4)[/tex]?

    I factorised [tex]a^2 (1 + b^4) + b^2(1 + a^4)[/tex] to [tex](a^2 + b^2)(1+a^2b^2)[/tex], but I don't really know where to go from here.
  2. jcsd
  3. Mar 19, 2005 #2
    Assuming a and b are real, what do you know must be true about the following expression?

  4. Mar 19, 2005 #3
    I get it now. Thanks!

    BTW, does anyone know a good site that teaches you factoring 'tricks'?
  5. Mar 19, 2005 #4

    matt grime

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    This isn't a factoring trick, but jsut a common observation that if you want to show

    X > Y

    then this is the same as showing

    X-Y > 0

    and what do we know is always positive?
  6. Mar 19, 2005 #5
    I don't see how you got this.
  7. Mar 19, 2005 #6
    Divine inspiration, perhaps?
  8. Mar 19, 2005 #7
    The general procedure is:

    Take the original inequality that you are trying to prove and move everything over to one side, so that you have (stuff) >= 0. Now rearrange (stuff) so that it consists entirely of squares of real numbers, added and multiplied together. Since such an expression is automatically greater than or equal to zero, the original inequality must be true.

    If it turns out that such a rearrangement can't be made, then you can't prove the inequality without more information about a and b, such as a > b or something similar. But in this example it was possible.
  9. Mar 20, 2005 #8
    I think you misundestood me. It's not the concept I don't understand, just the result. I don't see how you got to that expression.
  10. Mar 20, 2005 #9
    Oops, sorry about that. Start with the expression that we want to show is greater than or equal to zero.


    Since we are showing this is >= zero, it doesn't change anything to multiply the expression by a positive number, namely 2.


    Split up the first term:


    Now collect terms 1 and 2 together, and 3 and 4 together, and you end up with the expression you are looking for.
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