Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is this inequality true and provable?

  1. Aug 8, 2008 #1
    1. The problem statement, all variables and given/known data
    My question is whether the following inequality can be proven.

    2. Relevant equations

    3. The attempt at a solution
    I tried to write down the inequality in the form of it's primitives, where [tex]G\left(x\right)[/tex] is the primitive of [tex]g\left(x\right)[/tex] and [tex]H\left(x\right)[/tex] is the primitive of [tex]h\left(x\right)[/tex]. The inequality then becomes:


    But what next, or are there other means of getting a proof?
  2. jcsd
  3. Aug 8, 2008 #2
    Assuming [tex] a \leq b [/tex] and f is continuous on the interval [a,b], then

    [tex] \left|\int_a^bf\left(x\right)dx \right| \leq \int_a^b\left|f(x)\right|dx[/tex]

    which follows from the fact that [tex] f(x) \leq \left|f(x)\right| [/tex] and [tex] -f(x) \leq \left|f(x)\right| [/tex] and that

    If f,g are both continuous on the interval [a,b] and [tex] f(x) \leq g(x) [/tex] for all x in the interval. Then

    [tex] \int_a^b f(x)dx \leq \int_a^b g(x)dx [/tex]

    Rearranging and using the first inequality should give you the desired inequality.
  4. Aug 8, 2008 #3
    Oh, I see it now, it is indeed not that difficult.


    If we rearrange:


    Substituting [tex]f\left(x\right)=g\left(x\right)-h\left(x\right)[/tex] and using the first formula of snipez90, we get the proof.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook