Results 1 to 2 of 2

Math Help - Inequality involving added powers

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    72

    Inequality involving added powers

    Hello ... again.

    My current problem is this
    Prove
    \dfrac{a^{p+q}+b^{p+q}}{2}=\left(\dfrac{a^p+b^q}{2  }\right)\left(\dfrac{a^q+b^p}{2}\right)
    where a,b,p,q\in\mathbb{Z}

    My approach was to reduce this first into something else easier to prove

    a^{p+q}+b^{p+q}\geq a^pb^p+a^qb^q

    I've also noted that

    a^{p+q}+b^{p+q}\geq2
    a^{p+q}+a^pb^p+a^qb^q+b^{p+q}=(a^p+b^q)(a^q+b^p)\g  eq4
    and a^pb^p+a^qb^q\geq2

    Again, any help would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    I just realised that I made it an equation rather then an inequality. that's supposed to be \geq

    But I solved it

    I eventually observed that (a^p-b^q)(a^q-b^p)\geq0

    You then get a^{p+q}-a^pb^p-a^qb^q+b^{p+q}\geq0

    Move the negative terms over and that gives the simplified version of what I wanted to prove.

    a^{p+q}+b^{p+q}\geq a^pb^p+a^qb^q

    divide that by 2

    \frac{1}{2}a^{p+q}+\frac{1}{2}b^{p+q}\geq\frac{1}{  2} a^pb^p+\frac{1}{2}a^qb^q

    then add \frac{1}{2}a^{p+q}+\frac{1}{2}b^{p+q} to both sides

    a^{p+q}+b^{p+q}\geq \frac{1}{2}a^{p+q}+\frac{1}{2}a^pb^p+\frac{1}{2}a^  qb^q+\frac{1}{2}b^{p+q}

    Since (a^p+b^q)(a^q+b^p)=a^{p+q}+a^pb^p+a^qb^q+b^{p+q}

    We can write a^{p+q}+b^{p+q}\geq\frac{1}{2}(a^p+b^q)(a^q+b^p)

    Divide that again by 2 and you get \dfrac{a^{p+q}+b^{p+q}}{2}\geq\left(\dfrac{a^p+b^q  }{2}\right)\left(\dfrac{a^q+b^p}{2}\right)

    Sorry about that. It's just hard catching all my tex mistakes right away.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. equation involving addition of powers
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 28th 2011, 08:56 PM
  2. Simplifying quotients involving powers of e
    Posted in the Algebra Forum
    Replies: 4
    Last Post: August 14th 2011, 08:27 AM
  3. INequality with powers
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 22nd 2011, 03:02 AM
  4. Arithmetic involving powers.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 18th 2010, 09:20 PM
  5. Functions involving powers of logarithms
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: August 15th 2009, 06:19 AM

/mathhelpforum @mathhelpforum