Results 1 to 2 of 2

Thread: 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
    $\displaystyle \dfrac{a^{p+q}+b^{p+q}}{2}=\left(\dfrac{a^p+b^q}{2 }\right)\left(\dfrac{a^q+b^p}{2}\right)$
    where $\displaystyle a,b,p,q\in\mathbb{Z}$

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

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

    I've also noted that

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

    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 $\displaystyle \geq$

    But I solved it

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

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

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

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

    divide that by 2

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

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

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

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

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

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

    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: Nov 28th 2011, 08:56 PM
  2. Simplifying quotients involving powers of e
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Aug 14th 2011, 08:27 AM
  3. INequality with powers
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Feb 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: Aug 15th 2009, 06:19 AM

/mathhelpforum @mathhelpforum