• Sep 29th 2010, 04:08 AM
magus
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.
• Sep 29th 2010, 12:10 PM
magus
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.