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.