Equality part of Holder's inequality Given $ab=\dfrac{a^p}{p}+\dfrac{b^q}{q}$, where $p\ge 1$ and $\dfrac1p+\dfrac1q=1$, how does one show that $a^p=b^q$?

Using weighted A.M.-G.M.(A.M.-G.M.) we have,
$\displaystyle \frac{1}{p}(a^{p})+\frac{1}{q}(b^q)\ge (\frac{1}{p}+\frac{1}{q})((a^{p})^{\frac{1}{p}}(b^q)^{\frac{1}{q}})^{\frac{1}{\frac{1}{p}+\frac{1}{q}}}=ab$
We know that equality in case of A.M.-G.M. inequality holds iff the terms are equal.So $\displaystyle \frac{1}{p}(a^{p})+\frac{1}{q}(b^q)=ab$ would imply $a^p=b^q$
