Variance of negative binomial always greater than expected value

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I have this question.

When modelling counts Y defined on ${0,1,2,...} $ , it is common to use the Poisson distribution, when $ E[Y] = var[Y] $ is implied. Show that the negative binomial distribution may be a better choice if $var[Y] > E[Y]$.

The definitions of the NBIN are

$ \sigma^2=Var(x)=\dfrac{r(1-p)}{p^2} $

$ \mu=E(X)=\dfrac{r}{p} $

The problem is that for values of $ p > 0.5 $, then the $var[Y] < E[Y]$, not $>$ as the question is suggesting.

How does the above statement hold and how am I meant to prove this?