This is almost true in general. If $r$ is even, the function on the right-hand side of the inequality is in fact constant, so the inequality does not hold.
This answer is not exactly elegant nor completely elementary; it just establishes the required result. I'll give a proof by calculus. First, note that for $\Delta = 0$, we have equality, so it is sufficient to show that $\Delta = 0$ is a local maximum in the given interval.
Define $f_{p_0}(\Delta) = (p_0 + \Delta)(1-(p_0+\Delta)^r) + (p_0 - \Delta)(1-(p_0 - \Delta)^r)$. Then $\Delta = 0$ is a maximum if $f^{\prime}_{p_0}(0) = 0$ and $f^{\prime\prime}_{p_0}(0) < 0$, where $f^{\prime}_{p_0}$ and $f^{\prime\prime}_{p_0}$ denote the first and second derivative of $f_{p_0}$.
We get $f^{\prime}_{p_0}(\Delta) = (r+1) ((p_0-\Delta)^r - (p_0+\Delta)^r)$, so $f^\prime_{p_0}(0) = 0$. Also, $f^{\prime\prime}_{p_0}(\Delta) = -r(r+1) ((p_0-\Delta)^{r-1} + (p_0+\Delta)^{r-1})$. Since $p_0 \pm \Delta > 0$, $p_0 > 0$ and hence, $f^{\prime\prime}_{p_0}(0) = -2r(r+1)p_0^{r-1} < 0$. Thus, we know that there is some interval around $\Delta = 0$ such that $f_{p_0}$ has a local maximum.
Analyzing $f'_{p_0}(\Delta) = 0$ gives $(p_0+\Delta)^r = (p_0-\Delta)^r$, or equivalently, $\left( \frac{p_0+\Delta}{p_0-\Delta} \right)^r = 1$ for the case that $\Delta \neq p_0$. Since there are at most two solutions to $x^r =1$ in the real numbers, namely $x = 1$ and $x = -1$, we need to consider two cases:
- $p_0 + \Delta = p_0 - \Delta$. Then $\Delta = 0$, which is the case we already know.
- $p_0 + \Delta = \Delta - p_0$. Then every $\Delta$ is a solution, but $p_0 = 0$.
In the first case, $f_{p_0}$ has exactly one extremal point, so the maximum is a global maximum on $\Delta < p_0$.
In the second case, $f_{p_0}$ is constant. Note that the second case is only possible if $r$ is even; in that case, the exception mentioned above applies.
The case $\Delta = p_0$ need not be considered, since $0 < p_0 - \Delta$ by assumption.