Given a finite (multi)set of elements $\{x_1, \ldots, x_n\}$ the arithmetic mean $\mathsf{AM}$ is less than or equal to the maximum element call it $\max$. In otherwords, $\mathsf{AM} \leq \max$. But equality only happens when all elements in the population are equal. Another way to put this is when the average absolute deviation from the mean is $0$. Call the average absolute deviation from the mean $\mathsf{AAD}$. I would like to extend the $\mathsf{AM} \leq \max$ to include $\mathsf{AAD}$ as well.
To do so, I considered an extreme case: Every element but $1$ is $\max$ and one element has a minimal deviation (call it $\mathsf{minD}$). Thus, the average of this extreme case establishes the following inequality: $$\mathsf{AM} \leq \frac{\max + \max + \ldots + (\max - \mathsf{minD})}{n}$$ $$\mathsf{AM} \leq \frac{n\max}{n} - \frac{\mathsf{minD}}{n}$$ $$\mathsf{AM} \leq \max - \frac{\mathsf{minD}}{n}$$
I wanted to relate this back to $\mathsf{AAD}$ so I observed that $\mathsf{AAD}$ is also an average of a population, thus, by Samuelson's Inequality we have:
$$\mathsf{minD} \geq \mathsf{AAD} - \text{stddev}(\mathsf{AAD})\sqrt{n-1}$$
subbing back in we get
$$\mathsf{AM} \leq \max - \frac{\mathsf{AAD} - \text{stddev}(\mathsf{AAD})\sqrt{n-1}}{n}$$
Using inequality (3) from here we get an upperbound for $\text{stddev}(\mathsf{AAD})$ of
$$\text{stddev}(\mathsf{AAD}) \leq \sqrt{(\max - \mathsf{AAD})(\mathsf{AAD} - \min)}$$
Plugging back in we get:
$$\mathsf{AM} \leq \max - \frac{\mathsf{AAD} - \sqrt{(\max - \mathsf{AAD})(\mathsf{AAD} - \min)(n-1)}}{n}$$
But I would like to know if this bound can be improved and if the $\min$ term can be removed. Any guidance including hints would be appreciated.
