Suppose that we have $q$ positive integers $a_1, \ldots, a_q$ satisfying $a_1 \leq \cdots \leq a_q \leq q$. I'm interested in the possible types of behaviors for the function given by $$f(x) = (a_1^{x-1} + \cdots + a_q^{x-1})^{1/x},$$ where $x \in [2,\infty)$. In particular, I'm interested in the behavior at integer values of $x$ in that range, but I think that the continuous version of $f$ might be easier to handle.

It isn't hard to see that $\lim_{x \to \infty} f(x) = a_q$. I can show that if there is an $x_0 \in (2,\infty)$ with $f(x_0) > a_q$, then $f$ is decreasing at $x_0$, and furthermore I can show that $f$ is either eventually decreasing, increasing, or constant (this almost entirely comes from the number of $i$ with $a_i = a_q$).

Each example $(a_1,\ldots,a_q)$ that I've used also has the nice property that it is either always decreasing, always constant (this in fact only occurs if $q=a_q$ and $a_i = a_q$ for each $i$), or decreasing on $[2,x_0)$ and increasing on $(x_0,\infty)$ for some $x_0$ which depends on the $a_i$. Does anyone know of either a proof or counterexample of this property? Is any other behavior possible? Simply computing the derivative of $f$ doesn't seem to be too helpful, but perhaps I'm missing the correct viewpoint on this derivative.

**Edit**: added the restriction $a_q \leq q$. Without this restriction, $f(x)$ can be strictly increasing as well, e.g. $a_1=3,a_2=4$. An example of the decreasing/increasing is $a_1=1, a_2=2$.

Inequalities. The basic theorem is that the power mean of order $x$ increases with $x$, strictly so unless all the numbers you're taking the mean of are equal. In your case you might want to relate $f(x)$ to the $x$th order power mean of $a_1, \ldots, a_q$, weighted by $a_1^{-1}, \ldots, a_q^{-1}$. – Tom Leinster May 16 '12 at 16:09