Let $\tau(n)$ be the divisor function. Let $a$ be either a constant, or a function of $X$ that is slowly varying with $X,$ say $X/\log(X)<a(X)<X \log(X),$ for example. I want to lower bound sums of the following form
$$
\sum\_{1\leq n\leq X} a^{1-\frac{\tau(n)}{D}},\quad(1)
$$
and
$$
\sum\_{1\leq n\leq X: n\in I} a^{1-\frac{\tau(n)}{D}},\quad(2)
$$
where $I$ is an index set of roughly $n/2$ integers. Here, $0<D\leq X$ is also a function of $X$ that is slowly varying with $X.$


We can, of course, factor out an $a$.


I know the divisor function values for the interval $[1,X]$ obeys a kind of arcsine distribution law, and that should help obtain a bound, but I haven't been able to obtain one. Directly applying the AGM (arithmetic-geometric mean inequality) seems to be too weak.


**Edit:** For (1) we can apply the AGM 
$$
a X\left( X^{-1} \sum\_{1\leq n\leq X} a^{-\tau(n)/D}\right)\geq a X\left( \prod\_{1\leq n\leq X} a^{-\tau(n)/D}\right)^{1/X}
$$
which gives
$$
a X\left( a^{-\sum\_{1\leq n\leq X} \tau(n)/DX}\right)\approx aX \left( a^{-(X \log X +(2\gamma-1)X+o(\sqrt{X}))/DX}\right)= X a^{1-O(\frac{\log X}{D})}.
$$
As for 2, the question is how to obtain a worst case bound by using the distribution I alluded to in the original question, plus something else(?).



Lower Bound on "exponential" sum
I think I have an answer, whereby the same asymptotic order lower bound as in (1) is obtained for (2). *Please comment if there is an issue with the argument:*


Consider
$$
\sum\_{1\leq n\leq X: n\in I} a^{1-\frac{\tau(n)}{D}},\quad(2)
$$
where the question specified $I$ to be an index set of roughly $X/2$ integers. Thus I will take $\mid I\mid \in [(X-t)/2,(X+t)/2]$ for some fixed positive integer $t.$ By using the same argument as for (1) we obtain
$$
\sum\_{1\leq n\leq X: n\in I} a^{1-\frac{\tau(n)}{D}}\geq a \mid I\mid \left( a^{-\sum\_{1\leq n\leq X:n\in I} \tau(n)/(D \mid I\mid)}\right)\quad(3).
$$
Now focus on the sum in the exponent of (3). Let us assume that we choose $I$ to be an index set (there can be more than one) which achieves the maximal sum of divisors of $n\in I$ subject to the limits on $\mid I\mid$ given above.


If there were a function $f$ growing strictly faster than $O(X \log X)$ such that
$$
\sum\_{1\leq n\leq X: n\in I} \tau(n) \geq f(X),
$$
as $X$ grows, then this would imply that
$$
\sum\_{1\leq n\leq X} \tau(n) \geq f(X),
$$
since the divisor function is always positive. But this would contradict the fact that the full divisor sum is known to be $O(X \log X).$ Thus the partial sum is at most $O(X \log X).$


Now rewrite (3), using the order estimate from above and the fact that $\mid I\mid =\delta X$ where $\delta>0$ is constant. We get
$$
\sum\_{1\leq n\leq X: n\in I} a^{1-\frac{\tau(n)}{D}}\geq a \delta X \left( a^{-O(X \log X)/(D \delta X)}\right)\gg Xa^{1-O(\frac{\log X}{D})} .
$$


