The question asked here is a follow up to this question, which was answered by user GH from MO.

As there, let $q$ be a natural number, let $P = \phi(q)/q$ be the "probability" that a randomly chosen integer is relatively prime to q. Then, following Montgomery and Vaughan in On the distribution of reduced residues, the second moment of the number of reduced residues modulo $q$ in an interval of length $h$ about its mean, $hP$, can be stated as $$ M_2(q;h) = qP^2 \sum_{\substack{{r \mid q }\\{r > 1}}} \mu(r)^2 \left( \prod_{\substack{ {p \mid q }\\{p \nmid r} }} \frac{p(p-2)}{(p-1)^2} \right) r^2 \phi(r)^{-2} \left\{ \frac{h}{r}\right\}\left( 1 - \left\{ \frac{h}{r}\right\}\right). $$ Applying the inequality $\{\alpha\}(1 - \{\alpha\}) \leq \alpha$ then gives the upper bound $$\tag{1} M_2(q;h)\leq qhP. $$ But since $\{\alpha\}\leq 1$, we also have that $\{\alpha\}(1 - \{\alpha\}) \leq 0.25$, and hence that $$\tag{2} M_2(q;h) \leq 0.25 \, qP^2 \sum_{\substack{{r \mid q }\\{r > 1}}} \mu(r)^2 \left( \prod_{\substack{ {p \mid q }\\{p \nmid r} }} \frac{p(p-2)}{(p-1)^2} \right) r^2 \phi(r)^{-2}. $$

Question

Will replacing the inequality $\{\alpha\}(1 - \{\alpha\}) \leq \alpha$ with the inequality $\{\alpha\}(1 - \{\alpha\}) \leq 0.25$ lead to an improved upper bound for $M_2(q;h)$ compared to (1)? And is it possible to express the upper bound in (2) a bit more simplified or asymptotically?

The question asked here is a follow up to this question, which was answered by user GH from MO. [EDIT: The question is edited for clarification after receiving a comment on the original posting.]

As there, let $q$ be a natural number, let $P = \phi(q)/q$ be the "probability" that a randomly chosen integer is relatively prime to q. Then, following Montgomery and Vaughan in On the distribution of reduced residues, the second moment of the number of reduced residues modulo $q$ in an interval of length $h$ about its mean, $hP$, can be stated as $$ M_2(q;h) = qP^2 \sum_{\substack{{r \mid q }\\{r > 1}}} \mu(r)^2 \left( \prod_{\substack{ {p \mid q }\\{p \nmid r} }} \frac{p(p-2)}{(p-1)^2} \right) r^2 \phi(r)^{-2} \left\{ \frac{h}{r}\right\}\left( 1 - \left\{ \frac{h}{r}\right\}\right). $$ Applying the inequality $\{\alpha\}(1 - \{\alpha\}) \leq \alpha$ then gives the upper bound $$\tag{1} M_2(q;h)\leq qhP, $$ while a lower bound is given by $$ M_2(q; h) \geq qhP - qhPQ + O(qhP^2) $$ where $Q=\prod_{\substack{{p \mid q}\\{p>h}}} (1-1/p)$.

Since $\{\alpha\}\leq 1$, we also have that $\{\alpha\}(1 - \{\alpha\}) \leq 0.25$. This allows for a second upper bound, $$\tag{2} M_2(q;h) \leq 0.25 \, qP^2 \sum_{\substack{{r \mid q }\\{r > 1}}} \mu(r)^2 \left( \prod_{\substack{ {p \mid q }\\{p \nmid r} }} \frac{p(p-2)}{(p-1)^2} \right) r^2 \phi(r)^{-2}. $$

As pointed out by user Ofir Gorodetsky in the comment below, the upper bound in (1) is already optimal for those choices of $q$ such that the upper and lower bounds are of same order. For example, taking $q=\prod_{p\leq h^C} p$ for large enough $C$, we have that $M_2(q;h)\asymp qhP$.

Here I'd like to consider the rather opposite situation, when $C$ is small. Specifically, take $q=\prod_{p\leq h^C} p$, where $C<1$. Then we have that $Q=1$ and the lower bound takes the form $$ M_2(q; h) \geq O(qhP^2), $$ which leaves some more wiggle room between the upper and lower bounds, and I'd like to know whether the upper bound in (2) this case provides some improvement on (1).

Question

Given $q=\prod_{p\leq h^C} p$, where $C<1$. Does the upper bound for $M_2(q;h)$ stated in (2) provide a sharper bound compared to (1) in this case? And is it possible to express the upper bound in (2) a bit more simplified or asymptotically?