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?