The question asked here is a follow up to [this question](https://mathoverflow.net/questions/358982/on-the-distribution-of-reduced-residues-by-montgomery-and-vaughan-missing-ca), 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](https://www.jstor.org/stable/1971274?seq=1), 
 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?