Hello,

I too would be very curious about this bound and will think further about this. The nice thing about the $u_i(y)$, as I recall my advisor having told me, is that the estimates for geometric sums can be independent of their length. So, similar to an estimate for geometric sums found in Chapter 25 of H. Davenport, Multiplicative Number Theory, Third Edition, one has that

$$
u_i(y) \ll \min\left(N, \frac{1}{\|b_iy\|}\right).
$$ 

Following the perspective of this enlightening post, <a href="http://terrytao.wordpress.com/2011/12/31/montgomerys-uncertainty-principle/">Montgomery's uncertainty principle</a>, from the blog of Professor Tao, another nice way to think of it is that the $u_i(y)$ is the Fourier transform of a function $f: \mathbb{Z} \rightarrow \{0,1\}$ that avoids $p-1$ residue classes modulo $p$ for each prime $p$ dividing $b_i$.

A technique for the estimate you require which initially occurred to me was one I learnt from:

S. Baier, L. Zhao, <a href="http://arxiv.org/abs/math/0605563">Primes in Quadratic Progressions on Average</a>, Math. Ann., Vol. 338, 2007, No. 4, pp. 963-982.

The analogous situation to the bound you need is treated there at equations (4.2) and (4.3). But there is a difference between that problem and this one. Over there, one only has the case $b_i=1$ and the maximum value of $\sum_{m\leq N}e(m y)$ can be avoided by avoiding $(-Q/N,Q/N)$. But here, when $b_i>1$, the interval $(Q/N, 1/2)$ still contains values where $u_i(y)$ attains its maximum.

So one observes that by naively applying Hölder's inequality with $\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}=1$ to obtain

$$
\int_{Q/N}^{1/2}\left|u_1(y)u_2(y)u_3(y)\right|dy \ll \prod_{i=1}^3\left(\int_{Q/N}^{1/2}\left|u_i(y)\right|^{n_i}dy\right)^{1/n_i},
$$

where, for each $u_i(y)$, if one ignores that one is integrating from $Q/N$ and instead integrates from $0$, one has that

$$
\int_{0}^{1/2}\left|u_i(y)\right|^{n_i}dy \ll b_i\left(\int_{0}^{1/(b_iN)}N^{n_i} dy + \int_{1/(b_iN)}^{1/2b_i}\frac{1}{(b_iy)^{n_i}}dy\right) \ll N^{n_i-1}.
$$

which only gives $O(N^2)$ for the term you need to estimate. It can be observed that throwing away the part from $(0,Q/N)$ will not save much like this. I guess the key is to account for the fact that $u_1u_2u_3$ is periodic modulo $\frac{1}{b_1b_2b_3}$ and estimate as follows: 

$$
\int_0^1= b_1b_2b_3\left(\int_0^{Q/N}+\int_{Q/N}^{1/(b_1b_2b_3)}\right),
$$

whence 

$$
\int_0^{Q/N} = \frac{1}{b_1b_2b_3}\int_0^1 + O\left(\int_{Q/N}^{1/(b_1b_2b_3)}\right).
$$

Now apply Hölder's inequality to get

$$
\int_{Q/N}^{1/(b_1b_2b_3)}u_1(y)u_2(y)u_3(y)dy \ll \prod_{i=1}^3\left(\int_{Q/N}^{1/(b_1b_2b_3)}\left|\frac{1}{b_iy}\right|^{n_i}dy\right)^{1/n_i} \ll \frac{1}{b_1b_2b_3}\left(\frac{N}{Q}\right)^2.
$$