Estimating $\sum_{x_i < X} \prod_i \phi(x_i)/ \mathrm{lcm}(x_i)^a$ I would like to estimate from above  the following sum
$$
\sum_{1 \leq x_1 < X} .. . \sum_{1 \leq x_n < X}  \frac{\prod_{1 \leq i \leq n } \phi(x_i)}{\mathrm{lcm}(x_1, .., x_n)^a}.
$$
$\phi$ is the Euler totient function and $a$ is a positive integer less than $2n$. A trivial estimate would be $\ll X^{2n - a}$. Is there a way to get a better bound? Thank you!
 A: One can improve on $X^{2n-a}$ as long as $(a,n)\neq(1,1)$ (for $a=n=1$, the sum grows like $X/\zeta(2)$ so there's no room for improvement). Let us introduce $$f_n(m) := \# \{ (x_1,\ldots,x_n) : \mathrm{lcm}(x_1,\ldots,x_n) = m\},$$ which satisfies $f_n(m) \le \tau(m)^n \ll_{n,\varepsilon} m^{\varepsilon}$ where $\tau$ is the usual divisor function. First suppose that $a >n$. We have $$\frac{\prod_{i=1}^{n} \phi(x_i)}{\mathrm{lcm}(x_1,\ldots,x_n)^a} \le \frac{\prod_{i=1}^{n} \phi(x_i)}{\max_{1\le i \le n} x_i^n} \frac{1}{\mathrm{lcm}(x_1,\ldots,x_n)} \le \frac{1}{\mathrm{lcm}(x_1,\ldots,x_n)}$$
which gives the upper bound $$< \sum_{1 \le m < X^{na}} \frac{f_n(m)}{m} = X^{o(1)}.$$
This is optimal since we have the lower bound $\ge 1$. We may assume $n \ge a$ from now on. Your sum is $$< X^n \sum_{1 \le m < X^{na}} \frac{f_n(m)}{m^a}  \ll_{a,n} X^{n+o(1)}.$$
If $n\neq a$ this already beats $X^{2n-a}$.
We now sketch how one can do better than $X^{n+o(1)}$. In section 3 of R. R. Hall's ``The distribution of squarefree numbers'' (Reine Angew. Math. 394 (1989), 107–117), the author introduces `total decomposition sets', which help him study a sum related to yours with $a=2$ and $n \ge 2$ (see his Lemma 3). Modifying the proof of Lemma 3 slightly, we obtain the bound
$$\ll_{a,n} \begin{cases} X^{n-\frac{n(a-1)}{n-1}+o(1)} & \text{if }n > a,\\ X^{1+o(1)} & \text{if }n=a,\end{cases}$$
which beats $2n-a$ as long as $(a,n) \neq (1,1)$.
The dependence on $a,n$ can be made explicit but is quite horrible. To see that $n=a$ and $a=1$ are optimal consider the contribution of $x_1=x_2=\ldots=x_n$.
