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!



Estimating $\sum_{x_i < X} \prod_i \phi(x_i)/ \mathrm{lcm}(x_i)^a$
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 $x\_1=x\_2=\ldots=x\_n$.


