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!

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 $$x_1=x_2=\ldots=x_n$$.