Is $\int_{A_t} \frac{1}{|x_j|^{p}}\, dx \leq \sum_{j=1}^{d} \int_{A_{j,t}}\frac{1}{|x_j|^p}\, dx \leq \int_{A_t} \frac{1}{|x|^p} \,dx$? Let $t>0$  and  $A_t= \{ x\in \mathbb R^{d}: |x|>t\},$ $A_{j,t}=\{x\in A_t: |x_j|> |x_l| \ \text{for all} \    l\neq j \}.$

Question: (1)  Can we say $A_t \subset \cup_{j=1}^d A_{j,t}$? (Motivation for this is:$\int_{A_t} \frac{1}{|x_j|^{p}} \,dx\leq C \sum_{j=1}^{d} \int_{A_{j,t}}\frac{1}{|x_j|^p} \,dx$)
  (2) Can we say $  \int_{A_{j,t}}\frac{1}{|x_j|^p} \,dx \leq C_1 \int_{A_t} \frac{1}{|x|^p} \,dx$? (Assume $p$ is large/small enough  so integral in questions make sense. $C, C_1$ constant.)

Motivation: Roughly speaking this kind of trick has been used in handing oscillatory integral for $|x|>t$ in the paper I  have been reading. And  $\frac{1}{|x_j|}$ appears after integration by parts for the oscillatory integral.
My attempt: If $x\in A_t$, then $|x|>t$ with $x=(x_1,..., x_j,..., x_d)$, now we might needs to consider several cases to show $x\in A_{j,t}$.
 A: (1) Obviously not, since  if $|r| > \frac t{\sqrt{d}}$, then $(r,...,r) \in A_t$, but not in any $A_{j,t}$. More generally, any point $x$ will not be in the union if the largest coordinate magnitude is shared by two or more coordinates.
However, it is evidently true that $A_t \subset \bigcup_j \overline A_{j,t}$, since for any $(x_1, ..., x_d) \in A_t$, there are only finitely many coordinate magnitudes, so one of them must be largest.
(2) Yes. For $x \in \overline A_{j,t}$, $$|x|^2 = \sum_i x_i^2 \le \sum_i x_j^2 =  d|x_j|^2$$
Therefore $$|x|^p \le d^{p/2}|x_j|^p\\\frac 1{|x|^p} \ge \left(\frac1d\right)^{p/2}\frac1{|x_j|^p}$$
so
$$\int_{\overline A_{j,t}} \frac{1}{|x|^p} \,dx \ge \left(\frac1d\right)^{p/2}\int_{\overline A_{j,t}}\frac{1}{|x_j|^p} \,dx$$
By symmetry, the integral of $\frac{1}{|x|^p}$ over $\overline A_{j,t}$ is the same for all $j$, and since the various $\overline A_{j,t}$ intersect in sets of measure $0$, we have $$\int_{\overline A_{j,t}} \frac{1}{|x|^p} \,dx = \frac 1d\int_{A_t} \frac{1}{|x|^p} \,dx$$
Which gives $$\int_{\overline A_{j,t}}\frac{1}{|x_j|^p} \,dx \le d^{p/2 -1}\int_{A_t} \frac{1}{|x|^p} \,dx$$
for each $j$.
