= Solution
For <bond percolation>, the <Harris-FKG inequality> states that any two <increasing events> $A,B$ satisfy
$$
\mathbb P_p(A\cap B)\geq\mathbb P_p(A)\mathbb P_p(B).
$$
The same inequality holds for two <decreasing events>, either by reversing the coordinate order or by taking complements in the two-event identity. Intersections of <decreasing events> are <decreasing events>, so induction gives
$$
\mathbb P_p\left(\bigcap_{j=1}^n A_j^c\right)\geq\prod_{j=1}^n\mathbb P_p(A_j^c).
$$
Write $a=\mathbb P_p(A_1)=\cdots=\mathbb P_p(A_n)$ and $u=\mathbb P_p(\bigcup_jA_j)$. Then $1-u\geq(1-a)^n$. Taking the nonnegative $n$-th root and rearranging yields the <square-root trick for positively associated events>:
$$
\boxed{\mathbb P_p(A_1)\geq1-\left(1-\mathbb P_p\left(\bigcup_{j=1}^n A_j\right)\right)^{1/n}.}
$$
No independence among the events is needed.
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