Suppose $\theta$ is a measure preserving map on the probability space $(\O,\F,\P)$. Then for all $A\in\F$:
$$
\limsup_n\P(A\cap[\theta^n\in A])\geq\P(A)^2~.
$$
By Cauchy-Schwarz (or Jensen) we get:
$$
\E\Big(\sum_{k=1}^{n}I_{\theta^k\in A}\Big)^2
\geq\Big(\sum_{k=1}^{n}\E I_{\theta^k\in A}\Big)^2
=n^2\P(A)^2~.
$$
Now the left hand side can be written as
\begin{eqnarray*}
\sum_{j,k}\P(\theta^k\in A,\theta^j\in A)
&=&\sum_k\P(\theta^k\in A)+2\sum_{j < k}\P(\theta^k\in A,\theta^j\in A)\\
&=&n\P(A)+2\sum_{j < k}\P(A\cap[\theta^{k-j}\in A])\\
&\leq&n\P(A)+2{n\choose2}\sup_{1\leq k\leq n}\P(A\cap[\theta^k\in A])~.
\end{eqnarray*}
Therefore we conclude that
$$
2{n\choose2}\sup_{1\leq k\leq n}\P(A\cap[\theta^k\in A])
\geq n^2\P(A)^2-n\P(A)~.
$$
As $n$ goes to $\infty$ this implies:
$$
\sup_{k}\P(A\cap[\theta^k\in A])\geq\P(A)^2
$$
Replacing $\theta$ with $\theta^n$:
$$
\forall m\in\N:\quad
\sup_{k}\P(A\cap[\theta^{kn}\in A])\geq\P(A)^2
$$
which implies the asserted inequality.