= Solution
Let $a=\pi^2(\varepsilon^{-1}-1)$. From the supplied <second moment of a mixture likelihood ratio>, dropping a nonpositive term gives
$$
\chi^2(P_1\Vert P_0)\leq\frac{(1+a)^n}{d}\leq\frac{e^{na}}d\leq16\nu^2.
$$
The last step uses the assumed separation. It has a real square root only when $16\nu^2d\geq1$; otherwise no parameters satisfy it. The <chi-squared testing lower bound> follows from the <Cauchy-Schwarz inequality>: $\|P_1-P_0\|_{\mathrm{TV}}\leq\frac12\sqrt{\chi^2(P_1\Vert P_0)}\leq2\nu$. Every <statistical hypothesis testing> rule has sum of its <Type I error> and <Type II error> at least $1-\|P_1-P_0\|_{\mathrm{TV}}$. Its larger error is at least half the sum, hence
$$
\boxed{\inf_\psi\max\{P_0(\psi=1),P_1(\psi=0)\}\geq\tfrac12-\nu.}
$$
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