= Solution
Fix any <estimator> $\widehat\eta$ and put $A(x)=d(\widehat\eta(x),f)$, $B(x)=d(\widehat\eta(x),g)$ and $D=d(f,g)$. The <triangle inequality> for the <metric> gives $A+B\geq D$. Hence
$$
A^2+B^2\geq\frac{(A+B)^2}{2}\geq\frac{D^2}{2}.
$$
The larger of the two <risk functions> dominates their average. Using nonnegativity to replace both <probability density functions> by their minimum,
$$
\begin{aligned}
\sup_{\eta\in\mathcal F}P_\eta d(\widehat\eta,\eta)^2
&\geq\frac12\{P_fA^2+P_gB^2\}\\
&\geq\frac12\int\min(p_f,p_g)(A^2+B^2)\,d\mu\\
&\geq\frac{D^2}{4}\int\min(p_f,p_g)\,d\mu.
\end{aligned}
$$
The right side is independent of the <estimator>, so taking the infimum proves
$$
\boxed{\inf_{\widehat\eta}\sup_{\eta\in\mathcal F}P_\eta d(\widehat\eta,\eta)^2\geq\frac{d(f,g)^2}{4}\int\min(p_f,p_g)\,d\mu.}
$$
This <metric squared-loss two-point bound> uses overlap of the observation laws to quantify how hard it is to distinguish the two separated parameters.
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