Chi-squared testing lower bound ($\max \{\alpha ,\beta \}\ge \frac{1}{2}-\frac{1}{4}\sqrt{{\chi }^2(P\Vert Q)}$)

The Cauchy-Schwarz inequality gives $\Vert P-Q{\Vert }_{\mathrm{TV}}=\frac{1}{2}{\mathbb{E}}_Q|dP/dQ-1|\le \frac{1}{2}\sqrt{{\chi }^2(P\Vert Q)}$. Every statistical hypothesis testing rule has sum of its Type I error and Type II error at least $1-\Vert P-Q{\Vert }_{\mathrm{TV}}$, proving the displayed bound. A mixture model of alternatives yields a lower bound on the worst alternative error because a maximum dominates an average.