I would like to prove that the bound in the Pinsker inequality is tight in the sense that the ratio of $D(P||Q)$ and $d^2 (P, Q)$ can be arbitrarily close to $\frac{1}{2\mathrm{ln}2}$.
Pinsker inequality is:
$$D(P||Q) \leq \frac{1}{2\mathrm{ln}2} d^2 (P, Q),$$
where $d(P, Q)$ is the variational distance of distributions $P$ and Q, i.e., $$d(P, Q) = \sum_{a \in A} |P(a) − Q(a)|$$
I want to show that with $B = \{ a : P(a) \geq Q(a) \}, \widetilde{P} = (P(B), P(A − B)), \widetilde{Q} = (Q(B), Q(A − B))$, we have $D(P||Q) \leq D(\widetilde{P}||\widetilde{Q}), d(P, Q) = d(\widetilde{P}, \widetilde{Q})$. Hence it suffices to consider the case $A = \{0, 1\}$, i.e., to determine the largest $c$ such that $$p \log(\frac{p}{q})+(1-p)\log(\frac{1-p}{1-q})-4c(p-q)^2 \geq 0,$$ for every $0 \leq q \leq p \leq 1$, where that the base of the logarithm is $2$.
For $q = p$ the equality holds; further, the derivative of the left-hand side with respect to $q$ is negative for $q < p$ if $c \leq \frac{1}{2\mathrm{ln}2}$ while for $c > \frac{1}{2\mathrm{ln}2}$ and $p = \frac{1}{2}$ it is positive in the neighborhood of $p$.