Let $\mathscr P _0$ and $\mathscr P _1$ be two non-overlapping sets of probability distributions defined on $(\Omega,\mathcal{A})$. Consider the distance defined as $$D_u(P_0,P_1)=\int_\Omega \left(\frac{p_1}{p_0}\right)^u p_0 \mathrm{d}\mu<\infty.$$ Two distributions are chosen from each set $Q_0\in\mathscr P _0$ and $Q_1\in\mathscr P _1$ such that $$D_u(Q_0,Q_1)\geq D_u(P_0,P_1)\quad \forall (P_0,P_1)\in \mathscr P _0\times \mathscr P _1,\forall u\in[0,1]$$

Is it true that (A:)$$\int \min(q_0,t q_1)\mathrm{d}\mu\geq \int \min(p_0,t p_1)\mathrm{d}\mu\quad \forall (P_0,P_1)\in \mathscr P _0\times \mathscr P _1,\forall t$$or equaivalently (B:) $$Q_0\left[\frac{q_1}{q_0}>t\right]\geq P_0\left[\frac{q_1}{q_0}>t\right]\quad\ \forall P_0\in \mathscr P _0,\forall t$$ $$Q_1\left[\frac{q_1}{q_0}\leq t\right]\geq P_1\left[\frac{q_1}{q_0}\leq t\right]\quad\ \forall P_1\in \mathscr P _1,\forall t$$

Notes:

$\bullet$ One can consider A or B since both conditions are equivalent.

$\bullet$ $p_0$ and $p_1$ are densities of $P_0$ and $P_1$ and the same goes to $q_0$ and $q_1$ with $Q_0$ and $Q_1$.

What I know:

From Huber's paper (pages 260-261) Theorem 6.1 I know that if the distance is the $f$-divergence, i.e. $D_f$, then A and B are correct. Additionally, if A and B are correct, then $Q_0$ and $Q_1$ minimize $D_f$ (iff condition).

Huber considers $$Q_{jt}=(1-t)Q_{0t}+t Q_{1t}\\q_{jt}=(1-t)q_{0t}+t q_{1t}$$ and finds the first and second derivatives of $D_f(Q_{0t},Q_{1t})$. He then shows that the second derivative is $\geq 0$ (convex) and hence $(Q_{00},Q_{10})$ minimizes $D_f$ if and only if the first derivative evaluated at $t=0$ is $\geq 0$ for all $(Q_{01},Q_{11})\in(\mathscr P _0\times\mathscr P _1)$. He shows that this is really the case, hence the claim is true.

I think that this result can be strenghtened, i.e. if $(Q_0,Q_1)$ maximizes $D_u$ for all $u\in[0,1]$, then it should satisfy A or equivalently B. I dont know how to proceed.