Let $G$ be a finite game with set of players $N$ and strategy sets $(A_i)_{i\in N}$. For any player $i$ let his minmax-value in pure strategies be defined by $$ \underline v_i^p = \min_{a_{-i} \in A_{-i}} \max_{a_{i} \in A_i} u_i(a_i, a_{-i}), $$ and his minmax-value in mixed strategies by $$ \underline v_i^m = \min_{m_{-i} \in \Delta A_{-i}} \max_{m_{i} \in \Delta A_i} u_i(m_i, m_{-i}). $$
Question:
Is it always the case that $\underline v_i^m \le \underline v_i^p$?
The opposite inequality does not hold, because in the classical battle of the sexes with payoff matrix
$$ \begin{matrix} & \text{A} & \text{B} \\ \text{A} & 3,1 & 0,0 \\ \text{B} & 0,0 & 1,3 \\ \end{matrix} $$ player $1$'s min-max value in pure strategies is $1$ while it is $3 \over 4$ in mixed strategies.
This seems like an important question but so far I've failed to come up with a counterexample or a proof.
Thank you!