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!
$$v_i^{m}=\min_{m_{-i}\in \Delta A_{-i}} \max_{m_i\in \Delta A_i} u \leq \min_{m_{-i}\in \Delta A_{-i}} \max_{p_i\in A_i} u \leq \min_{p_{-i}\in A_{-i}} \max_{p_i\in A_i} u=v_i^p$$
The first inequality here holds because for any fixed strategy of the other players, the best thing to do for player $i$ is to pick a pure strategy. In fact, his payoff for any mixed strategy is the mean of the payoffs of pure strategies, and therefore cannot be larger than the largest pure payoff.
The second inequality is trivial.