So I have two finite sets $A,B$ and $u:A\times B\rightarrow \mathbb{R}$ a utility function. I am asked to give a certain proof but I don't need help with the whole thing, I just need help figuring out how to prove $\forall_{a\in A}\,\,\min_{b\in B} u(a,b)\leq \min_{b\in B}\max_{a\in A}u(a,b)$. Of course I can see that, for any fixed $a\in A$ and $b\in B$ we have $u(a,b)\leq \max_{a\in A}u(a,b)$ and I want to be able to argue that I can minimize both sides and still retain the inequality.
I feel like I'm close but I can't quite see my way there. I've been trying to reason as follows and find a way to formalize my reasoning: If $a\in A$ is fixed then $\min_{b\in B}u(a,b)$ is found by listing all utilities for choices of $b\in B$ and picking the least. On the other hand $\min_{b\in B}\max_{a\in A}u(a,b)$ is found by first trying each choice of $a\in A$ and seeing what maximum you can get for choices of $b\in B$, and then choosing the $b$ that results in a minimum of these maxima.
If we call $a^*,b^*$ any arguments such that $u(a^*,b^*)=\min_{b\in B}\max_{a\in A}u(a,b)$ then $b^*$ is the minimizer of the list of maximums for various $a\in A$. And so if $b'\ne b^*$ then there is some $a'\ne a^*$ such that $u(a^*,b^*)\leq u(a',b')$. But I feel like I'm getting lost.
[Edit: Actually, I'm starting to doubt that the theorem is correct. Take for example
$$u(0,1)=-1,\quad u(1,1)=1,\quad u(0,0)=0, \quad u(1,0)=2$$
and let $a=1$. Then $\min_{b\in B}u(a,b)=1$ but yet to find $\min_{b\in B}\max_{a\in A}u(a,b)$ we list all the possible maxima for various choices of $a$ which are $0,2$ respectively. Then we choose the minimum of these which is $0$ and get $\min_{b\in B}\max_{a\in A}u(a,b)=0<1=\min_{b\in B}u(a,b)$ which seems to break the theorem. Have I made a mistake?]
1. Proof that $\min_{b\in B} u(a,b)\leq \min_{b\in B}\max_{a\in A}u(a,b)$
Take $(a,b) \in A \times B$. You have $$u(a,b) \le \max\limits_{a \in A} u(a,b)$$ then $$\min\limits_{b \in B} u(a,b) \le \max\limits_{a \in A} u(a,b)$$ and finally $$\min\limits_{b \in B} u(a,b) \le \min\limits_{b \in B} \max\limits_{a \in A} u(a,b)$$
2. You made an error in computing $\min\limits_{b \in B} \max\limits_{a \in A} u(a,b)$
For $b=0$ you have $u(0,0)=0$ and $u(1,0)=2$ hence $\max\limits_{a \in A} u(a,0)=2$. In a similar way for $b=1$ you have $u(0,1)=-1$ and $u(1,1)=1$ hence $\max\limits_{a \in A} u(a,1)=1$. Finally $$\min\limits_{b \in B} \max\limits_{a \in A} u(a,b)=1$$