why minimax >= maximin
Given a function f(x,y), Min_y(Max_x(f(x, y))) >= Max_x(Min_y(f(x,y)))
For any y, we always pick the corresponding x0 that maximizes f(x, y), note x0 is a series of corresponding x values instead of a single value.
f(x0, y) >= f(x,y)
For any x, we always pick the y0 that minimizes f(x, y), again y0 is a series of corresponding y values.
f(x, y0) <= f(x, y)
therefore we have f(x0, y) >= f(x,y) > f(x,y0), note again the x0 and y0 are corresponding values with respect to x and y values
That means for any x, y, we have f(x0, y) >= f(x,y0)
Important, this is for any x and y pair, so on the left hand side if we replace y with the y' (now a single value) that minimizes f(x0, y), and replace the x with x' that maximizes f(x,y0)
we have:
f(x0, y') >= f(x',y0)
i.e. minmax >= maximin