Proof of Minmax Inequality (1)We want to proveminx∈Xmaxy∈YF(x,y)≥maxy∈Yminx∈XF(x,y)\min_{x \in X} \max_{y \in Y} F(x,y) \ge \max_{y \in Y} \min_{x \in X} F(x,y)x∈Xminy∈YmaxF(x,y)≥y∈Ymaxx∈XminF(x,y)Consider fixed values x0∈X,y0∈Yx_0 \in X, y_0 \in Yx0∈X,y0∈YDefineh(y0)=minx∈XF(x,y0)≤F(x0,y0)h(y_0) = \min_{x \in X} F(x,y_0) \le F(x_0, y_0)h(y0)=x∈XminF(x,y0)≤F(x0,y0)g(x0)=maxy∈YF(x0,y)≥F(x0,y0)g(x_0) = \max_{y \in Y} F(x_0, y) \ge F(x_0, y_0)g(x0)=y∈YmaxF(x0,y)≥F(x0,y0)Therefore∀x0∈X,y0∈Y,h(y0)≤g(x0)\forall x_0 \in X, y_0 \in Y, \\ h(y_0) \le g(x_0)∀x0∈X,y0∈Y,h(y0)≤g(x0)Somaxy0∈Yh(y0)≤minx0∈Xg(x0)\max_{y_0 \in Y} h(y_0) \le \min_{x_0\in X} g(x_0)y0∈Ymaxh(y0)≤x0∈Xming(x0)