let a=xy, b=1-x-y+xy,c=x+y-2xy then a+b+c=1,0<=x+y<=2 if 0<=x+y<2/3 or 2>= x+y>4/3,then a<=(x+y)^2/2<1/9 or b<1/9, then b+c>8/9 or a+c>8/9, A>4/9; if 2/3=<x+y<=4/3,then |x+y-1|<=1/3,then (x+y-1)^2<=1/9,then 2(x+y)>=(x+y)^2+8/9 2*c=2*(x+y)-4xy>=(x+y)^2-4xy+8/9>=8/9; A>=c>=4/9 so M(x,y)min=4/9,(x=y=1/3 or x=y=2/3)

juanbri wrote: We consider all pairs (x, y) of real numbers such that $0\leq x \leq y \leq 1$.Let $M (x,y)$ the maximum value of the set $$A=\{xy, 1-x-y+xy, x+y-2xy\}.$$Find the minimum value that $M(x,y)$ can take for all these pairs $(x,y)$. Let $ x,y\in [0, 1].$ Prove that $$\max \{xy,xy-x-y+1.x+y-2xy\}\geq \frac{4}{9}.$$