Min =1/2 when x =1/2
Another way to do it: , .
We must have , for some number , as well as x+y= 1.
So which says that .
If x= y, from x+y= 1 we have x= y= 1/2.
If then so we can divide by it to get 3(x+ y)- 1= 0 which says x+y= 1/3 contradicting x+y= 1.
Thus x= y= 1/2 is the only solution.
Yet another way, and probably simplest:
From x+ y= 1, y= 1- x. Putting that into [tex]x^3+ y^3+ xy[/itex] as Prove It suggested, we have . We can write that as and, completing the square, . Since a square is never negative, that will have its minimum value when x= 1/2.