If x and y are real, and x2+y2=1, compute the maximum value of (x+y)2.
Solution — click to reveal
By the Trivial Inequality, (x−y)2≥0. Then (x+y)2+(x−y)2≥(x+y)2.But (x+y)2+(x−y)2=x2+2xy+y2+x2−2xy+y2=2x2+2y2=2, so (x+y)2≤2.Equality occurs when x=y=21, so the maximum value is 2.