Problem:
What is the least possible value of (xy−1)2+(x+y)2 for real numbers x and y?
Answer Choices:
A. 0
B. 41​
C. 21​
D. 1
E. 2
Solution:
Because
(xy−1)2+(x+y)2=x2y2+x2+y2+1=(x2+1)(y2+1).
and both factors are at least 1, the least possible value of the expression is (D)1​. It occurs when x=0 and y=0.
The problems on this page are the property of the MAA's American Mathematics Competitions