Problem:
What is the least possible value of
(x+1)(x+2)(x+3)(x+4)+2019,
where x is a real number?
Answer Choices:
A. 2017
B. 2018
C. 2019
D. 2020
E. 2021
Solution:
Observe that
====(x+1)(x+4)(x+2)(x+3)+2019(x2+5x+4)(x2+5x+6)+2019[(x2+5x+5)−1][(x2+5x+5)+1]+2019(x2+5x+5)2−1+2019(x2+5x+5)2+2018.
Because (x2+5x+5)2≥0 for all x and equals 0 for x=2−5±5, it follows that the requested minimum value is (B)2018.
OR
Let r=x+25. Then
(x+1)(x+2)(x+3)(x+4)=(r−23)(r−21)(r+21)(r+23)=(r2−41)(r2−49)=(r2−45)2−1,
the minimum value of which is −1. Therefore the minimum value of the given expression is 2019−1=(B)2018.
The problems on this page are the property of the MAA's American Mathematics Competitions