Problem: For each positive number x, let
f(x)=(x+x1)3+(x3+x31)(x+x1)6−(x6+x61)−2
The minimum value of f(x) is
Answer Choices:
A. 1
B. 2
C. 3
D. 4
E. 6
Solution:
By observing that [x3+x31]2=x6+2+x61. one sees that
f(x)=(x+x1)3+(x3+x31)[(x+x1)3]2−[x3+x31]2=(x+x1)3−(x3+x31)=3(x+x1)
Since
0⩽(x−x1)2=x+x1−2.2⩽x+x1
and f(x)=3(x+x1) has a minimum value of 6 , which is taken on at x=1.