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​−x​1​)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.