Putnam 1998, Problem B-1

Find the minimum value of
\[
\frac{(x + 1/x)^6 - (x^6 + 1/x^6) - 2}{(x + 1/x)^3 + (x^3 + 1/x^3)}
\]
for \( x > 0 \).

Solution:
Notice that
\[
\frac{(x + 1/x)^6 - (x^6 + 1/x^6) - 2}{(x + 1/x)^3 + (x^3 + 1/x^3)} = \frac{(x + 1/x)^3 - (x^3 + 1/x^3)}{(x + 1/x)^3 - (x^3 + 1/x^3)} = 3(x + 1/x)
\]
(difference of squares). The latter is easily seen (e.g., by AM-GM) to have minimum value \( 6 \) (achieved at \( x = 1 \)).