What is the least value of $f(x) = x^{2} + \frac{16}{x^{2}}$, $x \ne 0$?

Explanation

1. 01

   $f'(x) = 2x - \frac{32}{x^{3}} = 0$ gives $x^{4} = 16$, so $x = \pm 2$; $f''(x) = 2 + \frac{96}{x^{4}} > 0$, a minimum.

2. 02

   $f(\pm 2) = 4 + 4 = 8$.