Two positive numbers add up to 16. What is the least possible value of the sum of their squares?

Explanation

1. 01

   $Q = x^2 + (16 - x)^2$, so $Q' = 2x - 2(16 - x) = 0$ at $x = 8$.

2. 02

   $Q'' = 4 > 0$, a minimum: $Q = 8^2 + 8^2 = 128$.