OlympiadHardInequalities10–12

For $x > 0$ , what is the minimum of $x + \frac{4}{x}$ ?

Para $x > 0$ , qual é o mínimo de $x + \frac{4}{x}$ ?

Answer options

Solution

By AM-GM: $x + \frac{4}{x} \ge 2\sqrt{x \cdot \tfrac{4}{x}} = 2\sqrt{4} = 4$ . Minimum when $x = 2$ .