Daily ChallengeHardInequalities

For positive reals $x, y$ with $xy = 1$ , what is the minimum of $x + y$ ?

Para reais positivos $x, y$ com $xy = 1$ , qual é o mínimo de $x + y$ ?

Answer options

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Solution

By AM-GM: $\frac{x+y}{2} \ge \sqrt{xy} = 1$ , so $x+y \ge 2$ . Min at $x=y=1$ .