OlympiadHardInequalities10–12
For positive reals x and y with x + y = 1, what is the minimum value of \left(1 + \frac{1}{x}\right)\left(1 + \frac{1}{y}\right)?
Para reais positivos x e y com x + y = 1, qual é o valor mínimo de \left(1 + \frac{1}{x}\right)\left(1 + \frac{1}{y}\right)?
Answer options
Solution
The product equals \frac{(1+x)(1+y)}{xy} = \frac{1 + (x+y) + xy}{xy} = \frac{2}{xy} + 1, which is smallest when xy is largest. With x + y = 1 that is xy = \frac{1}{4} at x = y = \frac{1}{2}, giving 8 + 1 = 9.