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    OlympiadHardInequalities10–12

    If a,b,c>0a, b, c > 0 and a+b+c=1a + b + c = 1, what is the minimum of 1a+1b+1c\frac{1}{a} + \frac{1}{b} + \frac{1}{c}?

    Se a,b,c>0a, b, c > 0 e a+b+c=1a + b + c = 1, qual é o mínimo de 1a+1b+1c\frac{1}{a} + \frac{1}{b} + \frac{1}{c}?

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    Solution

    By AM-HM or Cauchy-Schwarz: (a+b+c)(1a+1b+1c)≥9(a+b+c)\left(\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{c}\right) \ge 9. Since a+b+c=1a+b+c=1, min is 9.