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    Daily ChallengeHardInequalities

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

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

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    Solution

    By AM-HM or C-S: (a+b)(1a+1b)≥4(a+b)\left(\tfrac{1}{a}+\tfrac{1}{b}\right) \ge 4. Since a+b=1a+b=1, min is 4 when a=b=12a=b=\tfrac{1}{2}.