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    OlympiadHardInequalities7–9

    Power Mean Inequality

    Desigualdade das Médias Potenciais

    For positive reals aa and bb, prove that a2+b22≥(a+b2)2\frac{a^{2} + b^{2}}{2} \ge \left(\frac{a+b}{2}\right)^{2}.

    Expand and simplify. This is equivalent to (a−b)2≥0(a-b)^{2} \ge 0.

    Solution

    Step 1 of 6

    1. 1.a2+b22≥(a+b2)2\frac{a^{2} + b^{2}}{2} \ge \left(\frac{a+b}{2}\right)^{2}