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    2019高考数学最后一题

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    已知a, b, c为正数,且满足abc=1. 证明: \begin{align} \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \le a^2 + b^2 + c^2 \end{align} 和 \begin{align} (a+b)^3 + (b+c)^3+ (c+d)^3 \ge 24 \end{align}