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Last edited by lemondian 2023-12-2 09:35设数列${a_n}$是等差数列,$a_i>0(i=1,2,\cdots ,n)$，公差为$d$，且$0\leqslant d\leqslant 1$，则对于任意的正整数$k$，有$\displaystyle\sum_{i=1}^na_i^{\frac{1}{k}}\geqslant \dfrac{k}{kd+1}[a_na_{n-1}^{\frac{1}{k}}-(a_1-d)a_1^{\frac{1}{k}}]$. |
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