从一个古老的不等式multiply from i=1 to n a_i^(a_i)≥(multiply from i=1 to n a_n)_n^1 multiply from i=1 to n c_i的多种证明出发,将其加强为multiply from i=1 to n a_i^(a_i)≥(1/n multiply from i=1 to n a_i)^(multiply from...从一个古老的不等式multiply from i=1 to n a_i^(a_i)≥(multiply from i=1 to n a_n)_n^1 multiply from i=1 to n c_i的多种证明出发,将其加强为multiply from i=1 to n a_i^(a_i)≥(1/n multiply from i=1 to n a_i)^(multiply from i=1 to n a_i)。并用凸函数的工具给出一个简短的新证明。展开更多
文摘从一个古老的不等式multiply from i=1 to n a_i^(a_i)≥(multiply from i=1 to n a_n)_n^1 multiply from i=1 to n c_i的多种证明出发,将其加强为multiply from i=1 to n a_i^(a_i)≥(1/n multiply from i=1 to n a_i)^(multiply from i=1 to n a_i)。并用凸函数的工具给出一个简短的新证明。
