# [niv3] inégalité difficile,généralisation

• Generalization( Vasile Cirtoaje):
Let $n \ge 2$ be an integer,and let $k \le n - 1$ be a positive number.If ${a_1},{a_2}, \cdots ,{a_n}$ are positive numbers satisfying $\sqrt[n]{{{a_1}{a_2} \cdots {a_n}}} = p \ge {n^{\frac{1}{k}}} - 1$,then
$[\frac{1}{{{{\left( {1 + {a_1}} \right)}^k}}} + \frac{1}{{{{\left( {1 + {a_2}} \right)}^k}}} + \cdots + \frac{1}{{{{\left( {1 + {a_n}} \right)}^k}}} \ge \frac{n}{{{{\left( {1 + p} \right)}^k}}}]$

• Let ${x_1},{x_2}, \cdots ,{x_n} \ge 0$ be such that their product is 1,then
$\frac{1}{{{{\left( {1 + {x_1}} \right)}^k}}} + \frac{1}{{{{\left( {1 + {x_2}} \right)}^k}}} + \cdots + \frac{1}{{{{\left( {1 + {x_n}} \right)}^k}}} \ge \min { {1,\frac{n}{2^k}}}$

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