[niv3] inégalité difficile,généralisation


  • Math&Maroc

    Generalization( Vasile Cirtoaje):
    Let n2n \ge 2 be an integer,and let kn1k \le n - 1 be a positive number.If a1,a2,,an{a_1},{a_2}, \cdots ,{a_n} are positive numbers satisfying a1a2ann=pn1k1\sqrt[n]{{{a_1}{a_2} \cdots {a_n}}} = p \ge {n^{\frac{1}{k}}} - 1,then
    [1(1+a1)k+1(1+a2)k++1(1+an)kn(1+p)k][\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}}}]


  • Math&Maroc

    Let x1,x2,,xn0{x_1},{x_2}, \cdots ,{x_n} \ge 0 be such that their product is 1,then
    1(1+x1)k+1(1+x2)k++1(1+xn)kmin1,n2k\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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