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Last edited by hbghlyj 2020-1-3 22:03x,y,z为三角形的三边,试确定最小常数k,使$S=(k(xy+yz+zx)-x^2-y^2-z^2)(\frac1{x^2}+\frac1{y^2}+\frac1{z^2})$有最小值
用wolfram算出1.62021<k<1.62022
令x=y=z,S=9(k-1),由$(k(xy+yz+zx)-x^2-y^2-z^2)(\frac1{x^2}+\frac1{y^2}+\frac1{z^2})≥9(k-1)$解出$k≥\frac{(x^2+y^2+z^2)(\frac1{x^2}+\frac1{y^2}+\frac1{z^2})-9}{(xy+yz+zx)(\frac1{x^2}+\frac1{y^2}+\frac1{z^2})-9}$,只需求右边的最大值,然而用wolfram算出右边没有极大值,有极小值1.5
哪里错了? |
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