As follows~~~
Pure Mathematics - Inequality
2007-12-03 1:14 am
回答 (1)
2007-12-03 2:02 am
✔ 最佳答案
Let a1=a2=...=am=sin^2θ/m ; am+1=am+2=...=am+n=cos^2θ/na1+a2+...+am+n=1
Apply AM>=GM
[(a1+a2+...+am+n)/m+n]^(m+n)>=a1a2...am+n
[1/(m+n)]^(m+n)>=(sin^2mθcos^2nθ)/(m^m)(n^n)
sin^2mθcos^2nθ<=(m^m)(n^n)/[(m+n)]^(m+n)
when the equality holds
sin^2θ/m=cos^2θ/n
tan^2θ=m/n
or tanθ=sqrt(m/n)
收錄日期: 2021-04-25 16:54:21
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