若实数x,y满足x^2+y^2=1,2xy/(x+y-1)>=m恒成立,求m范围.用三角函数怎样解?

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若实数x,y满足x^2+y^2=1,2xy/(x+y-1)>=m恒成立,求m范围.用三角函数怎样解?
若实数x,y满足x^2+y^2=1,2xy/(x+y-1)>=m恒成立,求m范围.用三角函数怎样解?

若实数x,y满足x^2+y^2=1,2xy/(x+y-1)>=m恒成立,求m范围.用三角函数怎样解?
由式一令x=sint y=cost
带入式二化简得 [（sint-cost)^2-1]/(sint+cost-1)=sint+cost+1=根号2sin(t-a)+1>=-根号2+1
所以m

设x=cosa,y=sina,a属于【0,2pai】
而(cosa+sina-1)(cosa+sina+1)=(cosa+sina)^2-1=2sinacosa
于是2xy/(x+y-1)=2sinacosa/(cosa+sina-1)=cosa+sina+1>=1-根号2

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