Question

已知a^2+b^2=4 ，求 ab/(a+b-2)的最值

Answer 1

(a+b)^2-2ab=4
ab=(a+b)^2/2-2

ab/(a+b-2)
=[(a+b)^2/2-2]/(a+b-2)
=[(a+b)^2/2-(a+b)+(a+b)-2]/(a+b-2)
={[(a+b)/2](a+b-2)]+(a+b-2)}/(a+b-2)
=(a+b)/2+1

令a=2sinc b=2cosc

ab/(a+b-2)
=(2sinc+2cosc)/2+1
=sinc+cosc+1
=√2sin(c+π/4)+1
sin(c+π/4)属于[-1,1]
ab/(a+b-2)属于[1-√2,1+√2]
最大值为1+√2,最小值为1-√2.

Answer 2

(a+b)^2-2ab=4
ab=(a+b)^2/2-2

设y=ab/(a+b-2)
=[(a+b)^2/2-2]/(a+b-2)
=[(a+b)^2/2-(a+b)+(a+b)-2]/(a+b-2)
={[(a+b)/2](a+b-2)]+(a+b-2)}/(a+b-2)
=(a+b)/2+1

下面求a+b的范围.
设t=a+b
由于ab<=(a+b)^2/4,ab=(a+b)^2/2-2
所以得:t^2/2-2<=t^2/4
t^2/4<=2
t^2<=8
-2根号2<=t<=2根号2
所以,-根号2<=(a+b)/2<=根号2.
即1-根号2<=y<=1+根号2.

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附:公式ab<=(a+b)^2/4的证明:
由于(a-b)^2=a^2+b^2-2ab>=0
a^2+b^2>=2ab
二边同加上2ab:
(a+b)^2>=4ab
所以,ab<=(a+b)^2/4

Answer 3

三角换元
a=2sinx
b=2cosx
ab/(a+b-2)=4sinxcosx/(2sinx+2cosx-2)=2sinxcosx/(sinx+cosx-1)
=sin2x/(根2*sin(x+π/4)-1)
=-cos(π/2+2x)/(根2*sin(x+π/4)-1)
=-cos(2*(π/4+x))/(根2*sin(x+π/4)-1)
=2[sin(π/4+x)]^2-1/(根2*sin(x+π/4)-1)
=1+根2*cos(x+π/4)

当cos(x+π/4)=1时，即x=-π/4,有最大值1+根2
当cos(x+π/4)=-1时，即x=3π/4,有最小值1-根2

Answer 4

a^2+b^2=4
（a+b）^2-2ab=4
ab=1/2[（a+b）^2-4]
=1/2*(a+b+2)(a+b-2)
ab/(a+b-2)
=(a+b+2)(a+b-2)/2(a+b-2)
=(a+b+2)/2