Question

已知x>0,y>0,且1/x+2/（x+2y）=1,则x+y取最小值时对应的x值为

求解啊！！很急

Answer 1

设x+y=u
1/x + 2/(x+2y)
= 1/x + 2/(2u -x)

1/x + 2/(2u -x) = 1
（2u-x) + 2x = x(2u-x)
2u+x = 2ux -x^2
2u-2ux = -x^2 -x

2u(1-x) = -(x^2+x)
2u = (x^2+x)/(x-1)

因为2u>0, x^2+x>0, 所以x-1>0

即求(x^2+x)/(x-1)最小值。
=(x^2-x+2x)/(x-1)
= x+ 2x/(x-1)
= x+ (2x-2+2)/(x-1)
= x+2+2/(x-1)
= (x-1)+3 +2/(x-1)
>= 3 + 2√ [ (x-1) *2/(x-1)]
= 3+2√2
取=时， (x-1) = 2/(x-1)
(x-1)^2 =2
x-1 = √2
x = 1+√2

2u >= 3+2√2. u>=3/2 +√2. x+y最小值 3/2 +√2

Answer 2

令t=x+y x=t-y , t>0
1/(t-y)+2/(t+y)=1
t^2-3t+y-y^2=0
t=3/2±√[（y-1/2）^2+11/4]
根式里面≥11/4>9/4, t>0, 故
t=3/2+√[（y-1/2）^2+11/4]≥3/2+√11/2
故，当y=1/2时，t有最小t=3/2+√11/2
x=t-y=1+√11/2

Answer 3

由1/x+2/（x+2y）=1
得出y=(3x-x^2)/(2x-2)=-x/2+1+1/(x-1)
x+y=x-x/2+1+1/(x-1)=(1/2)(x-1)+1/(x-1)+3/2由均值不等式知道
(1/2)(x-1)=1/(x-1)时值最小
此时x=1+根号2