Question

已知x>0,y>0,x+y=1,求证:(1+1/x)(1+1/y)大于等于9

Answer 1

由 x+y=1
(1+1/x)(1+1/y) = [1+(x+y)/x][1+(x+y)/y] = (2 + y/x)(2 + x/y) =
4 + 1 + 2(x/y + y/x) = 5 + 2(x/y + y/x)
另一方面，
(x/y + y/x) = (x^2 + y^2)/xy >= 2
because x^2 + y^2 >= 2xy

so, 5 + 2(x/y + y/x) >= 5+4 = 9

Answer 2

先把原式变形
(1+1/x)(1+1/y)
=(XY+X+Y+1)/XY

(X-Y)的平方为X~2+Y~2-2XY
平方数>=0
故X~2+Y~2>2XY
(X+Y)~2=X~2+Y~2+2XY=1
XY<=1/4

再看原式
(XY+X+Y+1)/XY (其中X+Y=1)
=1+2/XY
又XY<=1/4
故原式>=9

Answer 3

(1+1/x)(1+1/y)=1+1/x+1/y+1/xy,
x>0,y>0,x+y=1,则0<x<1,0<y<1
1/x+1/y+1/xy=1/x+1/(1-x)+1/x(1-x)
=2/x(1-x)
(x-1/2)2=x2-x+1/4>=0,x-x2<=1/4,
2/x(1-x)>=8,
所以(1+1/x)(1+1/y)>=9

Answer 4

(1+1/x)(1+1/y)=1+1/x+1/y+1/xy=1+(x+y+1)/xy=1+2/xy
又x+y=1,故xy≤1/4，所以1+2/xy≥1+2*4=9
故:(1+1/x)(1+1/y)≥9