- 证明题设x,y为正数,且x+y=1,用反证法证明:[1/(x^2
- 设x,y为正数,且x+y=1,用反证法证明:
[1/(x^2)-1,1/(y^2)-1]≥9
- 证明:假设[1/(x^2)-1,1/(y^2)-1]<9,则
1/(xy)^2 - 1/x^2 - 1/y^2<8
由x+y=1,1/xy=(x+y)/xy,带入上式,(x^2+y^2+2xy)/x^2y^2 - 1/x^2 - 1/y^2=1/x^2+1/y^2+2/xy- 1/x^2 - 1/y^2<8,即1/xy<4,又x,y为正数,xy>1/4.
且可知x+y≥2√xy(x>0,y>0),xy≤1/4.
则假设为假命题。[1/(x^2)-1,1/(y^2)-1]≥9成立