若x,y属于R,且x^2+y^2=1,求证:3/4<=(1-xy)(1+xy)<=1

(x+y)^2>=0
x^2+y^2+2xy>=0
x^2+y^2>=-2xy
xy>=-(x^2+y^2)/2=-1/2

(x-y)^2>=0
x^2+y^2-2xy>=0
x^2+y^2>=2xy
xy<=(x^2+y^2)/2=1/2

所以-1/2<=xy<=1/2
所以0<=x^2y^2<=1/4
-1/4<=-x^2y^2<=0
1-1/4<=1-x^2y^2<=1+0
3/4<=(1-xy)(1+xy)<=1

x,y属于R,
x^2+y^2=1>=2xy
xy<=1/2

(1-xy)(1+xy)=1-(xy)^2<=1 因为(xy)^2>=0
1-(xy)^2>=3/4
得证

(x+y)^2>=0
x^2+y^2+2xy>=0
x^2+y^2>=-2xy
xy>=-(x^2+y^2)/2=-1/2

(x-y)^2>=0
x^2+y^2-2xy>=0
x^2+y^2>=2xy
xy<=(x^2+y^2)/2=1/2

由此得出-1/2<=xy<=1/2
由此得出0<=x^2y^2<=1/4
-1/4<=-x^2y^2<=0
1-1/4<=1-x^2y^2<=1+0
-/4<=(1-xy)(1+xy)<=1

以证