Question

设x>0,y>0,z>0,且x^2+y^2+z^2=1 (1)求证：x^2/(1+9yz)+y^2/(1+9xz)+z^2/(1+9xy)>=1/4

Answer 1

证明：由柯西不等式：
[(1+9xy)+(1+9yz)+(1+9zx)][x^2/(1+9xy)+y^2/(1+9yz)+z^2/(1+9zx)]>=(x+y+z)^2
上式也即x^2/(1+9xy)+y^2/(1+9yz)+z^2/(1+9zx)]>=(x+y+z)^2/[3+9(xy+yz+zx)]
注意到：因为x^2+y^2+z^2=1,所以(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=1+2(xy+yz+zx)
于是x^2/(1+9xy)+y^2/(1+9yz)+z^2/(1+9zx)]>=[1+2(xy+yz+zx)]/[3+9(xy+yz+zx)]
只要证明[1+2(xy+yz+zx)]/[3+9(xy+yz+zx)]>=1/4即可
化简后上式等价于：xy+yz+zx<=1
由于x^2+y^2+z^2=1，所以上式可齐次化为:
xy+yz+xz<=x^2+y^2+z^2
<=>(x-y)^2+(y-z)^2+(z-x)^2>=0
显然成立。
于是原不等式得证。。