证明:
设f(x,y,z)=1/(x+y)+1/(y+z)+1/(z+x)
由对称性不妨设0(1-xy)/(x+y)=z}
2+2z^2=t+1/t,而由均值不等式t>=2,而我们知道函数t+1/t当x>=1时是单调递增的.所以f(x,y,z)>=t+1/t>=5/2
证毕
证明:
设f(x,y,z)=1/(x+y)+1/(y+z)+1/(z+x)
由对称性不妨设0(1-xy)/(x+y)=z}
2+2z^2=t+1/t,而由均值不等式t>=2,而我们知道函数t+1/t当x>=1时是单调递增的.所以f(x,y,z)>=t+1/t>=5/2
证毕