设a>1,b>1,求证:(a^2)/(b-1)+(b^2)/(a-1)>=8

设a>1,b>1,求证:(a^2)/(b-1)+(b^2)/(a-1)>=8
设a>1,b>1,求证:(a^2)/(b-1)+(b^2)/(a-1)>=8

设a>1,b>1,求证:(a^2)/(b-1)+(b^2)/(a-1)>=8
令x=a-1,y=b-1；
x>0,y>0
(a^2)/(b-1)+(b^2)/(a-1)
=(a^3+b^3-a-b)/[(a-1)(b-1)]
将a=x+1,b=y+1代入得：
(a^3+b^3-a-b)/[(a-1)(b-1)]
=[x^3+y^3+3(x^2+y^2)+2(x+y)]/xy
由于
x^3+y^3>=2(xy)^1.5
x^2+y^2>=2xy
x+y>=2(xy)^0.5
那么
[x^3+y^3+3(x^2+y^2)+2(x+y)]/xy
>=2(xy)^0.5+6+4(xy)^(-0.5)
这一步到下一步有点问题,你再看一下题目吧,这里用到[x+a/x>=2根号a]
>=8