不等式最值问题

f(x）=4x+(1/(x-1））+1

=-1/(1-x)-4(1-x)+4+1,(因为1-x>0,所以要换成1-x)
=-[1/(1-x)+4(1-x)]+5
<=-2根号[1/(1-x)*4(1-x)]+5
=-4+5
=1

即最大值是：1

f(x)=4x+[1/(x-1)]+1
=4(x-1)+[1/(x-1)]+5
=-{4(1-x)+[1/(1-x)]}+5
<=-2*2+5=1
所以函数f(x）=4x+(1/(x-1））+1的最大值是1

f(x)=-[4(1-x)+1/(1-x)]+5<=5-2sqrt(4(1-x)*1/(1-x))=5-2*2=1
等号当且仅当4(1-x)=1/(1-x)
x=1/2时成立。