当x>1时,求y=2x^2-2x+1/x-1的最小值?

题目

当x>1时,求y=2x^2-2x+1/x-1的最小值?

答案

y=[2x(x-1)+1]/(x-1)
=2x(x-1)/(x-1)+1/(x-1)
=2x+1/(x-1)
=2(x-1)+1/(x-1)+2
x>1,x-1>0
所以2(x-1)+1/(x-1)>=2√[2(x-1)*1/(x-1)]=2√2
所以y>=2√2+2
所以最小值=2√2+2