已知x>1,求证：x-(x²/2)-lnx-1\2

已知x>1,求证：x-(x²/2)-lnx-1\2
已知x>1,求证：x-(x²/2)-lnx-1\2

已知x>1,求证：x-(x²/2)-lnx-1\2
证明：令f(x)=x-(x²/2)-lnx
f(x)'=1-x-1/x
当x＞1时,f(x)'＜0
所以f(x)=x-(x²/2)-lnx在(1,+无穷)单调递减
即f(x)＜f(1)
(1)=1-1/2-0
=1/2
所以f(x)＜1/2
即可知：x-(x²/2)-lnx-1\2