Proof: when x > 0, the inequality LNX > = 1-1 / X

Proof: when x > 0, the inequality LNX > = 1-1 / X

prove:
Let f (x) = LNX + 1 / X-1
f'(x)=1/x-1/x^2=(x-1)/x^2
When x > = 1, f '(x) > = 0, f (x) is an increasing function