Show that if $x>0$, then $\ln(x)\geq 1-\frac{1}{x} $ Show that if $x>0$, then 

$$ \ln(x)\geq 1-\dfrac{1}{x}. $$

I tried a few things but so far nothing has worked, I could use a hint.

You can use mean value theorem for $\ln$.


*

*$x = 1$, then equality holds.

*$0 < x < 1$:
$$\ln x \ge \frac{x-1}{x} \\
\frac{\ln x - \ln 1}{x- 1} \le \frac{1}{x} \\
\frac{1}{c} \le \frac{1}{x} \\
c \ge x$$
Which indeed is true, because $c \in (x,1)$. Change from $\ge$ to $\le$ is due to division by $(x-1)$ which is smaller than $0$.

*$x>1$ similarly as in 2.:
$$\ln x \ge \frac{x-1}{x} \\
\frac{\ln x - \ln 1}{x- 1} \ge \frac{1}{x} \\
\frac{1}{c} \ge \frac{1}{x} \\
c \le x$$
here $c \in (1, x)$ so the inequality is true as well.

