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.
 A: This answer doesn't differ from the others in principle. But it differs in the details. With enough care, you can make the inequality look more obvious and the calculations more simple. Since $\ln \frac{1}{x} = - \ln x$, the inequality you are trying to prove can be rewritten as $$\ln \frac{1}{x} \leq \frac{1}{x} - 1.$$
We can introduce a new variable $t = \frac{1}{x}$. Saying that $x > 0$ is the same as saying that $t > 0$. So, we need to prove for every $t > 0$ that $$\ln t \leq t - 1.$$
It is more convenient to solve the problem when it is formulated like this. You can now more easily visualise the functions in your head.
If you look at point $t = 1$, there is an equality $\ln 1 = 1 - 1$. Also, $\ln'(1) = 1$, so the graph $y = \ln t$ is tangent to the line $y = t - 1$ at point $t = 1$. Since $\ln$ is a concave function, its whole graph must then lie under line $y = t -1$, so $\ln t \leq t -1$, qed.
Of course, if you expand the argument from convexity and make it rigorous, it boils down to using the mean value theorem, just like in the other answers.
A: Hint: Find minimum of $\ln(x)-(1-1/x)$, show that for $x$ less than $x$ of minimum it is monotonically decreasing and for $x$ more than $x$ of minimum it is monotonically increasing.
A: I know this proof is pretty much a first derivative test, but alternatively I feel the following directly gives the form of OP's inequality.
One case, when $t\ge1$, 
$$\begin{align}
\frac1t \ge& \frac1{t^2}\\
\int_1^x \frac{dt}{t} \ge& \int_1^x \frac{dt}{t^2} &\text{for }x\ge1\\
\ln x \ge& -\frac1x+1
\end{align}$$
Another case is when $t<1$.
A: 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.

A: Let $x>1$ and $f:[1,x]\to \mathbb{R}$ be the function defined by
$$f(t)=ln (t).$$
By the mean value theorem, there exists $1<c<x$, such that
$$lnx- ln 1 = f(x)  -f(1)=f'(c)(x-1)= \frac{1}{c}(x-1)\geq \frac{1}{x}(x-1).$$
Using the same arguments, we prove for $0<x\leq 1$.
A: Consider:
$f(x) = \frac{x-1}{x}-\ln(x)$
$f'(x) = \frac{1}{x^2}-\frac{1}{x} = \frac{1-x}{x^2}$
Case A: $0 < x \leq 1$
$f'(x)\geq0$ $\implies$ $f(x)$ is non-decreasing in $(0,1]$ and hence:
$f(x) \leq f(1)\ \forall x \in (0,1] $ with the equality holding only at $x=1$ 
For $ x \in (0,1]$:
Thus, $f(x) \leq 0 \ $ and hence $\frac{x-1}{x}-\ln(x) \leq 0$
Case B: 1 < x
$f'(x) <0$ and hence f(x) is strictly decreasing in $(1, \infty)$
For $x \in (1, \infty)$:
$f(x) < f(1)$ $\implies$ $\frac{x-1}{x}-\ln(x) < 0$
