Solving logarithmic inequality $\ln(x)\geq1-\frac{1}{x}$ I need to solve the following logarithmic inequality:

Show that $\ln(x)\geq1-\frac{1}{x}$ when $x>0$

My attempt
$$\ln(x)\geq1-\frac{1}{x}$$
$$\ln(x)\geq\ln(e)-\ln(e^{\frac{1}{x}})$$
$$\ln(x)\geq\ln(\frac{e}{e^{\frac{1}{x}}})$$
$$x\geq\frac{e}{e^\frac{1}{x}}$$
$$x{e^\frac{1}{x}}\geq e$$
I think I could do the last inequality by solving for $x$ in $x{e^\frac{1}{x}}=e$ (which gives $1$ according to WolframAlpha) and then checking all three cases. The problem is that I haven't officially learned how to solve exponential equations and I was wondering if there is another way to solve this.
 A: Consider:  $f(x) = \frac{x-1}{x}-\ln(x)$
Then, $f'(x) = \frac{1}{x^2}-\frac{1}{x} = \frac{1-x}{x^2}$
Case 1: $0 < x \leq 1$
$f'(x)\geq0$ $\implies$ $f(x)$ 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]$: $f(x) \leq 0 \ $ and hence $\frac{x-1}{x}-\ln(x) \leq 0$
Case 2: $1 < x$
$f'(x) <0$ and hence $f(x)$ is strictly decreasing in $(1, \infty)$
For $x \in (1, \infty)$: $f(x) < f(1)$ $\implies$ $\frac{x-1}{x}-\ln(x) < 0$.  Hope it helps.
A: Let $f(x)=x\ln x-x+1$, we want to show that 
$$f(x)\geq 0 , x>0$$
Let's look at its derivative(s), we have:
\begin{align}
f'(x) &= \ln x\\
f''(x) &= \frac{1}{x}\\
\end{align}
Now by putting $f'(x)=0$ we see that $\ln x=0$ or $x=1$. We also have $f''(1)=1>0$, therefore we have a local minimum in $x=1$ which is $f(1)=0$. Also $f'(x)=\ln x$ is negative for $x<1$ so it means the $f(x)$ is decreasing for $x<1$. Similarly by observing that $f'(x)>0$ is positive for $x>1$ we conclude that $f(x)$ is increasing for $x>1$. Therefore overall we have
$$f(x)\geq 0 ,\text{ for }x>0$$
A: Let $u=\frac1x$. Then the inequality is
$$
-\log(u)\ge1-u
$$
That is,
$$
\log(u)\le u-1
$$
which is simply
$$
\log(1+t)\le t
$$
A: Obvious with the following corollary of the Mean value theorem:

Let $f, g$ differentiable functions defined on an interval $I$, $a\in I$. Suppose that for all $x\in I$, except perhaps at $a$, we have $f'(x)<g'(x)$. Then
  $$f(x)-f(a)<g(x)-g(a)\enspace\forall x>a,\qquad f(x)-f(a)> g(x)-g(a)\enspace\forall x<a.$$

Just apply this corollary to $f(x)=\ln x$, $\;g(x)=-\dfrac1x$, $\;a=1$.
