Prove that $\forall x > 0, x - 1 \ge \ln(x)$ 
Prove that $\forall x > 0, x - 1 \ge \ln(x)$ .

Here is my proof:
We prove the inequality on two intervals, $(0,1]$ and $[1,+\infty)$.
First the easier one, $[1,+\infty)$. 
Notice that at $x=1$, the inequality holds true. Notice also that $\forall x\in [1,+\infty)$, $\frac{1}{x} \le 1 \le x$. But this is the same as: 
$$\forall x\in [1,+\infty), \frac{d[ \ln{x}]}{dx} \le 1 \le \frac{d[x-1]}{dx}$$
But we know that if $f,g$ are two functions such that, for some $x_0$, $f(x_0) \ge g(x_0)$ and $\forall x\ge x_0, f'(x)\ge g'(x),$ then $\forall x \ge x_0 , f(x) \ge g(x). $ Therefore our inequality holds on the interval $[1,+\infty)$.
Now for the second interval. Consider the functions $f(x) = -x -1$ and $g(x) = \ln(-x)$. These are essentially the reflections of our two previous functions in the y-axis. We will consider these functions only on the interval $[-1, 0)$. Notice that at $x=-1$, we have $f(x) = g(x)$. Furthermore, on our interval $\frac{1}{x} \le -1 \le x$. But this is exactly the same as: 
$$\forall x \in [-1, 0), g'(x) \le f'(x)$$
By the same theorem as before, we have:
$$\forall x \in [-1, 0), \ln(-x) \le -x-1$$
Now let $y=-x$. Then $y \in (0,1]$ and the previous statement is equivalent to: 
$$\Leftrightarrow \forall y\in (0,1], \ln(y) \le y-1$$
Hence we have shown that the equality holds on $(0,+\infty].$
Is the proof complete? I am especially curious about the validity of the last step, where we let $-x = y$. Is that a valid step?
Is there any way to shorten the proof?
 A: It is correct, but there is no need to use two intervals. Let $g(x)=x-1-\ln x$. Then $g(1)=0$ and $g'(x)=1-\frac1x$, which is greater that $0$ if $x>1$ and smaller than $0$ if $x\in(0,1)$. So, $g$ has a minimum at $1$, and that minimum is $0$.
A: We need to prove that $e^{x-1}\geq x$ and since $y=x$ is a tangent to the graph of $f$, 
where $f(x)=e^{x-1}$ and $f$ is a convex function, we are done!
A: For $x>0$ let $f (x)=x-1-\ln (x) $.
$f $ is differentiable at $(0,+\infty) $.
by MVT
$$f (x)-f (1)=(x-1)f'(c) $$
$$\implies f (x)=\frac {(x-1)(c-1)}{c} $$
with
$0 <x<c<1$ or
$1<c <x.$
in both cases , we have $(x-1)(c-1)\ge 0$
hence $f (x)\ge 0$. Done.
A: The function $f(x):=\ln(x)$ is concave in $(0,+\infty)$. Therefore any tangent line lies above the graph of the function. Hence for any positive numbers $x$ and $x_0$ 
$$\frac{(x-x_0)}{x_0}+\ln(x_0)=f'(x_0)(x-x_0)+f(x_0)\geq f(x)=\ln(x).$$
In particular, for $x_0=1$, we get the desired inequality
$$x-1\geq \ln(x).$$
P.S. If $f\in C^2(a,b)$ is concave  then $f''\leq 0$ and, by Taylor's Theorem, for $x_0,x\in(a,b)$, there is $c$ between $x$ and $x_0$ such that
$$f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(c)}{2}(x-x_0)^2\leq f(x_0)+f'(x_0)(x-x_0).$$
