for each $x>1 , \frac{x-1}{x}\ < \ln x < x-1$ I tried to prove this with differentiation:
when $x >$ 1, all 3 functions are positive and when $x = 1$, all 3 reaches zero. And the derivatives are varying like 
$$\frac{\mathrm{d}(\frac{x-1}{x})}{\mathrm{d}x}\ < \frac{\mathrm{d}(\ln x)}{\mathrm{d}x}\ < \frac{\mathrm{d}(x-1)}{\mathrm{d}x}$$
Are they enough to say that above inequality is true?
 A: If you have the inequality for the derivatives for all $x\gt0$ and the inequality you want at a given $x_0$, then you will have shown the inequality you want for all $x\ge x_0$. However, the inequality you want is meaningless for $x=0$, so proceeding as you are, it seems difficult to show the inequality for all $x\gt0$.
However, if you use the fact that at $x=1$, you have $\frac{x-1}x=\log(x)=x-1$, then, using the inequality for the derivatives, you should be able to prove the inequality for all $x\ge1$. Then looking at the inequality for $1/x$, You get the inequality for $x\le1$.

Pre-Calculus Proof
Bernoulli's Inequality, which can be proven by induction for positive integer exponents, says that for $x\ge-n$,
$$
1+x\le\left(1+\frac xn\right)^n\tag{1}
$$
and since
$$
\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x\tag{2}
$$
we have for all $x\in\mathbb{R}$,
$$
1+x\le e^x\tag{3}
$$
Plugging in $-x$ into $(3)$ and taking reciprocals, we get
$$
e^x\le1+\frac x{1-x}\tag{4}
$$
Substituting $x\mapsto\frac x{1+x}$ into $(4)$ and combining with $(3)$, gives for $x\gt-1$,
$$
e^{x/(1+x)}\le 1+x\le e^x\tag{5}
$$
Substituting $x\mapsto x-1$ in $(5)$ and taking logs, yields for $x\gt0$,
$$
\frac{x-1}x\le\log(x)\le x-1\tag{6}
$$
A: You are "correct", but your proof is not rigorous in the sense that you seem to have simply been given a method without any explanation as to why the property works. Consider the following: define two functions
$$ f(x) = x - 1 - \ln(x), \ g(x) = \ln(x) + \frac{1 - x}{x}$$
First note that 
$$ f(1) = 0, \ g(1) = 0$$
The reason I have mapped these values is because next we are going to prove that both $f$ and $g$ are genuinely increasing functions, and as such all values of $f$ and $g$ such that $x > 1$ will be positive! Now to show that both functions are increasing we note that
$$ f'(x) = 1 - \frac{1}{x} > 1 - \frac{1}{1} = 0 \implies \forall x > 1 : f'(x) > 0$$
and for $g$ we have
$$ g'(x) = \frac{1}{x} - \frac{1}{x^2} = \frac{1}{x} \left(1 - \frac{1}{x} \right) >\frac{1}{x} \left(1 - \frac{1}{1}\right) = 0 \implies \forall x > 1 : g'(x) > 0$$
Now we use the mean value theorem and note that for $a,b \in \mathbb{R} : 1 < a < b$ we will have
$$ \exists c \in (a,b) : f'(c) = \frac{f(b) - f(a)}{b - a}$$
Now using the above we may evaluate this to
$$ \frac{f(b) - f(a)}{b - a} > 0 \implies f(b) - f(a) > 0 \iff f(b) > f(a)$$
Now that when evaluating the inequality we used $b - a > 0$ the same goes for $g$. The choice of $a$ and $b$ above is arbitrary so long as $1 < a < b$ hence we may conclude that $f$ and likewise $g$ are genuinely increasing functions. Now consider any $x > 1$. Earlier we had that $f(1) = 0$ clearly $f(x) > f(1)$ by monotonicity. So now we can conclude that
$$ f(x) > f(1) \implies f(x) > 0 \implies x - 1 - \ln(x) > 0$$
which gives us the inequality
$$ \ln(x) < x - 1$$
The exact same argument can be fleshed out for $g$ and then we will get the full inequality! I hope this helped!
A: Note that value of all the functions at 1 is same(i.e 0).
Now taking $f(x) = x - 1 - \ln x$.
$f'(x) = 1 - 1/x$
Note that $f'(x) > 0$ for all $x > 1$.
Thus $f(x)$ is strictly increasing in the domain $(1,\infty)$
Also $f(1) = 0$
Thus $f(x) > 0$ for all x in the domain $(1,\infty)$
Thus $x-1 > \ln x$ for all x in the domain $(1,\infty)$
You can do it similarly for the function $f(x) = \ln x - \frac{x-1}{x}$
A: The inequality is very crude so it should have a simple proof. If $t \in (1,x)$ then $\frac 1x < \frac 1t < 1$. Now integrate on the interval $[1,x]$ to get $$\frac{x-1}{x} < \ln x < x-1.$$
A: It's correct: if you have two continuous functions $f$ and $g$ defined over $[1,\infty)$ and differentiable on $(1,\infty)$, then $f'(x)<g'(x)$ for all $x\in(1,\infty)$ and $f(1)\le g(1)$ implies $f(x)<g(x)$ for all $x\in(1,\infty)$.
Indeed, the function $F(x)=g(x)-f(x)$ has positive derivative on $(1,\infty)$, so it's increasing; since $F(1)\ge0$, we have $F(x)>0$ for $x>1$.
(Of course, the interval can be arbitrary.)
It's obvious that we can extend this to three functions as in your setting, just by looking at pairs. If
$$
f(x)=\frac{x-1}{x},\quad g(x)=\log x,\quad h(x)=x-1
$$
we have
$$
f'(x)=\frac{1}{x^2},\quad g'(x)=\frac{1}{x},\quad h(x)=1
$$
and so, indeed,
$$
f'(x)<g'(x)<h'(x)
$$
for all $x>1$, while $f(1)=g(1)=h(1)=0$. Thus, for all $x>1$,
$$
f(x)<g(x)<h(x)
$$
