Question: > 
> Prove that for every $x>0$:
>  $$\ln x \ge \frac{x-1}{x}$$
> 
> 
> 

What I did:

$$f(x) = \ln x, \text{ } g(x) = \frac{x-1}{x} $$

$$f(1) = g(1) = 0 $$

So it's enough to prove that $$ f'(x) \ge g'(x)$$
I proved this for every $x>1$.
How can I prove this for every $0<x\le1$ ?

Answer: 
You approach is good, but there is another which is worthy of consideration. Define $h : (0,\infty) \rightarrow \mathbb{R}$ by
\begin{equation}
h(x) = x \text{ln}(x) - x + 1.
\end{equation}
and show that this function is non-negative if and only if $f(x) \ge g(x)$. Then subject $h$ to a standard functional analysis in order to determine its range.
Compute the limits as $x$ tends to $0$ and $\infty$. Moreover, the derivative of $h$ is easily seen to be $\text{ln}(x)$. You should find that $h$ is always non-negative.

Edit: word "positive" replaced with the correct word "non-negative".
HINT: set $$f(x)=\ln(x)-\frac{x-1}{x}$$ then we get $$\lim\_{x \to 0+}f(x)=+\infty$$ further is $$f'(x)=\frac{x-1}{x^2}$$ and $$f''(x)=-\frac{x-2}{x^3}$$
can you proceed?