Prove $\ln x \ge \frac{x-1}{x}$ 
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 $0<x\le1$ ?
 A: I thought it might be instructive to present a way forward that does not rely on calculus, but rather elementary analysis only.
In THIS ANSWER and THIS ONE, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequality 
$$e^x\ge 1+x \tag 1$$
Setting $x=-z/(z+1)$ into $(1)$ and taking the logarithm of both sides reveals
$$\log(1+z)\ge \frac{z}{z+1} \tag 2$$
for $z>-1$.  Finally, substituting $x=1+z$ in $(2)$ yields the coveted inequality
$$\frac{x-1}{x}\le \log x$$
for $x>0$.
A: 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?
A: 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".
