Show that $\ln (x) \leq x-1 $ Show that $\ln (x) \leq x-1 $
I'm not really sure how to show this, it's obvious if we draw a graph of it but that won't suffice here. Could we somehow use the fact that $e^x$ is the inverse? I mean, if $e^{x-1} \geq x$ then would the statement be proved?

Denote $f(x)=\log{x}-x+1$. We have $f'(x)=\frac{1}{x}-1$.
For $0\lt x\leq 1$ we have $f'(x)\geq 0$ so $f$ is increasing from $-\infty$ to $0$
For $x\gt 1$ we have $f'(x)\lt 0$ so $f$ is decreasing from $0$ to $-\infty$
And we have proven $f(x)\leq 0$ Q.E.D
