$g(x)=\frac{x}{m}-1+\log m$, $g(x) \ge log x$

Question

Let $m \in \mathbb N$ and $g(x)=\frac{x}{m}-1+\log m$, if $m-\frac{1}{2} \le x<m+\frac{1}{2}$. How could I show that $g(x) \ge \log x$?

Answer 1

If you now that $$\forall y>0\ \ \ \log(y)\leq y-1\tag{1}$$

then apply $(1)$ for $y=\frac{x}{m}$, if you don't know $(1)$ you must prove it by studying the variations of $h(x)=x-1-\log(x)$