I have to prove that $|\log(1 + \frac{n-x}{x})| \geq \frac{|h|}{2x}$ where $n = [x] + h$ and $3 \leq |h| \leq \frac{x}{4}$. Here $[x]$ is the integer part of $x$.
I'm thinking since $\frac{n-x}{x}$ is roughly $\frac{h}{x} < 1$ we can look at the taylor series for $\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + \dots > x - \frac{x}{2}$. Hence if $x = \frac{n-x}{x}$ it follows that $|\log(1 + \frac{n-x}{x})| > |\frac{n-x}{2x}| ~ \frac{|h|}{2x}$.
Is this correct? Are the steps rigorous? If not could someone help me fill in the details or point to a strategy.