Bound on log 2
Dependencies: None
Let $x \ge 0$. Then \[ \ln(1+x) \ge \frac{2x}{2+x} \] \[ \ln\left(\frac{1}{1-x}\right) \ge \frac{2x}{2-x} \]
Proof
Let $f(x) = \ln(1+x) - \frac{2x}{2+x}$. \[ f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2} \] So for $x \ge -1$, $f'(x) \ge 0$.
Since $f(0) = 0$ and $f'(x) \ge 0$, $f(x) \ge 0$ for $x \ge 0$ and $f(x) \le 0$ for $x \le 0$.
Dependency for:
Info:
- Depth: 0
- Number of transitive dependencies: 0