Question

Prove $ln(1+x)$ is larger than $\frac{ta{n}^{-1}x}{1+x}$, for x > 0.

Answer 1

Let
$f\left(x\right)=ln(1+x)-\frac{ta{n}^{-1}x}{1+x}$
Now
$f^{\prime }\left(x\right)=\frac{1}{1+x}-\frac{1}{(1+x^2)(1+x)}+\frac{ta{n}^{-1}x}{(1+x{)}^2}$
Now
$f^{\prime }\left(x\right)>0$ implies
$\frac{1}{1+x}-\frac{1}{(1+x^2)(1+x)}>-\frac{ta{n}^{-1}x}{(1+x{)}^2}$
$\frac{x^2}{(1+x)(1+x^2)}>-\frac{ta{n}^{-1}x}{(1+x{)}^2}$
$\frac{1}{1+x}[\frac{x^2}{x^2+1}+\frac{ta{n}^{-1}\left(x\right)}{1+x}]>0$
Now for x>0, f'(x) is positive.
Hence $f\left(x\right)>0$ for x>0.
Or
$ln(1+x)-\frac{ta{n}^{-1}\left(x\right)}{1+x}>0$ for x>0
$ln(1+x)>\frac{ta{n}^{-1}\left(x\right)}{1+x}$ for x>0.