Prove with the mean value theorem that $x-\frac{x^2}{2} < \ln(1+x)$

Question

Prove with the mean value theorem that $x-\frac{x^2}{2}<\ln(1+x)<x$ in $(0,\infty)$

Approach $f(x) := \ln(1+x) $ with the mean value theorem in $[0,x]$

$\frac{1}{1+\xi}= \frac{\ln(1+x)-0}{x-0}$

$\frac{1}{1+\xi}$ takes the biggest value when $\xi$ is $0$

and so $1 <\frac{\ln(1+x)}{x}$ multiply with x and you get $x<\ln(1+x)$

I can't prove the second part.

Answer 1

For MVT

$$\ln (1+x)+\frac{x^2}{2}-(\ln 1+0)=x\cdot\left(\frac{1}{1+c}+c\right)>x \quad c\in(0,x)$$

Indeed

$$\frac{1}{1+c}+c=\frac{c^2+c+1}{1+c}>1$$

Answer 2

There is no need for the MVT, the claim is a straightforward consequence of the integral representation for the logarithm function: $$\log(1+x)=\int_{0}^{x}\frac{du}{1+u}\leq\int_{0}^{x}1\,du = x $$

$$\log(1+x)=x-\int_{0}^{x}\frac{u}{1+u}\,du\geq x-\int_{0}^{x}u\,du = x-\frac{x^2}{2}.$$