By taking $f(x) = \ln(1 + x)$ and using the inequality $\frac{1}{1 + x} < \frac{1}{1 + c} < 1$ for $c \in (0, x)$, applying LMVT gives $\frac{x}{1 + x} < \ln(1 + x) < x$.
By taking $f(x) = x - \ln(1 + x)$ and using the inequality $1 - \frac{1}{1 + x} > 1 - \frac{1}{1 + c} > 0$ for $c \in (0, x)$, applying LMVT gives $\frac{x^2}{1 + x} > x - \ln(1 + x) > 0$.
But I am unable to find a function to prove $\frac{x^2}{2} > x - \ln(1 + x) > \frac{x^2}{2(1 + x)}$.
I know it can easily be solved using Taylor's theorem, but I want to solve it using LMVT. Any help is appreciated.
$\begin{array}\\ f(x) &=x - \ln(1 + x)\\ &=x - \int_0^x \dfrac{dt}{1+t}\\ &=\int_0^x (1-\dfrac1{1+t})dt\\ &=\int_0^x \dfrac{1+t-1}{1+t}dt\\ &=\int_0^x \dfrac{t}{1+t}dt\\ &\ge\int_0^x \dfrac{t}{1+x}dt \qquad \text{(since }1+t \le 1+x)\\ &=\dfrac1{1+x}\int_0^x tdt\\ &=\dfrac{x^2}{2(1+x)}\\ f(x) &=\int_0^x \dfrac{t}{1+t}dt\\ &\le\int_0^x tdt \qquad \text{(since } 1+t \ge 1)\\ &=\dfrac{x^2}{2}\\ \end{array} $