$0 \leq u - \ln(1+u) \leq u^2$ for $\mid u\mid < \frac{1}{2}$

We have that

$$\psi(u)= u - \ln(1+u) \implies \psi'(u)= 1 - \frac{1}{1+u} \implies \psi''(u)=\frac1{(1+u)^2}\ge 0$$

and since $\psi(0)=\psi'(0)=0$ we have $u - \ln(1+u) \ge 0$.

And since

$$f(u)=u^2-u+ \ln(1+u) \implies f'(u)=2u-1+\frac{1}{1+u}\implies f''(u)=2-\frac1{(1+u)^2}\ge 0$$

and since $f(0)=f'(0)=0$ we have $u^2-u+ \ln(1+u)\ge 0$.