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

Question

I want to prove without calculator that :

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

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My attempt :

$$ \psi(u)= u - \ln(1+u)$$ $$\psi'(u)= 1 - \frac{1}{1+u} = \frac{u}{1+u} $$

so $\psi$ decreasing for $u <0$ and increasing for $u >0$ and $\psi(0) =0$ so $\psi \geq 0$

For the second inequality:

$$ \phi(u)=u^2- u - \ln(1+u)$$ $$ \phi'(u)= 2u -1 + \frac{1}{1+u} = \frac{u(1+2u)}{1+u}$$

So $\phi$ is decreasing for $u<0$ and increasing for $u>0$, and $\phi(0)=0$ so $\phi \geq 0$

Answer 1

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 $u^2-u+ \ln(1+u)\ge 0$.

Answer 2

For the error in proof of first inequality pointed out in comments section :

you can prove it by checking that $\psi(u)$ is decreasing for $u < 0$ , $\psi(0) = 0$ and $\psi(u)$ is increasing for $u > 0 $