$\left| \ln(1+u) - u + \frac{u^2}{2} \right|\leq |u| ^3 $ for $| u | \leq \frac{2}{3}$

Question

Shall we prove that : $\left| \ln(1+u) - u + \frac{u^2}{2} \right|\leq |u| ^3 $ for $| u | \leq \frac{2}{3}$

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My attempt :
Let $s$ be the sign function

$f(u)= u^3 - \ln(1+u) + u - \frac{u^2}{2}$
$f'(u) = s(u) 3 u^2 - \frac{1}{1+u} +1 - u$
If $u >0$, $s(u)=1$
$f'(x) = \frac{ 3 u^2 (1+u) -1 + 1 +u - u - u^2}{1+u} = \frac{3u^3+2u^2}{1+u} = \frac{u^2 (3u+2)}{1+u}$
$f'(u)$ for $u >0$
if $u \leq 0$, $s(u) =-1$ and $f'(u) = \frac{-u^2 (3u+4)}{1+u} <0$
So $f$ is decreasing then increasing, the minimum is $f(0)=0$ so the right inequality is proved.
For the left inequality, let define $\psi(u) = \ln(1+u) -u + \frac{u^2}{2} +s(u) u^3$. For $u>0$, $s(u)=1$ , and $\psi'(u)= \frac{4u^2}{1+u} >0$
For $u<0$, $\psi'(u)= -\frac{2u^2}{1+u} >0$
$\psi$ is decreasing then increasing, the min is $\psi(0)=0$. It proves the left inequality.

Answer 1

Hint: By Taylor's formula with Lagrange remainder $$ \log (1 + u) = u - \frac{{u^2 }}{2} + \frac{{u^3 }}{{3(1 + \xi )^3 }}, $$ where $0\leq \xi\leq u$ for $u\geq0$ and $u\leq\xi\leq 0$ for $-1<u<0$.