Prove $\ln (1+x) \leq x - x^2/4 $ for $x \leq 1$ using Taylor's theorem

Question

RTP: $\ln (1+x) \leq x - \frac{1}{4} x^2$ for $x \leq 1$

I am trying to prove this specifically using Taylor theorem. Here is what I have so far:

$\ln (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3(1+\xi)^3}$, where $\xi \in (0, x)$

If $x > 0$ then $\frac{x^3}{3(1+\xi)^3} \leq \frac{x^3}{3}$

If $x < 0$ then $\frac{x^3}{3(1+\xi)^3} \leq \frac{x^3}{3(1+x)^3}$

For $x>0$, we have $\ln (1+x) \leq x - \frac{x^2}{2} + \frac{x^3}{3}$

Now I would like to show that $- \frac{x^2}{2} + \frac{x^3}{3} \leq - \frac{x^2}{4}$, but simply graphing this shows that it isn't true.

Similarly, for $x <0$, we have $\ln (1+x) \leq x - \frac{x^2}{2} + \frac{x^3}{3(1+x)^3}$, which is again not true for all $x < 0$.

Could anyone help me prove this inequality? Thank you!

Answer 1

For $x>-1$, we have $$ \log (1 + x) = x - x^2 \int_0^1 {\frac{t}{{1 + xt}}{\rm d}t} . $$ If $-1<x\le 1$, then $$ \int_0^1 {\frac{t}{{1 + xt}}{\rm d}t} \ge \int_0^1 {\frac{t}{{1 + t}}{\rm d}t} = 1 - \log 2 > \frac{1}{4}. $$