Q:

Show that $\ln(1+x)\leq x-{1\over 2}x^2+{1\over 3}x^3$.

Prove that for $x\in (0,\infty)$, $\ln(1+x)\leq x-{1\over 2}x^2+{1\over 3}x^3$.
I'm a little bit stuck, but I think I have the right idea. Any hints or solutions are greatly appreciated. Here is what I have.
If we assume that $\ln(1+x)>x-{1\over 2}x^2+{1\over 3}x^3$ and let $f(x)=\ln(1+x)-x+{1\over 2}x^2-{1\over 3}x^3>0$, then $f'(x)={1\over{1+x}}-1+x-x^2={-x^3\over 1+x}<0$ for all $x>0$. That is, $f'(x)<0$ implies that $f(x)$ decreasing which is a contradiction.
Does this make sense?

A:

You could convert yours into a direct proof easily. As you have, let
$$f(x) = \ln(1+x) - x + \dfrac{x^2}2 - \dfrac{x^3}3$$
We then have
$$f'(x) = \dfrac1{1+x} - 1 + x - x^2 = \dfrac{-x^3}{1+x} < 0 \text{ since }x \in (0,\infty)$$
Hence, $f(x)$ is a decreasing function, which means
$$f(x) < f(0) \implies \ln(1+x) - x + \dfrac{x^2}2 - \dfrac{x^3}3 < 0 \implies \ln(1+x) < x - \dfrac{x^2}2 + \dfrac{x^3}3$$

