Taylor's Remainder $x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) Prove that 
$\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) <x-\frac{x^2}{2}+\frac{x^3}{3}$
My attempt: I can prove this by taking one side at a time and assuming $\displaystyle f(x) = \log (1+x)-x+\frac{x^2}{2}-\frac{x^3}{3}$ 
and then proving that it is a decreasing function and then the same for the other side.
But I am looking for a better solution using Lagrange's Mean Value Theorem or using Taylor's remainder.

For the inequality on the right, note that the Lagrange form of the remainder gives
$$\log (1+x) = x- \frac{x^2}{2} +\frac{x^3}{3} -\frac{1}{4(1+c)^4},$$
where $0<c<x.$ Since the last term in negative, we get the strict inequality. In fact we get the strict inequality for $-1<x<0$ as well, by the same argument.
The inequality on the left seems easier for $-1<x<0$ to me. Here we get
$$\log (1+x) = x-x^2/2 + x^3/3 + \cdots = -(|x|+|x|^2/2 + |x|^3/3+ |x|^4/4\cdots )$$ $$> -(|x|+|x|^2/2 + (|x|^3/3)(1+|x| + |x|^2 + \cdots))$$ $$ = -(|x|+|x|^2/2 + (|x|^3/3)/(1-|x|) = x-x^2/2 + (x^3/3)/(1+x).$$
So there is good motivation for looking at $x-x^2/2 + (x^3/3)/(1+x)$ as below $\log (1+x)$ for all $x\in (-1,\infty), x\ne 0.$ At this point I would do exactly as you did in showing this is true for $x>0.$
