Proving inequality on functions $x-\frac{x^2}{2}<\ln(1+x)To prove: $$x-\frac{x^2}{2}<\ln(1+x)<x-\frac{x^2}{2(1+x)},\quad\forall x>0$$
I have used Taylor series expansion at 0 for both the inequalites. The greater than by expanding $\ln(1+x)$ and the less than by expanding $\int \ln(1+x)\,dx$ at 0. 
Is there a cleaner / more elegant way of achieving the same?

Another way to do the lower bound, for example, would be to consider $f(x) = x - \dfrac{x^2}{2} - \ln(1+x)$. $f(0) = 0$.
But also $f'(x) = 1 - x - \dfrac{1}{1+x} = \dfrac{(1-x)(1+x) - 1}{1+x} = \dfrac{1 - x^2 - 1}{1+x} = \dfrac{-x^2}{1+x} < 0$.
So since $f(0) = 0$ and $f' < 0$, $f$ is monotonically and strictly decreasing. Thus $f(x) \leq 0$, and if $x > 0$ we have that $f(x) < 0$. And this says exactly that $x - x^2/2 < \ln(1 + x)$.
