<s>[INST] Proving the inequality: $[x-(x^2)/2 < \ln(1+x) < x]$ , $x>0$ Prove the inequality:
$[x-(x^2)/2 < \ln(1+x) < x]$ ,    $x>0$ 

The right side is easy, I used taylor expansion to show that $e^x > 1+x$ since
$e^x = 1 + x + x^2/2 + x^3/3!  +\cdots $
The left side I expanded $\ln(1+x)$ to $x-x^2/2 + x^3/3 - x^4/4 +\cdots$ 
moved $(x-x^2/2)$ to the left and got this:
$$0 < x^3/3 - x^4/4 \cdots $$
The only thing I can think of here is that since the first term is positive and this series goes on forever the first term will always stay positive since the other terms cancel out...
Not sure this is mathematically coherent or even correct. 
thanks. 
 [/INST] For the left inequality, since you know that $x-\frac{x^2}{2}=\ln(1+x)$ when $x=0$ you need only check the derivatives: $\frac{d}{dx}(x-\frac{x^2}{2})=1-x$ while $\frac{d}{dx}(\ln(1+x))=\frac{1}{1+x}$, and so
$$\frac{d}{dx}(\ln(1+x))-\frac{d}{dx}(x-\frac{x^2}{2})=\frac{1}{1+x}-(1-x)=\frac{1-(1+x)(1-x)}{1+x}=\frac{x^2}{(1+x)}$$
which is positive for positive $x$, thus for $x>0$, $\ln(1+x)$ is always growing faster than $x-\frac{x^2}{2}$ so it must be greater than $x-\frac{x^2}{2}$.
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