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$x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x)0$

Here I can only see that the right side of second inequality i.e. $x-\frac{x^2}{2}+\frac{x^3}{3}$ comes in the expansion of $\log(1+x)$.

We have done the Lagrange's mean value theorem and intermediate value theorem, do these have anything to do with the inequality.

Kindly provide some hint.

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    what kind of $\log$ is this? – 2017-02-21
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    Integrate the equation $$\frac{1}{1+t}=1-t+t^{2}-\frac{t^{3}}{1+t} $$ between $0$ and $x$. – 2017-02-21

2 Answers 2

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defining $$f(x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\ln(1+x)$$ for $x=0$ we get $$f(0)=0$$ and $$f'(x)=1-x+x^2-\frac{1}{x+1}=\frac{(1-x+x^2)(1+x)-1}{x+1}=\frac{x^3}{1+x}>0$$ for $x>0$ thus our function is monotonously increasing and we get $$f(x)>0$$ for all $$x>0$$

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    can we adapt this easily to get the lower bound for $\log(1 + x)$ – 2017-02-21
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I expand my comment into an answer.

We have for $t > 0$ $$\frac{1}{1 + t} = 1 - t + \frac{t^{2}}{1 + t}$$ and hence if $0 < t < x$ then $$1 - t + \frac{t^{2}}{1 + x} < \frac{1}{1 + t} < 1 - t + t^{2}$$ and integrating the above with respect to $t$ on interval $[0, x]$ we get $$x - \frac{x^{2}}{2} + \frac{x^{3}}{3(1 + x)} < \log(1 + x) < x - \frac{x^{2}}{2} + \frac{x^{3}}{3}$$

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    @MartinR: Thanks for catching typo. Its fixed now. – 2017-02-21
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    http://math.stackexchange.com/a/1548163/42969 :) – 2017-02-21
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    @MartinR: its the same answer repeated. i was searching for it, but couldn't find it so wrote again. – 2017-02-21