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Show that: $ \frac{x^2}{2(x + 1)} < x − \ln(x + 1) < \frac{x^2}{ 2} , x > 0$

One method is considering a function $f(x) = x − \ln(1 + x) − \frac{x^2}{2}$ and showing its derivative $f'(x) < 0$. Similarly, considering another function $g(x) = x − \ln(1 + x) − \frac{x^2}{ 2(x + 1)}$ and showing its $g'(x) > 0$.

Is there some other method that is not lengthy?

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    I suppose that's $\frac{x^2}{2(x+1)}$ on the left? – 2017-02-10
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    don't forget the dollar signs when you type the maths! – 2017-02-10
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    Another tip: \$\frac{x^2}{2(x+1)}\$ results in $\frac{x^2}{2(x+1)}$ – 2017-02-10

3 Answers 3

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Note that $$ \int_0^x\frac{t}{t+1}dt=x-\ln(x+1). $$ Since $f(t)=\frac{1}{t+1}$ is decreasing in $[0,x]$, one has $$ \frac{x^2}{2(x+1)}=\int_0^x\frac{t}{x+1}dt<\int_0^x\frac{t}{t+1}dt<\int_0^xtdt=\frac12x^2 $$ and hence $$ \frac{x^2}{2(x+1)}0.$$

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    +1 At first it seems that writing the expression as integral of something is the same as the OP's taking the derivative. However, the clever difference is in finding useful estimates instead of blind calculation – 2017-02-10
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You can use limits for the natural log (which can basically be seen after Taylor expansion, proof is here) to see that

$\ln(x+1) > x - x^2/2$ and $(x+1)\ln(x+1) < x + x^2/2$

The first inequality immediately gives

$x-\ln(x+1) < x^2/2$ and

and the second one gives

$(x+1) (x-\ln(x+1)) > x + x^2 - (x + x^2/2) = x^2/2$

so we have the two required limits.

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You can consider \begin{align} f(x)&=\frac{x^2}{x+1} & f'(x)&=1-\frac{1}{(x+1)^2} \\ g(x)&=x-\ln(x+1) & g'(x)&=1-\frac{1}{x+1} \\ h(x)&=\frac{x^2}{2} & h'(x)&=x \end{align} and note that, for $x>0$, $$ f'(x)

Not so different from your method, but a bit faster.