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I looking for help with proving the following inequality. Any relevant logarithmic identities would be great. Tried differentiating and taking limits and I'm lost as to how to approach this.

$\frac 1{x+1}<\ln\left(\frac{x+1}x\right)<0.5\left(\frac 1x+\frac 1{x+1}\right)$

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    relatedhttps://math.stackexchange.com/questions/652581/showing-fracx1x-log1xx-for-all-x0-using-the-mean-value-theorem2018-02-09

2 Answers 2

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Let us recall a property of integrals, so simple one could think nothing good could come out of it, and yet:

For every interval $(a,b)$ with $a\leqslant b$ and functions $u$ and $v$ such that $u\geqslant v$ on $(a,b)$, $\int_a^bu(x)\mathrm dx\geqslant\int\limits_a^bv(x)\mathrm dx.$ In particular, if $u\geqslant c$ on $(a,b)$, for some constant $c$, then $\displaystyle\int_a^bu(x)\mathrm dx\geqslant c(b-a)$.

In the present case, one may start from the identity $\log((x+1)/x)=\int\limits_x^{x+1}u(t)\mathrm dt$ with $u(t)=1/t$ and note that $u(t)\geqslant1/(x+1)$ for every $t$ in the interval $(x,x+1)$. This should give one inequality.

For the other inequality, one may prove that $v(s)=u(x+1/2+s)+u(x+1/2-s)$ is an increasing function of $s$ for $s$ in $(0,1/2)$ and note that $\log((x+1)/x)=\int_0^{1/2}v(s)\mathrm ds$.

Can you reach that point and finish the proof from there?

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    See expanded version.2012-11-06
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For LHS:use this well konw inequality $\dfrac{x}{x+1}\le\ln{(1+x)},x>-1\Longrightarrow \ln{\left(\dfrac{x+1}{x}\right)}=\ln{\left(1+\dfrac{1}{x}\right)}\ge\dfrac{\dfrac{1}{x}}{1+\dfrac{1}{x}}=\dfrac{1}{x+1}$ For RHS: we only prove $2\ln{(1+x)} let $g(x)=2\ln{(1+x)}-\dfrac{x^2+2x}{x+1}\Longrightarrow g'(x)=\dfrac{2}{x+1}-1-\dfrac{1}{(x+1)^2}=\dfrac{-x^2}{(x+1)^2}\le 0$ so $g(x)\le g(0)=0$ so $\ln{(1+x)}\le\dfrac{1}{2}(x+\dfrac{x}{x+1})$ so $\ln{\left(\dfrac{x+1}{x}\right)}\le 0.5\left(\dfrac{1}{x}+\dfrac{1}{x+1}\right)$