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

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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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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Hi did, how does u(t) > 1/(x+1) relate the integral being more then that as well. You kinda lost me there... – blinduck Nov 6 '12 at 16:09
See expanded version. – Did Nov 6 '12 at 16:37

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 $$2\ln{(1+x)}<x+\dfrac{x}{x+1}=\dfrac{x^2+2x}{x+1}$$ 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 $$\ln{\left(\dfrac{x+1}{x}\right)}\le 0.5\left(\dfrac{1}{x}+\dfrac{1}{x+1}\right)$$

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