# Showing x 1 + x </mrow> </mfrac> &lt; log &#x2061;<!-- ⁡

Showing $\frac{x}{1+x}<\mathrm{log}\left(1+x\right) for all $x>0$ using the mean value theorem
I want to show that
$\frac{x}{1+x}<\mathrm{log}\left(1+x\right)
for all $x>0$ using the mean value theorem. I tried to prove the two inequalities separately.
$\frac{x}{1+x}<\mathrm{log}\left(1+x\right)⇔\frac{x}{1+x}-\mathrm{log}\left(1+x\right)<0$
Let
$f\left(x\right)=\frac{x}{1+x}-\mathrm{log}\left(1+x\right).$
Since
$f\left(0\right)=0$
and
${f}^{\prime }\left(x\right)=\frac{1}{\left(1+x{\right)}^{2}}-\frac{1}{1+x}<0$
for all $x>0$, $f\left(x\right)<0$ for all $x>0$. Is this correct so far?
I go on with the second part: Let
$f\left(x\right)=\mathrm{log}\left(x+1\right)$. Choose $a=0$ and $x>0$ so that there is, according to the mean value theorem, an ${x}_{0}$ between $a$ and $x$ with
${f}^{\prime }\left({x}_{0}\right)=\frac{f\left(x\right)-f\left(a\right)}{x-a}⇔\frac{1}{{x}_{0}+1}=\frac{\mathrm{log}\left(x+1\right)}{x}$
Since
${x}_{0}>0⇒\frac{1}{{x}_{0}+1}<1.$
$⇒1>\frac{1}{{x}_{0}+1}=\frac{\mathrm{log}\left(x+1\right)}{x}⇒x>\mathrm{log}\left(x+1\right)$
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razdiralem
By Definition of log(which already a kind of mean value theorem ) we have For any $x>0$ We have
$\frac{x}{x+1}={\int }_{0}^{x}\frac{dt}{x+1}\le {\int }_{0}^{x}\frac{dt}{t+1}=\mathrm{ln}\left(x+1\right)={\int }_{0}^{x}\frac{dt}{t+1}\le {\int }_{0}^{x}\frac{dt}{1}=x$
Thus,
$\frac{x}{x+1}\le \mathrm{ln}\left(x+1\right)\le x$
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Lena Bell
Substitute $x=y-1$ to get
$1-{y}^{-1}<\mathrm{log}y
for all $y>1$. Now note
${\int }_{1}^{y}{t}^{-2}dt<{\int }_{1}^{y}{t}^{-1}dt<{\int }_{1}^{y}dt$
for $y>1$