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I would like to prove the following inequality without integration; could you help?

$$\frac {1}{x+1}\leq \ln (1+x)- \ln (x) \leq \frac {1}{x}, \quad x > 0. $$

I can however differentiate this.

Thanks in advance.

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  • $\begingroup$ Is it allowed to use an argument of form "$f(0)=0$ and $f'(x)\ge0$ for all $x\ge0$ so therefore $f(x)\ge0$ for all $x\ge0$, perhaps using Rolle's theorem? $\endgroup$ – kimchi lover Mar 13 '19 at 22:31
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    $\begingroup$ $e^{1/x}\ge 1+\frac 1 x$ implies $\frac 1 x \ge ln(1+x)-ln(x)$ $\endgroup$ – J. W. Tanner Mar 13 '19 at 22:32
  • $\begingroup$ No i cant use roll for this :/ $\endgroup$ – user653975 Mar 13 '19 at 22:38
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    $\begingroup$ Here is a question about the right inequality: Show that $\log(x+1)-\log(x)<\frac{1}{x}$ for $x >0$. (Maybe somebody can find some other related posts.) $\endgroup$ – Martin Sleziak Mar 14 '19 at 4:03
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Let $f(x):=\ln(x)$, then the Mean Value Theorem (Differentiation) says that there exists some $\xi\in (x,x+1)$ $$\ln (1+x)- \ln (x) = \frac{\ln (1+x)- \ln (x)}{(1+x)-x}= \frac{f (1+x)- f (x)}{(1+x)-x} =\frac{df}{dx}(\xi) $$

As $\frac{df}{dx}(x)= 1/x$ and as the function $1/x$ is monotonic we know that $$\frac {1}{x+1}\leq\frac{df}{dx}(\xi)\leq \frac {1}{x}$$

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  • $\begingroup$ Could you develop the first part? Thks $\endgroup$ – T.D. Mar 13 '19 at 22:59
  • $\begingroup$ @T.D. Inserted a link to the Mean Value Theorem. Not sure, if that's what you were thinking of ..? $\endgroup$ – Maksim Mar 13 '19 at 23:04
  • $\begingroup$ Oh dear sorry, the name of this theorem is different from my country. $\endgroup$ – T.D. Mar 13 '19 at 23:08
  • $\begingroup$ As op cant use rolle thm i think he cannot use a corollary $\endgroup$ – T.D. Mar 13 '19 at 23:11
  • $\begingroup$ I see, I already wondered what "can't use roll" would mean. Yes - then the proposed answer is not admitted.. $\endgroup$ – Maksim Mar 13 '19 at 23:13
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Lemma: For all real $x, e^x \ge 1 + x$.

Proof of lemma: Consider $f(x)=e^x-x-1.$ Its first derivative ($e^x-1$) is $0$ when and only when $x=0$, and its second derivative ($e^x$) is positive for all real $x, $ so $e^x-x-1$ has a global minimum when $x=0\; (f(0)=0$) and $e^x-x-1\ge 0$ for all real $x.$

From the lemma, $\exp\left({\frac 1x}\right)\ge 1+ \frac 1 x,$ which implies (take logarithm of both sides, noting that preserves order) $ \frac 1 x \ge \ln\left(1 + \frac 1 x\right)=\ln\left(\frac{x+1}x\right)=\ln(x+1)-\ln(x),$ which is one part of the desired inequality.

Also from the lemma, $\exp\left(\frac {-1}{1+x}\right)\ge 1-\frac 1 {1+x}=\frac x {1+x}$ implies (taking reciprocal of both sides, reversing the order) $\exp\left(\frac {1}{1+x}\right)\le \frac {1+x} {x}$ so (again, taking logarithm of both sides) $ \frac 1 {1+x} \le \ln \frac{1+x}{x}=\ln (1+x) - \ln(x),$ the other part of the desired inequality.

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If we write $t=\frac1x$ the inequalities become $${t\over t+1}\leq\ln(1+t)\leq t, t>0$$ Observe that the three expressions are equal when $t=0,$ and then show, by differentiating, that $t-\ln(1+t)$ and $\ln(1+t)-{t\over1+t}$ are increasing functions.

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    $\begingroup$ Did you mean become $$\frac t {t+1}?$$ And what happened to $\ln(x)$? $\endgroup$ – J. W. Tanner Mar 13 '19 at 22:43
  • $\begingroup$ @J.W.Tanner Yes, I did mean ${t\over1+t},$ thanks. $\ln(1+x)-\ln{x}=\ln\left({1+x\over x}\right)=\ln(1+1/x)=\ln(1+t)$ $\endgroup$ – saulspatz Mar 13 '19 at 22:47
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The right inequality.

We need to prove that $f(x)\geq0,$ where $$f(x)=\frac{1}{x}-\ln(1+x)+\ln{x}.$$ Indeed, $$f'(x)=-\frac{1}{x^2}-\frac{1}{1+x}+\frac{1}{x}=-\frac{1}{x^2(1+x)}<0,$$ which says $$f(x)\geq\lim_{x\rightarrow+\infty}f(x)=\lim_{x\rightarrow+\infty}\left(\frac{1}{x}+\ln\frac{x}{1+x}\right)=0.$$ The left inequality.

We need to prove that $g(x)\geq0,$ where $$g(x)=\ln(1+x)-\ln{x}-\frac{1}{1+x}.$$ We have $$g'(x)=\frac{1}{1+x}-\frac{1}{x}+\frac{1}{(1+x)^2}=-\frac{1}{x(1+x)^2}<0,$$ which says $$g(x)\geq\lim_{x\rightarrow+\infty}g(x)=0$$ and we are done!

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