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Why is $\ln\left(1+t\right)\geq\frac{t}{1+t}$ for every $t>-1$?

I have tried letting $f(x)=\ln(1+t)-\frac{t}{1+t}$ but I'm stuck from here. What do I do next? I am supposed to use the mean value theorem, I think I've done it completely wrong.

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    $\begingroup$ What does the mean value theorem say? something about derivatives? Have you taken the derivative? $\endgroup$
    – ziggurism
    Commented Apr 4, 2020 at 2:56
  • $\begingroup$ Oops, I goofed. SOrry. I plotted $\log_{10}$. $\endgroup$
    – copper.hat
    Commented Apr 4, 2020 at 2:59

4 Answers 4

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If $t\ge 0$, then $$\log (1+t) = \int_1^{1+t}\frac{dx}x =\int_0^t \frac{du}{1+u} \ge \int_0^t\frac{du}{1+t} = \frac{t}{1+t}.$$

An analogous argument could be used when $-1<t<0$. If it's too messy, I would consider doing $1+t\mapsto z$.

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  • $\begingroup$ +1: Nice. ${}{}$ $\endgroup$
    – copper.hat
    Commented Apr 4, 2020 at 3:10
  • $\begingroup$ Very elegant ! (+1) $\endgroup$ Commented Apr 4, 2020 at 3:17
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Consider the function $$f(t)=\ln\left(1+t\right)-\frac{t}{1+t}$$ $$f'(t)=\frac{t}{(t+1)^2} >0 \quad \forall t \, > 0$$ So, at least $f(t)$ is an increasing function for $t>0$ and $f(0)=0$ means that $$\ln\left(1+t\right)\geq \frac{t}{1+t}\quad \forall t \, \geq 0$$ On the other side $t=0$ is an extremum and $$f''(t)=\frac{1-t}{(t+1)^3}$$ shows that this is a minimum value. So it is true as long as $\ln(1+t)$ is defined in the real domain.

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    $\begingroup$ @nyz. Thanks for the edit but there is only one logarithm (as far as I now, at least). All other are (again at least to me) just artifacts. Cheers :-) $\endgroup$ Commented May 8, 2023 at 12:10
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To atone for my sins...

Let $f(x) = \ln (1+x)-{x \over 1+x}$, note (meaning prove it) that $\lim_{x \downarrow -1} f(x) = \infty$, $\lim_{x \downarrow 1} f(x) = \infty$ and $f'(x) = 0$ has exactly one solution at $x= 0$ where $f(x) = 0$. Hence $x=0$ is the minimiser on $x>0$.

Hence $f(x) \ge 0$ for all $x \in (-1,\infty)$.

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The fundamental inequality satisfied by logarithm function is $$\log x\leq x-1\,\forall x>0\tag{1}$$ and it can be proved using any chosen definition of $\log x$. In fact combined with the functional equation $$\log xy=\log x+\log y\, \forall x, y>0\tag{2}$$ the above inequality characterizes the logarithm function uniquely.

Let's replace $x$ by $1/x$ in above inequality $(1)$ to get $$-\log x\leq \frac{1-x}{x}$$ or $$\log x\geq \frac{x-1}{x}$$ Putting $x-1=t$ we get $$\log(1+t)\geq\frac{t}{1+t}$$ and note that $x>0$ implies $t>-1$ so that above inequality is valid for all $t>-1$.

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