2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ What does the mean value theorem say? something about derivatives? Have you taken the derivative? $\endgroup$
    – ziggurism
    Apr 4 '20 at 2:56
  • $\begingroup$ Oops, I goofed. SOrry. I plotted $\log_{10}$. $\endgroup$
    – copper.hat
    Apr 4 '20 at 2:59
6
$\begingroup$

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 $1+t\mapsto z$.

$\endgroup$
2
  • $\begingroup$ +1: Nice. ${}{}$ $\endgroup$
    – copper.hat
    Apr 4 '20 at 3:10
  • $\begingroup$ Very elegant ! (+1) $\endgroup$ Apr 4 '20 at 3:17
1
$\begingroup$

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 $x \in (-1,\infty)$.

$\endgroup$
0
$\begingroup$

Consider the function $$f(t)=\log\left(1+t\right)-\frac{t}{1+t}$$ $$f'(t)=\frac{t}{(t+1)^2} >0 \quad \forall t \, > 0$$ So, at least $f(t)$ is an increasing function for $t>0$ and $f(0)=0$ means that $$\log\left(1+t\right)\geq \frac{t}{1+t}\quad \forall t \, \geq 0$$ On the other side $t=0$ is an extremum but $$f'(t)=\frac{1-t}{(t+1)^3}$$ shows that this is a minimum values. So, it is true as long as $\log(1+t)$ is defined in the real domain.

$\endgroup$
0
$\begingroup$

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 $t>-1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.