# Why is $\ln\left(1+t\right)\geq\frac{t}{1+t}$?

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.

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

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 $$-1. If it's too messy, I would consider doing $$1+t\mapsto z$$.

• +1: Nice. ${}{}$ Apr 4 '20 at 3:10
• Very elegant ! (+1) Apr 4 '20 at 3:17

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

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.

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 $$t>-1$$.