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. 
 A: 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 $1+t\mapsto z$.
A: 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 $x \in (-1,\infty)$.
A: 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.
A: 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 $t>-1$.
