What is a nice way to prove that : $\frac{t}{t+1} \le 1-e^{-t}\le \frac{2t}{1+t}$

Solution 1:

$f(t) = (1+t)e^{-t} + t$ is strictly increasing since $f'(t) =1-te^{-t} =e^{-t}(e^{t}-t)>0$ for all $t > 0$. This holds as $t < 1 + t \le e^t$ .

Therefore, $$1=f(0)\le f(t) =(1+t)e^{-t} + t $$ for all $t>0$ which, together with $(1+t)e^{-t}\le 1$ since $1+t \le e^t$, implies that $$0\le 1-(1+t)e^{-t} \le t$$

so adding $t$ on both sides, we get $$t\le (1+t)(1-e^{-t})\le 2t.$$

As $1+t-e^{-t}-te^{-t}=(1+t)(1-e^{-t})$, the result follows on dividing the last inequality by $t+1.$

Solution 2:

If we set $$ g(t) = (t+1)\frac{1-e^{-t}}{t} $$ we have that $\lim_{t\to 0^+}g(t)=\lim_{t\to +\infty}g(t)=1$ and $g(t)\geq 1$ for any $t>0$. $g(t)$ attains its maximum value close to $t=2$, and such maximum value is way less than $2$, it is around $1.3$.

Indeed, the maximum value of $g(t)$ is attained at the only positive solution of $e^{-t}=\frac{1}{1+t+t^2}$.
If we introduce $h(t)=(t+1)\frac{1-\frac{1}{1+t+t^2}}{t}=1+\frac{t}{1+t+t^2}$ we may easily check that $h(t)\leq \frac{4}{3}$.

It follows that $$\boxed{ \forall t>0,\qquad \frac{t}{t+1}\leq 1-e^{-t} \leq\color{red}{\frac{4}{3}}\cdot\frac{t}{t+1}.} $$