What is a nice way to prove that : $\frac{t}{t+1} \le 1-e^{-t}\le \frac{2t}{1+t}$ I am trying to prove that for $t>0$$$\frac{t}{t+1} \le 1-e^{-t}\le \frac{2t}{1+t}$$
I know that Simplest or nicest proof that $1+x \le e^x$
taking $$e^{-x} \le \frac{1}{1+x} \implies 1-e^{-x}\ge \frac{x}{x+1}$$
Now I don't know to prove the other inequality.
 A: $f(t) = (1+t)e^{-t} + t$ 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 $t+1.$
A: 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
$$\boxed{ \forall t>0,\qquad \frac{t}{t+1}\leq 1-e^{-t} \leq\color{red}{\frac{4}{3}}\cdot\frac{t}{t+1}.} $$
A: You already proved the first part, 
$$ 1-e^{-t}\ge\frac{t}{t+1}. \tag{1}$$
For the second part, noting that, for $t\ge0$,
$$ e^t=1+t+\frac12t^2+\cdots\ge\frac12+t+\frac12t^2=\frac12(1+t)^2$$
one has
$$ e^{-t}\le \frac{2}{(1+t)^2}. $$
Integrating from $0$ to $t$, one has
$$ \int_0^te^{-s}ds\le \int_0^t\frac{2}{(1+s)^2}dt $$
or
$$ 1-e^{-t}\le\frac{2t}{t+1}. \tag{2}$$
From (1) and (2), one has
$$ \frac{t}{t+1}\le1-e^{-t}\le\frac{2t}{t+1}. $$
A: The second inequality is trivial if $t \geqslant 1$, so we may
assume that $0 < t < 1$. In fact, we only assume $0 < t < 2$.
Put $u = t/2$, so that $0 < u < 1$. The curve $y = \tfrac{1}{x}$
lies above the tangent at $(1, 1)$ and below the secant from
$(1-u, \tfrac{1}{1-u})$ to $(1+u, \tfrac{1}{1+u})$, therefore,
calculating the areas of two trapezia,
$$
2u < \int_{1-u}^{1+u}\frac{dx}{x} < \frac{2u}{1-u^2}.
$$
Using the first of these inequalities (the second was only mentioned
for interest),
\begin{gather*}
t < \log\left(1+\frac{t}{2}\right)-\log\left(1-\frac{t}{2}\right) =
\log\frac{2+t}{2-t},
\\\text{i.e.}\quad
e^t < \frac{2+t}{2-t},
\quad\text{i.e.}\quad
1 - e^{-t} < \frac{2t}{2 + t},
\end{gather*}
which is stronger than the required inequality (and is trivially
true when $t \geqslant 2$, so holds for all $t > 0$).
A: $\frac{t}{t+1} 
\le 1-e^{-t}\le 
\frac{2t}{1+t}
$
is the same as
$t
\le (1+t)(1-e^{-t})
\le 2t
$.
Since
$(1+t)(1-e^{-t})
=1+t-(1+t)e^{-t}
$,
this is the same as
$0
\le 1-(1+t)e^{-t}
\le t
$.
The left side is
$(1+t)e^{-t}
\le 1
$
or
$1+t
\le e^t$
which is well known.
The right side is
$1-(1+t)e^{-t}
\le t
$
or
$1-t
\le (1+t)e^{-t}
$.
If $t \ge 1$,
this is obviously true.
If $0 \le t 
\lt 1$,
this is
$e^t
\le \dfrac{1+t}{1-t}
$
and even more than this is true
because
$e^t \le \dfrac1{1-t}
$
for $0 \le t \lt 1$
by comparing terms in the power series
($\dfrac1{k!} \le 1$).
