Proof $e^x - 1 \le e x$ for $x \in [0, 1]$ In reading a recent paper, I came across the inequality: $e^x - 1 \le e x$ for $x \in [0, 1]$.
I tried to prove this using (the reverse) Bernoulli's inequality i.e. $(1 + y)^r \le 1 + ry$, for $r \in [0, 1], y \in [-1, \infty)$, per here. Here using substitutions $y = e - 1 \ge -1$ and $r = x \in [0, 1]$.
This yielded:
\begin{align*}
(1 + (e - 1))^x &\le 1 + x(e - 1) \\
\implies e^x &\le ex + (1 - x) \\
\iff e^x - 1 &\le ex - x \\
\implies e^x - 1 &\le ex \quad \text{since $x \in [0, 1] \Leftrightarrow - x \in [-1, 0]$}
\end{align*}
2 questions: First, are there any issues with the above proof?
Second, can anyone suggest another way to prove this? There are
many such lemmas used in this paper, so thought to check.
 A: For any $x\in[0,1]$ we have: $$(1+ex-e^x)'=e-e^x\geq0$$
Thus, $$1+ex-e^x\geq1+e\cdot0-e^0=0.$$
You can use Bernoulli for the following.
$$(1+ex)^{\frac{1}{x}}\geq e$$ but to check before the original inequality for $x=0.$
A: For $0\le x\le1$ we have
$$e^x-1=\int_0^xe^t\,dt\le\int_0^xe\,dt=ex$$
since $e^t$ is an increasing function of $t$ (and $x\le1$).
A: Here is a proof using mean value theorem. The inequality is trivial for $x=0$. For $0 < x \leq 1, $ define
$$f: [0,x] \to \Bbb{R}:  y \mapsto ey - e^y +1$$
Then $f'(y) = e-e^y \geq 0$ for all $y \in [0,x]\subseteq [0,1]$.  By the mean value theorem, there is $\xi\in (0,x)$ with
$$\frac{f(x)-f(0)}{x-0}= f'(\xi) \geq 0$$
But then $$0 \le\frac{f(x)-f(0)}{x-0}= \frac{ex-e^x+1}{x}$$
and thus $0 \leq ex-e^x+1 \iff  e^x-1 \leq ex$ as desired.
A: The perhaps most versatile inequality about the exponential is
$$\tag1 e^t\ge 1+t$$
for all $t\in \Bbb R$.
For $x\ge0$ we can apply $\int_0^x\mathrm dx$ to both sides and obtain
$$\tag2 e^x\ge 1+x+\frac12x^2.$$
From this,
Hence (at least for $0\le x\le 1$)
$$\begin{align} (1+\tfrac32x)e^{-x}&\ge (1+\tfrac32x)(1-x+\tfrac12x^2)\\
&=1+\frac12x-x^2+\frac34x^3\\&=1+\frac x4(2(1-x)^2+x^2)\\&\ge1.\end{align}$$
By multiplication with $e^x$ ($>0$) and rearranging,
$$e^x\le 1+\frac32x.$$
Your lemma then follows from $e>\frac32$.
