Show $ex \leq e^x$ for all $x \in \mathbb{R}$ So far all I have is this:
Let $f$ be a function where $f(x)=ex-e^x\leq 0$
$f'(x)=e-e^x \leq 0$, so $f$ is decreasing.
I'm stuck here. Can someone help me with the next steps?
 A: Hint: $e^x$ is convex, hence stays above its tangent at $x=1$.
A: First of all, $f'(x)$ isn't always less than $0$. It's less than/equal to $0$ at  $x=1$ however. At $x=1$, $f(x)=0$, and if it's decreasing it will always be less than $0$. For $x<1$, it shouldn't be hard to show.
A: Let
$f(x) := e^{x} - ex$ for all $x \in \mathbb{R}.$ We claim that $f \geq 0$ on $\mathbb{R}.$
Note that $f(1) = 0,$
that
$f'(x) > 0$ for all $x > 1$,
and that $f'(x) < 0$ for all $x < 1.$
Thus $f$ is increasing on $]1, +\infty[$ so that $f > 0$ on $]1, +\infty[.$
Since $f$ is decreasing on $]-\infty, 1[$
and is continuous on $\mathbb{R},$ it follows that $f > 0$ on $]-\infty, 1[.$
Including the case where $f(1) = 0,$ we thus have $f \geq 0$ on $\mathbb{R}.$
A: There are $3$ cases to consider:
$a):$ $x < 0 \Rightarrow ex < 0 < e^x \Rightarrow ex < e^x$.
$b):$ $ 0 \leq x \leq 1$, $f(x) = ex - e^x \to f'(x) = e - e^x \geq 0 \to f(x) \leq f(1) = e - e = 0 \Rightarrow ex - e^x \leq 0 \Rightarrow ex \leq e^x$.
$c): $ $ 1 < x $: $f(x) = ex - e^x \to f'(x) = e - e^x < 0 \to f(x) < f(1) = e - e = 0 \to ex < e^x$.
