Prove that $ex \leq e^x$ for all $x \in \mathbb{R}$ This is easy to prove for negative $x$ but what about positive $x$? Should I use MVT?
 A: You should prove that:
$$0 \leq e^{x-1}-x=f(x)$$
Let's calculate derivative of $f(x)=e^{x-1}-x$, it's:
$$f'(x)=e^{x-1}-1$$
Note that $f'(x)<0$ for $x<1$ and $f'(x)>0$ for $x>1$. So $f(x)$ reaches the smallest value at $x=1$, but we can calculate that $f(1)=0$, so
$$f(x) \geq f(1)=0$$
for all $x \in \mathbb{R}$.
A: This inequality is equivalent to $e^{x-1}\ge x$, which, after substitution $y:=x-1$ is equivalent to
$$e^y\ge 1+y.$$
You can find many proofs for this one:


*

*Simplest or nicest proof that $1+x \le e^x$

*Prove that $e^x\ge x+1$ for all real $x$

*Proof of $e^x - 1 \geq x$ for ${x: -1 \leq x < 0}$
A: Define
$$f(x):=ex-e^x\implies f'(x)=e-e^x=e(1-e^{x-1})=0\iff x=1$$
But also
$$f''(1)=-e^1<0$$
so that at $\;x=1\;$ we have a maximum point... End the argument now.
A: Use the convexity of $\exp$ at the point 1 to get:
$$
\exp x \ge \exp' 1 \times (x - 1) + \exp 1  = ex
$$
A: Note the tangent to $y=e^x$ at $x=1$ is
$$ y-e = e(x-1), $$
and the Taylor series for $e^x$ about $x=1$, with remainder, is
$$ e^x -ex = \int_1^{x} (x-t)e^{t} \, dt, $$
and the integrand is clearly positive for $x>1$. For $x<1$, you can rearrange integrand and limits to show the integral is still positive.
