Find the minimum value of $xe^x$ without calculus Recently I have been interested in finding minimum value of various functions without calculus, as a challenge.
How can we find the minimum value of $xe^x$ without calculus? Assume the limit or series definition of $e$.
If you use Calculus, you would differentiate $f(x) = xe^x$ and find that the only stationary point is $x = -1$. Since
$$ f(-1) = e^{-1}, \ \ \ \lim_{x\to -\infty} f(x) = 0, \ \ \ \lim_{x\to +\infty} f(x) = +\infty,$$
we conclude that $e^{-1}$ is the minimum of $xe^x$.
However, I am not able to find the stationary point of this function without calculus.
 A: This answer tries to find the minimizer of $xe^x$.
Obviously the minimum is attained when $x<0$. Substituting $u=-x$, it suffices to minimize $-ue^{-u}$ when $u>0$, or equivalently maximize $ue^{-u}$.
Hopefully you know $\ln$ is monotonic (which should follow by monotonicity of $e^x$ if you define $\ln x$ as the inverse of $e^x$), so it suffices to maximize $\ln(ue^{-u})=\ln u-u$ for $u>0$.
We know (see below) $e^t\ge1+t$ for all real $t$ (equality at $t=0$), so substituting $t=\ln u$, we get $u\ge 1+\ln u$, i.e. $\ln u-u\le-1$ with equality at $u=1$. This immediately implies the minimum of $xe^{-x}$ is $-e^{-1}$, at $x=-1$.
EDIT: Here's a proof of $e^t\ge 1+t$ without calculus, expanded from this answer. First, note Bernoulli's inequality $(1+x)^n\ge 1+nx$ (where $n>0,n\in\mathbb Z,x\ge-1$) can be proven without calculus (only induction). Now for any fixed $x$, $\lim_{n\to\infty}\frac xn=0$, so when $n$ is sufficiently large, $\frac xn>-1$. Hence for all but finitely many $n$, $(1+\frac xn)^n\ge1+x$. Taking the limit $n\to\infty$, we now have $e^x\ge1+x$ as limit preserves non-strict inequalities (and ignores finitely many starting terms).
