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This basic bound on $e^x$ occasionally pops up in my reading as a relatively tight bound for small positive values of $e^x$, but I have been unable to prove it/find a proof of it. One paper I read says that it follows from "elementary calculus," but I have not yet seen how.

What I've tried:

Near cousins like $1 + x \le e^x$ are easy to prove by differentiating the difference $f(x) = e^x - (1 + x)$ and showing that it has a global minimum at $f(0) = 0$.

When I try this approach for $e^x \le 1 + x + x^2$ (when $x <$ some constant $C$ that presumably works out to 1.79 or thereabouts), however, I arrive at

$$f'(x) = 1 + 2x - e^x = 0,$$

which I suppose I could solve with Lambert's W—but I was expecting something cleaner.

This question shows how to prove a similar inequality whose critical points also lead to a product logarithm expression (namely $e^x > 1 + x^2$) by carefully inspecting the second derivative. But that solution doesn't seem to apply here.

Edit:

This question shows that $e^x \le 1 + x + x^2$ for all $x < ln(2)$, which helps. An answer there solves $f'(x) = 0$ by guessing $x=0$ and showing that it holds (which is cleaner than using Lambert's W).

The puzzle I still can't solve is how to show that the bound holds more generally, up to $x < 1.79$.

I can see that this is true by plotting $f(x)$ in a graphing calculator. Showing that it holds for the portion of the graph where $f(x)$ is positive and $x > ln(2)$ and (thus) $f''(x)$ is negative still has me stumped!

Graph of f(x) and its derivatives

SigmaX
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    Check this: https://math.stackexchange.com/q/1357437/42969 – Martin R Apr 13 '22 at 15:05
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    @Snaw: I hesitated to close this as a duplicate because the interval is different ($x < 1.79$ vs $x< \ln(2)$). It could be that the same estimates work on the larger interval, I did not check it. – Martin R Apr 13 '22 at 15:09
  • @MartinR: Thanks! That gets me part way there—question edited. – SigmaX Apr 13 '22 at 15:51

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$f''(x)$ is positive for $x<\ln 2$ and negative for $x>\ln 2$. So $f'(x)$ is [strictly] increasing for $x<\ln 2$ and [strictly] decreasing for $x>\ln 2$.

Now it is easy to see $f'(0)=0$ and $f'$ has a root $x_1$ between $1$ and $2$ (because it changes the sign).

So $f(0)=0$, $f$ is [strictly] increasing until $x_1$ and then [strictly] decreasing. If $x_0\approx 1.79$ is the second root of $f$, then it is easy to see that $f(x)\ge 0$ on the interval $[0,x_0]$.

But of course, there is no closed form for $x_0$, so you cannot completely get rid of numerical approximations.

Momo
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  • Great analysis. Yeah, I've been wondering where that value of $1.79$ comes from. Perhaps I need to be content with proving the bound for $x < \ln(2)$ in closed form, and accept that an approximation is unavoidable for the higher values :-/. – SigmaX Apr 13 '22 at 15:59
  • @SigmaX If you assume/prove $e<2.8$, you can easily prove by hand the bound for $x<\frac{3}{2}$ by showing that $f\left(\frac{3}{2}\right)>0$ – Momo Apr 13 '22 at 16:13