The inequality $$\boxed{e^3 > 20}$$ is occasionally useful, including in the answer I wrote for this question that comes from a GRE subject exam.
This bound is relatively tight: $$e^3 = 20.08553\!\ldots ,$$ a relative error of $< \frac{1}{200}$, which means establishing the inequality might be a little delicate. In a comment under the linked answer, TheSimpliFire posed the following natural question:
What is an efficient way to prove the inequality $e^3 > 20$ by hand?
(I would have guessed that this had been asked before, but neither the internal search nor searchonmath turned up any duplicates.)
A naive method is to use the series truncation
$$e = \sum_{k = 0}^\infty \frac{1}{k!} > 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} = \frac{120 + 120 + 60 + 20 + 5 + 1}{160} = \frac{163}{60} .$$
Then, it suffices to prove that $\left(\frac{163}{60}\right)^3 > 20$, which is equivalent to $4\,330\,747 > 4\,320\,000$. This last step could even be outsourced to an enthusiastic primary school student, but it involves cubing a three-digit prime and so is slightly tedious.
One might try to refine this method by looking for rationals that are easier to cube, but the only rational numbers satisfying $\sqrt[3]{20} < q < e$ with denominator $< 60$ are $\frac{106}{39}, \frac{125}{46}, \frac{144}{53}$. It's again straightforward to show that the cube of any of these $> 20$, but doing so is no faster than cubing $\frac{163}{60}$ and one then has the additional burden of showing the number is $< e$.
One could also search for integrals analogous to the classic Dalzell integrals for the difference $e^3 – 20$ (or to the difference corresponding to some other inequality equivalent thereto), by which I mean evidently positive definite integrals equal to that difference.
For example, some experimentation yields the definite integral
\begin{align}
&\int_1^2 – \frac{(x – 1) (2 – x) p(x) \,dx}{20 x (x^2 + 1)} \\
&\qquad = \int_1^2 \left(-\frac{1}{2} x^3 + \frac{63}{20} x^2 – \frac{153}{20} x + 9 – \frac{3}{x} – \frac{2 x}{x^2 + 1} \right) dx \\
&\qquad = 3 – \log 20 ,
\end{align}
where $p(x) = 10 x^4 – 33 x^3 + 44 x^2 – 45 x + 30$. Computing gives that all of the coefficients of $p(x + 1)$ are positive, so $p$ is strictly positive for $x \geq 1$, and thus the integrand is strictly positive on $(1, 2)$. So, the integral is positive, that is, $3 > \log 20$, which is equivalent via exponentiation to $e^3 > 20$. This is again elementary, but not terribly fast.
Remark Incidentally this latter method lets us extract cheap but relatively sharp rational bounds on $\log 20$: Since $2 < x (x^2 + 1) < 10$ on the interval of integration, our integral is bounded by polynomial integrals:
$$\int_1^2 – \frac{(x – 1) (2 – x) p(x) \,dx}{20 \cdot 10} < \int_1^2 – \frac{(x – 1) (2 – x) p(x) \,dx}{20 x (x^2 + 1)} < \int_1^2 – \frac{(x – 1) (2 – x) p(x) \,dx}{20 \cdot 2} .$$
Integrating gives
$$\frac{163}{84000} < 3 – \log 20 < \frac{163}{16800},$$
and rearranging gives the bounds
$$2.99027\!\ldots = \frac{251185}{84000} < \log 20 < \frac{251837}{84000} = 2.99805\!\ldots .$$
Best Answer
Similar to your last proof, I found a positive function whose integral is $e^3-20$.
For $$f(x)=\frac{1}{186}(x-1)^2(x-2)^4e^x\ge0$$
we have $$\int_{0}^{3}f(x)dx=e^3-20$$