Proving $\pi \gt e+\frac{1}{e} \gt \pi-\frac{1}{\pi} \gt e$

Question

I created this problem for myself as a fun exercise. I want to prove the following statement:

$$\pi \gt e+\dfrac{1}{e} \gt \pi-\dfrac{1}{\pi} \gt e$$

I found that the following upper/lower bounds for $e$ and $\pi$ are "good enough" to establish the above statement as true:

$$\dfrac{30}{11} \gt e \gt \dfrac{8}{3}$$

$$\dfrac{13}{4} \gt \pi \gt \dfrac{25}{8}$$

The upper/lower bounds for $e$ are easily proved by considering the series representation of $e^x$, and calculating partial sums for $x=1$ and $x=-1$.

However, I'm at a loss for how to establish the upper/lower bounds for $\pi$. I could approach it like Archimedes and use inscribed/circumscribed polygons (I believe it requires at least a $10$-gon and $18$-gon in this case). Is there an easier way to get these upper/lower bounds on $\pi$?

EDIT:

I've also included the "alternative-proofs" tag because I am open to proofs of any kind, especially those which are elegant or particularly simple, and don't require knowledge of $e$ or $\pi$ to high precision.

Answer 1

The approximations $ 3.14 < \pi < 3.34 $ and $ 2.70 < e < 2.76 $ suffice when doing the computation with two decimals.

Answer 2

I have a promising idea but still have to work the details.
There is a class of integrals that connect $\pi$ and $e$ that come from: $$ \int_{0}^{+\infty}\frac{\cos(x)}{1+x^2}=\frac{\pi}{2e}\tag{1} $$

Now we may apply integration by parts multiple times, reaching: $$ \frac{\pi}{e} = \int_{0}^{+\infty}\frac{p(x)(1-\cos x)}{(1+x^2)^k}\,dx \tag{2}$$ with $p(x)$ being some polynomial, then apply the Cauchy-Schwarz inequality to the RHS of $(2)$: $$\begin{eqnarray*} \frac{\pi^2}{e^2}&=&\left(\int_{0}^{+\infty}\frac{p(x)(1-\cos x)}{(1+x^2)^{k}}\,dx\right)^2\\&\leq& \left(\int_{0}^{+\infty}\frac{p(x)(1-\cos x)}{(1+x^2)^{k-1}}\,dx\right)\cdot\left(\int_{0}^{+\infty}\frac{p(x)(1-\cos x)}{(1+x^2)^{k+1}}\,dx\right)\tag{3}\end{eqnarray*}$$ leading to a product of two integrals that can still be evaluated in terms of $\pi$ and $e$.

That should give arbitrarily accurate approximations for the ratio $\frac{\pi}{e}$ and prove all the wanted inequalities. As I said, I will keep working on this approach. We also have: $$ \int_{0}^{+\infty}\frac{\sin x}{x(1+x^2)}\,dx = \frac{\pi(e-1)}{2e}$$ where the integrand function in the LHS is waiting to be decomposed with respect to an orthogonal base of $L^2(\mathbb{R}^+)$ - I just have to understand with respect to which inner product.