Prove that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator.
I did in the following way. Are there other ways?
Proof : Let $f(x)=e\pi\frac{\ln x}{x}$. Then,
$$e^{\pi}-{\pi}^e=e^{f(e)}-{e}^{f(\pi)}\tag1$$
Now,
$$f'(x)=\frac{e\pi(1-\ln x)}{x^2},\quad f''(x)=\frac{e\pi (2\ln x-3)}{x^3},\quad f'''(x)=\frac{e\pi (11-6\ln x)}{x^4}.$$
Since $f'(x)\lt 0$ for $e\lt x\lt\pi$, one has $f(e)\gt f(\pi)$. By Taylor's theorem, there exists a point $c$ in $(e,\pi)$ such that
$$\begin{align}f(\pi)&=f(e)+(\pi-e)f'(e)+\frac{(\pi-e)^2}{2}f''(c)\\&\gt f(e)+(\pi-e)\cdot 0+\frac{(\pi-e)^2}{2}\cdot \frac{e\pi(2\ln e-3)}{e^3}\\&=f(e)-\frac{\pi(\pi-e)^2}{2e^2}\tag2\end{align}$$
because $f'''(x)\gt 0\ (e\lt x\lt \pi)$ implies $f''(c)\gt f''(e)$.
By the mean value theorem and $(2)$,
$$e^{f(e)}-e^{f(\pi)}\lt (f(e)-f(\pi))e^{f(e)}\lt \frac{\pi (\pi-e)^2}{2e^2}\cdot e^{\pi}=\frac{e\pi(\pi-e)^2}{2}\cdot e^{\pi-3}\tag3$$
Since $e^x\lt \frac{1}{1-x}\ (0\lt x\lt 1)$ and $0\lt \pi-3\lt 1$,
$$e^{\pi-3}\lt\frac{1}{4-\pi}\tag4$$
From $(1)(3)(4)$,
$$e^{\pi}-{\pi}^e\lt \frac{e\pi(\pi-e)^2}{2}\cdot\frac{1}{4-\pi}\lt\frac{3\times\frac{22}{7}\left(\frac{22}{7}-2.718\right)^2}{2}\cdot\frac{1}{4-\frac{22}{7}}\lt 0.993\lt 1$$
Best Answer
Consider $f(x,y) = x^y - y^x$ where $x, y \approx 3$. Increasing $x$ should bring $f$ down and increasing $y$ should bring $f$ up. In general, without any real knowledge of $f$:
$$ 1 > f(x+ \epsilon_1, y + \epsilon_2) \approx f(x, y ) + \epsilon_1 \frac{\partial f}{\partial x} + \epsilon_2 \frac{\partial f}{\partial y} $$
where the first inequality is something we are trying to prove for $x = e \approx 2.718$ and $y = \pi \approx 3.141$.
Let's try $x = \frac{11}{4}$ and $y = \frac{13}{4}$. Then we have taken off too much slack:
$$ \left( \frac{11}{4} \right)^{\frac{13}{4}} - \left( \frac{13}{4} \right)^{\frac{11}{4}} \approx 1.214$$
Using the continued fractions of $e$ and $\pi $ does give less than one and check with a calculator:
$$ \left( \frac{8}{3} \right)^{\frac{22}{7}} - \left( \frac{22}{7} \right)^{\frac{8}{3}} \approx 0.621$$
Even your proof requires a calculator in the last step... but we did not use the exact values of $\pi, e$.
How good is this approximation? Somehow we should compute how quickly $f$ changes with $x$ and $y$:
$$ \frac{\partial f}{\partial x} = y \, x^{y-1} - (\ln y ) y^x \approx 3 \times 3^{3-1} - (\ln 3) 3^3 \approx -27(\ln \frac{e}{3}) \approx -2.7 $$
The $y$ partial derivative is similar, except it is positive instead of negative.
It can be proven, the continued fraction error of $e$ and $\pi$ are inversely proportional to the denominators:
$$ \left| e - \frac{8}{3} \right| < \frac{1}{3^2} \text{ and } \left| \pi - \frac{22}{7} \right| < \frac{1}{7^2}$$
These would be our estimates of $\epsilon_1$ and $\epsilon_2$. Then using a tiny bit of multivariable calculus:
$$ \epsilon_1 \frac{\partial f}{\partial x} + \epsilon_2 \frac{\partial f}{\partial y} \approx \frac{1}{3^2} \times 2.7 + \frac{1}{7^2} \times (-2.7) < 0.355 $$
This should be enough to establish $|f(\pi,e)| < 1$.
This used a calculator in many places but not the exact value of the constants $\pi$ and $e$.