Which is greater, $\left(\frac{e}{2}\right)^\sqrt{3}$ or $(\sqrt{2})^{\pi/2}$? (no calculators)

Question

From a math contest in 1985:

Determine which of the following is greater: (no calculators)

$$\left(\frac{e}{2}\right)^\sqrt{3} \, \hspace{3mm} \text{or} \hspace{3mm} \, (\sqrt{2})^{\pi/2}$$

Hints are welcome, but I'm totally lost and would really appreciate a solution.

Answer 1

This comparison is potentially pretty tricky inasmuch as the numerical values are pretty close, within $2\%$:

$$\left(\frac{e}{2}\right)^{\sqrt{3}} = 1.701\ldots, \qquad (\sqrt 2)^{\pi / 2} = 1.723\ldots$$

The following method is pretty quick-and-dirty, and I doubt that it's the most elegant way. But it doesn't use anything more advanced than knowledge of a few well-known Taylor series, and it doesn't involve integers with more than two digits.

Hint Since $\log$ is an increasing function, we may as well take the logarithms of both sides and compare quantities with the awkward exponents in more tractable places, so we're comparing $\sqrt{3}(1 - \log 2)$ and $\frac{\pi}{4} \log 2$. Subtracting these and collecting terms in $\log 2$ gives $$\left(\frac{\pi}{4} + \sqrt{3}\right) \log 2 - \sqrt{3},$$ and the claim that $(\sqrt 2)^{\pi / 2} > \left(\frac{e}{2}\right)^{\sqrt{3}}$ is equivalent to showing that this quantity is positive. At this point we can see that this amounts to establishing sufficiently tight lower bounds for $\pi, \log 2$.

One efficient way to produce such bounds is to use the first terms of the Maclaurin series

$$\operatorname{arctanh} x \sim x + \frac{1}{3} x^3 + \cdots \qquad \textrm{and} \qquad \arcsin x \sim x + \frac{1}{6} x^3 + \cdots,$$ the identities $$\log 2 = \log\left(\frac{1 + \frac{1}{3}}{1 - \frac{1}{3}}\right) = 2 \operatorname{arctanh} \frac{1}{3} \qquad \textrm{and} \qquad \pi = 6 \arcsin \frac{1}{2},$$ and the facts that all of the coefficients of both of these series are nonnegative. These are good choices in part because both series converge relatively quickly at the indicated values.

Solution Using the first two terms of each series gives \begin{align*}\log 2 &= 2 \operatorname{arctanh} \frac{1}{3} > 2 \left[\left(\frac{1}{3}\right) + \frac{1}{3}\left(\frac{1}{3}\right)^3\right] = \frac{56}{81}, \\ \pi &= 6 \arcsin \frac{1}{2} > 6 \left[\left(\frac{1}{2}\right) + \frac{1}{6}\left(\frac{1}{2}\right)^3\right] = \frac{25}{8} ,\end{align*} giving the bound $$\left(\frac{\pi}{4} + \sqrt{3}\right) \log 2 - \sqrt{3} > \left[\frac{1}{4}\left(\frac{25}{8}\right) + \sqrt{3}\right] \frac{56}{81} - \sqrt{3} = \frac{25}{4 \cdot 81}(7 - 4 \sqrt{3}).$$ So, it suffices to show that $7 > 4 \sqrt{3}$, but squaring both sides shows that this is equivalent to $49 > 48$.

Answer 2

Edit: used same strategy correctly but had to take a $\log_2$ at the end to get the answer.

Assume $(\frac{e}{2})^{\sqrt3} = (\sqrt2)^{\frac{\pi}{2}}$

Square both sides

$$(\frac{e}{2})^{2\sqrt3} = (2)^{\frac{\pi}{2}}$$

Raise both sides $\sqrt3$

$$(\frac{e}{2})^6 = (2)^{\frac{\pi\cdot \sqrt3}{2}}$$

Multiply both sides by $2^6$

$$e^6 = (2)^{\frac{\pi\cdot \sqrt3+12}{2}}$$

Take the $6th$ root of both sides

$$e = 2^{\frac{\pi\cdot \sqrt3+12}{12}}$$

Taking $\log_2$ both sides

$$\frac{\log(e)}{\log(2)} = {\frac{\pi\cdot \sqrt3+12}{12}} = \frac{3.14\cdot 1.73+12}{12}= 1.45$$

Then $1.44 < 1.45$

So $(\frac{e}{2})^{\sqrt3} < (\sqrt2)^{\frac{\pi}{2}}$

Answer 3

Both $\frac{\pi}{4}$ and $\log(2)$ can be simply approximated in terms of Beuker integrals:

$$ \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}=\frac{22}{7}-\pi,\qquad \int_{0}^{1}\frac{x^3(1-x)^3}{1+x}\,dx = 16\log(2)-\frac{621}{56} $$ hence in order to prove that $$ \frac{\pi}{4}\log(2) > \sqrt{3}(1-\log 2), $$ i.e. that $\left(\sqrt{2}\right)^{\pi/2}>\left(\frac{e}{2}\right)^{\sqrt{3}}$, it is enough to show that $$ \left(\frac{22}{7\cdot 4}\cdot\frac{621}{896}\right)^2-3\cdot\left(1-\frac{621}{896}\right)^2>\frac{1}{100} $$ since the relative error of both the approximations $\pi\approx\frac{22}{7},\log(2)\approx \frac{621}{896}$ is less than five parts in ten thousands. The last inequality is tedious to check but utterly straightforward.

A tighter inequality can be derived through Cauchy-Schwarz: $$ \int_{0}^{1}\frac{\log(1+x)}{(1+x^2)^2}\,dx \int_{0}^{1}\log(1+x)\,dx \geq \left(\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\,dx\right)^2 $$ leads to $$ 4\pi-24\log(2)-12\pi\log(2)+48\log^2(2)+8\pi\log^2(2)-\pi^2\log^2(2)\geq 0.$$