What is bigger: $\sqrt2^{\sqrt3^\sqrt3}$ or $\sqrt3^{\sqrt2^\sqrt2}$?

Solution 1:

Here is an outline of an approach that should work with enough diligence and without calculator. The inequality is equivalent to comparing: $$\dfrac{3^{\sqrt{3}}}{2^{\sqrt{2}}}\,\, ?\,\, \dfrac{\ln^2 3}{\ln^2 2}.$$ For, $\ln 2:$ $$\ln 2 = 1-\frac 12+\frac 13 - \frac 14 + \frac 15-...$$ so you can get rational lower and upper bounds with arbitrary precision by hand. For $\ln 3:$ $$\ln 3 = -\ln\frac 13 = -\ln\left(1-\frac 23\right) = \frac 23+\frac 29+\frac{8}{81}+...$$

For the square roots exponents, one can again use Taylor series: $$(1+x)^{\frac 12} = \sum_{n=0}^\infty \binom{\frac 12}{n}x^n,$$ which again yields rational approximation for $\sqrt{2}$ directly. For $\sqrt{3}$, just rewrite it as: $$\sqrt{3} = 2\sqrt{1-\frac 14} = 2\left(1 - \sum_{n=0}^\infty\dfrac{2}{(n+1)2^{2n+1}}\binom{2n}{n}\right). $$ But this will most likely be ridiculous, when you get a fine enough, rational bounds for the exponents and I suspect it will take an hour or two meticulous, hand-checked algebra.