As the title implies: what is bigger $\sqrt2^{\sqrt3^\sqrt3}$ or $\sqrt3^{\sqrt2^\sqrt2}$. Specifically I am interested in working this out without actually calculating the values. So far I have tried applying order preserving operations on both and seeing if the comparison will become clearer but this has so far been unyieldy because I am stuck at the following point:
$\sqrt2^{\sqrt3^\sqrt3}$ or $\sqrt3^{\sqrt2^\sqrt2}$
$e^{\sqrt3^\sqrt3\ln\sqrt2}$ or $e^{\sqrt2^\sqrt2\ln\sqrt3}$
${\sqrt3^\sqrt3\ln\sqrt2}$ or ${\sqrt2^\sqrt2\ln\sqrt3}$
${\sqrt3^\sqrt3\ln2}$ or ${\sqrt2^\sqrt2\ln3}$
And at this point I have explored a few options but nothing has made it clear. Have I been pursuing the correct root (if you pardon the pun) and how should I proceed.
Update:
$\ln({\sqrt3^\sqrt3\ln2})$ or $\ln({\sqrt2^\sqrt2\ln3})$
$\frac{\sqrt3}{2}\ln3 +\ln({\ln2})$ or $\frac{\sqrt2}{2}\ln2 +\ln({\ln3})$
Update $2$: At this point it seems more approapriate to deal with the comparison as in inequality
The question is now if the following statement is correct: $\frac{\sqrt3}{2}\ln3 +\ln({\ln2}) > \frac{\sqrt2}{2}\ln2 +\ln({\ln3})$
$\sqrt3 \ln3 +2\ln({\ln2}) > \sqrt2\ln2 +2\ln({\ln3})$
$\sqrt3 \ln3 - 2\ln({\ln3}) > \sqrt2\ln2 - 2\ln({\ln2})$
To check this I look at $f(x)= \sqrt{x} {\ln(x)} - 2\ln(\ln(x))$
$f'(x)=\frac{1}{2\sqrt x}\ln x +\frac{1}{\sqrt x}-\frac{2}{x\ln x} $ Setting this to $0$ gives an equation I do not know how to solve and would be curious to know how to solve it (other than through numerical methods) if that is even possible.
Experimenting on my calculator suggests there is a point where this derivative is 0 at around 2.4 i.e between 2 and 3. Therefore this is inconclusive in determining which side is larger and thus whether the inequality is correct. Where do we go from here?
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.