Prove that ${\sqrt{2}^{(\sqrt{3}^\sqrt{3})}}$ is greater than ${\sqrt{3}^{(\sqrt{2}^\sqrt{2})}}$ I have already met this type of problem with $e^\pi$ $>$ $\pi^e$. I also wondered if someone would please explain when $a^b$ $>$ $b^a$ if $b>a$ please. It is clear that the "powers dominate over bases rule" is not a rigorous rule at all.
Thank you. 
 A: I can answer your $a^b$ versus $b^a$ question.  Take $a,b$ positive:
$a^b > b^a$ if $\frac{a^b}{b^a} >1$.
We can work with that by turning all exponentials into base $e$:
$$
\frac{a^b}{b^a} = \frac{e^{b\ln a}}{e^{a\ln b}}=e^{b\ln a - a\ln b}>1\\
\implies b\ln a - a \ln b >0 \implies b\ln a > a \ln b \\
\implies \frac{\ln a}{a} > \frac{\ln b}{b}  
$$
Now the function $f(x) = \frac{\ln x}{x}$ is an increasing function for $x<e$, and a decreasing function for $x>e$.  $f(x) = 0$ at $x=1$ and $\lim_{x \to \infty}f(x) = 0$.  That said, we have:


*

*If  $b>a>e$ then $\frac{\ln a}{a} > \frac{\ln b}{b}$ and $a^b > b^a$. For example, $3^4 > 4^3$.

*If $a<b<e$ then $\frac{\ln a}{a} < \frac{\ln b}{b}$ and $a^b < b^a$. For example, $2^{2.5} < (2.5)^2$.

*If $a<1<b$ then $\frac{\ln a}{a} < \frac{\ln b}{b}$ and $a^b < b^a$. For example, $\left(\frac12\right)^2 < 2^\frac12$.

*If $1 < a < e < b$ we cannot say much about the relative values of $a^b$ and $b^a$.

Your title question is much more interesting; I have not found a good reason why 
$$
\sqrt{2}^\left( \sqrt{3}^\sqrt{3}\right) > \sqrt{3}^\left( \sqrt{2}^\sqrt{2}\right) 
$$
although I have squeezed that into 
$g(3) > g(2)$
with $$g(x) = \frac12\sqrt{x}\ln x - \ln (\ln x)$$
The difficulty is that while $g(x)$ forms a concave upward curve with a minimum at about 2.4, the values $2$ and $3$ are on opposite sides of the minimum, kind of analogous to the  $1 < a < e < b$ case in the above answer.  So I don;t see why $g(x)$ is greater at $2$ than at $3$.
I hope this question is left open and active until somebody comes up with a good reason or a reference in the literature.  
A: 
EDIT: As pointed out by Mark Fischler in the comments, I made a miscalculation and the answer below is not correct.

This is an answer to the title question of why $\sqrt{2}^{({\sqrt{3}}^{\sqrt{3}})}>\sqrt{3}^{({\sqrt{2}}^{\sqrt{2}})}$. Consider the function
$$f(x,y)=\ln\left(x^{\displaystyle\left(y^y\right)}\right)=y^y\ln(x)$$
Let $p=(\sqrt{2},\sqrt{3})$ and $q=(\sqrt{3},\sqrt{2})$. We wish to determine whether or not $f(p)>f(q)$. To that end, consider the line segment $L$ joining them, and $g:[0,1]\longrightarrow \mathbb{R}$ given by
$$g(t)=f(t\cdot p + (1-t)\cdot q)$$
that is, consider the function $f$ along $L$. Notice that $g(0)=f(q)$ and $g(1)=f(p)$. We will show that $g'(t)$ is always positive, which implies that $g$ is strictly increasing, proving the result.
Indeed, let
\begin{align}
a(t)=t\sqrt{2}+(1-t)\sqrt{3}\\
b(t)=t\sqrt{3}+(1-t)\sqrt{2}
\end{align}
so that $a'(t)=-b'(t)=\sqrt{2}-\sqrt{3}<0$ for all $t \in [0,1]$. Then we have that
$$g(t)={b(t)}^{b(t)}\cdot\ln(a(t))$$
and hence
$$g'(t)=\frac{b'}{a(t)}\cdot\left({b(t)}^{b(t)}\cdot\Big(a'+a(t)\cdot\ln(a(t))\Big)\cdot (\ln(b(t)+1)\right)$$
Since $\frac{b'}{a(t)}>0$ for all $t \in [0,1]$, we need only prove the other term is always positive. Now, ${b(t)}^{b(t)}$ and $(\ln(b(t)+1)$ are also always positive, so it suffices to prove that
$$h(t)=a'+a(t)\cdot\ln(a(t))$$
is always positive. The derivative of $h$ is
$$h'(t)=a'\cdot(\ln(a(t))+1)$$
which is always negative (because $a'<0$), so $h$ is strictly decreasing. On the other hand, we have that:
\begin{align}
h(1)&=(\sqrt{2}-\sqrt{3})+\sqrt{2}\cdot(\ln(\sqrt{2})+1)\\
&=\underbrace{(2\sqrt{2}-\sqrt{3})}_{>0}+\underbrace{\sqrt{2}\cdot\ln(\sqrt{2})}_{>0}
\end{align}
So that $h(t) \geq h(1)>0$ for all $t$, which concludes the proof.
