For $ a>b>1$, prove that $a^{b^a}>b^{a^b}$ Given that $a>b>1$, show that $a^{(b^a)}>b^{(a^b)}$.
I've proved that the case $b\geq e$ holds as follows:
$$\ln\ln a^{(b^a)}-\ln\ln b^{(a^b)}=a\ln b+\ln\ln a-b\ln a-\ln\ln b$$
Let $t=a-b>0$. Define function 
$$f(t)=(b+t)\ln b+\ln\ln (b+t)-b\ln(b+t)-\ln\ln b$$
By direct computation
$$f'(t)=\frac{g(t)}{(b+t)\ln(b+t)}$$
where $g(t)=\ln b [(b+t)\ln(b+t)]-b\ln(b+t)+1$. 
If $b\geq e$, then $g(t)\geq t\ln(b+t)+1>t+1>0$.
And it follows that $f(t)$ is increasing on $[0,\infty)$ and therefore $f(t)\geq f(0)=0$, for all $t>0$.
However, when $b$ is close to $1$, the function $f(t)$ may first increase, then decrease, and eventually increase. And it's hard to estimate the minimum. Having graphed it, I found that that there seemed to be no counterexample.
Is there any inequality or multivariate optimization techniques I can use here?
 A: EDITTED SOLUTION
Consider the function for $b \leq x \leq a$
$$f(x) = x \ln(a+b - x) + \ln(\ln x)$$
$$f'(x) = \ln(a+b - x) + \frac{x}{a+b -x} + \frac{1}{x \ln x}$$
This is an increasing function, as all terms are positive
Hence if $a > b$, then $f(a) > f(b)$
Now, we rewrite that as
$$a\ln b + \ln(\ln a) \gt b\ln a + \ln(\ln b)$$
This will simplify to our original inequality
A: This is from an old thread of de.sci.mathematik I participates, see https://groups.google.com/forum/#!searchin/de.sci.mathematik/hoppe/de.sci.mathematik/Ciz3I81R1Rs/sRJZSiAt1T4J.  It shows that for any two positive numbers $x$ and $y$ it follows that
$$x<y\Rightarrow x^{y^x}<y^{x^y}.$$
It took nearly three weeks in those times to solve the problem. Here we go!
Let $D\subset\mathbb R^2$ and $f\colon D\to\mathbb R$ a real-valued function defined on $D$.  Define $\colon D\to\mathbb R$ by 
$$h(x,y):=f\bigl(x,f(y,x)\bigr).$$
Then if $x<y$ it follows that $h(x,y)<h(y,x)$ if the following (nearly trivial) conditions hold:
(1) $f$ is strictly monotone in its first argument, that is, either $f(y,x)<f(x,y)$ or $f(y,x)>f(x,y)$.
(2) There is a real number $a$ such that $h(x,y)<a<h(y,x)$.
Now $f(x,y):=x^y$, defined for positive $x$ and $y$, satisfies (1) if $1<x<y$ and $y^x<x^y$ or $0<x<y<1$.  The condition (2) is satisfied with $0<x<1<y$ choosing $a=1$.
Answer to the OP's question:
The remaining (non-trivial) case is $1<x<y$ and $y^x>x^y$.  In that case there exists a real number $s>1$ with $y=x^s$.  From $y^x>x^y$ we have $sx^{1-s}>1$.
Notice that
$$
x^{y^x}<y^{x^y}\iff x^{sx}<sx^{x^s}.
$$
Now let us invoke Bernoulli, recalling that $x>0$ and $s>1$: as $x^s=\bigl(1+(x-1)\bigr)^s>1+s(x-1)$
we conclude
$$
\frac{sx^{x^s}}{x^{sx}} = sx^{x^s - sx} > sx^{1+s(x - 1)  - sx}= sx^{1 - s} > 1.
$$
A: The inequality is equivalent to
$$\ln\ln a-\ln\ln b>b\ln a-a\ln b.$$
Now let
$$y=\ln b>0,\ x=\frac{\ln a}{\ln b}>1.$$
It's sufficient to show that
$$\ln x>y\left(xe^y-e^{xy}\right)\ \ (x>1,y>0).\tag{1}$$
Put $f(x,y)=xe^y-e^{xy}$. Then
$$\frac{\partial f(x,y)}{\partial y}=x\left(e^y-e^{xy}\right)<0.$$
Hence $f$ is decreasing for $y$. Put $t=\frac{\ln x}{x-1}$ to get
$$f(x,t)=0,\ f(x,0)=x-1.$$
(i) If $y\geqslant t$, $yf(x,y)\leqslant 0<\ln x$.
(ii) If $y<t$, $yf(x,y)<tf(x,0)=\ln x$.
Thus the validity of (1) completes the proof.
