For $ a>b>1$, prove that $a^{b^a}>b^{a^b}$

Question

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?

Answer 1

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 $$ \frac{sx^{x^s}}{x^{sx}} = sx^{x^s - sx} > sx^{1+s(x - 1) - sx}= sx^{1 - s} > 1. $$

Answer 2

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

Answer 3

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.