Given $a>b>2$ both positive integers, which of $a^b$ and $b^a$ is larger?

Question

Given $a>b>2$ both positive integers, which of $a^b$ and $b^a$ is larger?

I tried an induction approach. First I showed that if $b=3$ then any $a \geq4$ satisfied $a^b<b^a$.

Then using that as my base case I tried to show that given any pair of positive integers $a,b$ satisfying $a>b>2$ and $a^b<b^a$, then $(a+1)^{b+1}<(b+1)^{a+1}$ - but that is where I got stuck.

Any help would be appreciated.

Answer 1

The result follows easily using calculus. Here's an elementary approach, which uses the fact that $a, b$ are positive integers.

Consider $n \geq 3$. Then $$(n+1)^n=\sum_{i=0}^{n}{\binom{n}{i}n^{n-i}}=1+n^2+\sum_{i=0}^{n-2}{\binom{n}{i}n^{n-i}}<n^n+\sum_{i=0}^{n-2}{n^n}=n^{n+1}$$ since $\binom{n}{i} \leq n^i$, and $1+n^2<n^n$ for $n \geq 3$.

Therefore $n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$ for $n \geq 3$. This immediately implies that $a^{\frac{1}{a}}<b^{\frac{1}{b}}$, so $a^b<b^a$.

Answer 2

Hint:

Which is larger, $b\ln a$ or $a\ln b$?

Which is larger, $\dfrac{\ln a}{a}$ or $\dfrac{\ln b}{b}$?

How does $\dfrac{\ln x}{x}$ behave as $x$ increases? Looks like a job for the derivative.

Answer 3

Suppose for a moment: $$a^b = b^a.$$ Taking natural log: $$ b\ln a = a\ln b, $$ which is $$ \frac{\ln a}{a} = \frac{\ln b}{b} . $$ Now consider the function: $$ f(x) = \frac{\ln x}{x}, $$ where $$ f'(x) = \frac{1 - \ln x}{x^2} < 0 \quad \text{ if }\;x>e. $$ I believe you could take it from here.

Answer 4

Another elementary approach:
First notice that it suffices to prove that $n^{n+1}>(n+1)^n$ for $n\ge 3$. We divide both sides by $n^n$ to turn it into $n>(1+1/n)^n$. As a last step, show by induction that $(1+1/n)^n\le n$ for $n\ge3$.

The case $n=3$ is clear. Now, if $(1+1/k)^k\le k$, then $(1+1/(k+1))^{k+1}\le (1+1/(k))^{k+1}\le k(1+1/(k))\le k+1$.

Q.E.D.
Inform me if anything needs improvements. Thanks in advance.