Let $a,b\in\mathbb N, a\gt b\gt 2$. Prove that $b^{a}\gt a^{b},\forall a,b$.

Question

Let $a,b\in\mathbb N, a\gt b\gt 2$. Prove that $b^{a}\gt a^{b},\forall a,b$.

This is a relatively new question, so this may or may not be true. Does anyone know if this is true or not?

Answer 1

I would start by noticing that \begin{align*} b^{a} \geq a^{b} \Longleftrightarrow a\ln(b) \geq b\ln(a) \Longleftrightarrow \frac{\ln(b)}{b} \geq \frac{\ln(a)}{a} \end{align*} Thus the given problem reduces to prove the function $\ln(x)/x$ is decreasing for $x \geq 3$. Indeed, this is the case:

\begin{align*} f(x) = \frac{\ln(x)}{x} \Longrightarrow f'(x) = -\frac{\ln(x)}{x^{2}} + \frac{1}{x^{2}} = \frac{1 - \ln(x)}{x^{2}} \leq 0 \end{align*} and we are done.

Hopefully this helps.

Answer 2

Let $f(x) = \frac{\ln x}{x}, x>0$. Then $f'(x) = \frac{1-\ln x}{x^2}$. As you can see for $x>e$ we have $f'(x) < 0$ so (due to continuity) $f(x)$ increasing for $x\ge e$. Since $a>b>2$ and $a, b$ integers we have $a>b>e$. So:

$$f(b)>f(a)\implies\frac{\ln b}{b}>\frac{\ln a}{a}\implies \ln b^a > \ln a^b\implies b^a > a^b$$

Answer 3

You can prove in a purely arithmetical way, by induction on $a$.

• Initialisation: suppose $a=b+1$ (and $b>2$). To be proved: $$b^{b+1}>(b+1)^b\iff b>\Bigl(1+\frac1b\Bigr)^b$$ Using the binomial formula, we have $$\Bigl(1+\frac1b\Bigr)^b=1+\sum_{k=1}^b\binom bk\frac1{b^k}=1+\sum_{k=1}^b\frac{1}{k!}\underbrace{\frac{b(b-1)\dots(b-k+1)}{b^k}}_{{}<\,1}<1+b-1=b$$
• Inductive step: suppose $b^a>a^b$ for some $a>b>2$. By the inductive hypothesis, we have $b^{a+1}>b\cdot a^b$, so it is enough to prove $$b\cdot a^b>(a+1)^b\iff b>\Bigl(1+\frac1a\Bigr)^{b}.$$ Now observe that, since $b<a$, : $$\Bigl(1+\frac1a\Bigr)^{b}<\Bigl(1+\frac1b\Bigr)^{b}< b$$ by the initialisation step.