PROBLEM 26. $a^{b}$ and $b^{a}$

It is often not difficult to determine which is greater: $a^{b}$ or $b^{a}$. It is obvious that

$$
2^{3} < 4^{3}
$$

However, when considering two numbers, one of which is between 2 and 3, and the other between 3 and 4, this question can be challenging. Which is greater:

$$
e^{\pi} \text { or } \pi^{e} ?
$$
Solution. For positive $x$ we have

$$
e^{x}=1+x+\frac{x^{2}}{2}+\ldots>1+x
$$

Since $\pi>e, \pi / e>1$ and $x=\pi / e-1>0$. Thus,

$$
\begin{gathered}
e^{\pi / e-1}>1+\left(\frac{\pi}{e}-1\right), \quad \frac{e^{\pi / e}}{e}>\frac{\pi}{e} \\
e^{\pi / e}>\pi, \quad e^{\pi}>\pi^{e} .
\end{gathered}
$$

As an easy exercise, show that the function $e^{x}-1-x$ has a unique minimum at zero (when $x=0$). Thus, for all real values of $x$ we have

$$
e^{x} \geqslant 1+x
$$

Using this inequality, George Pólya constructed the following beautiful proof of the famous arithmetic mean-geometric mean inequality. Let $a_{1}, a_{2}, \ldots, a_{n}$ be some positive real numbers, and let $A$ and $G$ be their arithmetic mean and geometric mean, respectively, i.e.

$$
A=\frac{1}{n}\left(a_{1}+a_{2}+\ldots+a_{n}\right), \quad G=\left(a_{1} \cdot a_{2} \cdot . a_{n}\right)^{1 / n}
$$

By setting $x$ to $\left(a_{i} / A\right)-1, i=1,2, \ldots, n$, we obtain $n$ inequalities

$$
\begin{aligned}
& e^{\left(a_{1} / A\right)-1} \geqslant \frac{a_{1}}{A} \\
& e^{\left(a_{2} / A\right)-1} \geqslant \frac{a_{2}}{A} \\
& \cdot \cdot \cdot \cdot \cdot \\
& e^{\left(a_{n} / A\right)-1} \geqslant \frac{a_{n}}{A}
\end{aligned}
$$

44

Multiplying them, we get

$$
e^{\frac{a_{1}+a_{2}+\ldots+a_{n}}{A}-n} \geqslant \frac{a_{1} a_{2} \ldots a_{n}}{A^{n}}
$$

which is simply

$$
e^{n-n} \geqslant \frac{G^{n}}{A^{n}} \text { or } 1 \geqslant \frac{G^{n}}{A^{n}}
$$

from which it follows that $A \geqslant G$.

Note that $A=G$ only if equality holds in all $n$ inequalities. This corresponds to the case where $a_{i} / A-1=0$ in all cases. This shows that $A=G$ only when all $a_{i}$ are equal (to the number $A$).