(From the book The Cauchy-Schwarz Master Class by J. Michael Steele, Exercise 2.10)
For nonnegative \(a_1, a_2, \ldots, a_n\) and \(n \ge 2\)
$$ a_n \bigg(\frac{a_1+ a_2+ \ldots + a_{n-1}}{n-1} \bigg)^{n-1} \le \bigg(\frac{a_1+ a_2+ \ldots + a_n}{n} \bigg)^n $$
Proof.
Let \(y_i = \frac{a_1+ a_2+ \ldots + a_i}{i}\) for \(i \ge 2 \).
Then the inequality can be written as \(\frac{a_n}{y_{n-1}} \le \bigg(\frac{y_n}{y_{n-1}} \bigg)^n \).
On the other hand, \( n \cdot y_n – (n-1) \cdot y_{n-1} = a_n \), hence \( n \cdot \frac{y_n}{y_{n-1}} – (n-1) = \frac{a_n}{y_{n-1}} \).
Therefore, \( n \cdot \frac{y_n}{y_{n-1}} – (n-1) \le \bigg(\frac{y_n}{y_{n-1}} \bigg)^n \).
If we let \(y = \frac{y_n}{y_{n-1}} \), this inequality can be written as \( n \cdot y – (n-1) \le y^n \).
So it suffices to show that \( n (y-1) \le y^n – 1 = (y-1)(y^{n-1}+ \ldots +1) \).
It is easy to prove this inequality by considering two cases \(y \ge 1 \) or \(y \lt 1 \).