SolFeb2205 HL

Claim: For any positive numbers $a_1 \le a_2 \le \cdots \le a_n$ , $a_1^{a_1}\cdots a_n^{a_n} \ge \left( a_1 \cdots a_n \right)^{(a_1 + \cdots + a_n)/n}$ .

Proof: We solve this using an averaging argument. Taking logs and dividing both sides by $a_1 + \cdots + a_n$ , it suffices to show that

$p_1 X_1 + \cdots + p_n X_n \ge\frac{1}{n} X_1 + \cdots + \frac{1}{n} X_n,$

where

$p_i = \frac{a_i}{a_1 + \cdots + a_n}$

and $X_i = \log a_i\,$ . Since the $a_i\,$ are non-decreasing, so are the $p_i\,$ and the $X_i\,$ . All we need show is that any such weighted average, where the weights increase as the numbers increase, is bounded below by the arithmetic mean:

Since the $p_1 \le \cdots \le p_n$ and $X_1 \le \cdots \le X_n$ , we can apply one of Chebyshev's inequalities (http://planetmath.org/encyclopedia/ChebyshevsInequality.html) to this.

$p_1 X_1 + \cdots + p_n X_n \ge(\frac{1}{n} X_1 + \cdots + \frac{1}{n} X_n)(p_1 + \cdots + p_n)$

Since $p_1 + \cdots + p_n = 1\,$ the claim is immediately established.

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