# An “AGM-GAM” inequality

For positive real numbers $x_1,x_2,\ldots,x_n$ and any $1\leq r\leq n$ let $A_r$ and $G_r$ be , respectively, the arithmetic mean and geometric mean of $x_1,x_2,\ldots,x_r$.

Is it true that the arithmetic mean of $G_1,G_2,\ldots,G_n$ is never greater then the geometric mean of $A_1,A_2,\ldots,A_n$ ?

It is obvious for $n=2$, and i have a (rather cumbersome) proof for $n=3$.

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–  picakhu Jul 3 '12 at 12:30

## 2 Answers

It's a special case ($r=0$, $s=1$) of the mixed means inequality $$M_n^s[M^r[\bar a]]\le M_n^r[M^s[\bar a]], \quad r,s\in \mathbb R,\ r<s,$$ where $M^s$ is the power mean with exponent $s$, see Survey on Classical Inequalities, p. 32, theorem 2.

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Thanks Andrew, and I noticed that the result actually appears in this exact formulation as theorem 1 on page 31. –  tipshoni Jul 3 '12 at 15:33

Here's a proof for $n=2$. Apply the Cauchy-Schwarz inequality to the vectors $(\sqrt{a},\sqrt{b})$ and $(1/2,1/2)$ to get $${\sqrt{a}+\sqrt{b}\over 2}\leq\sqrt{a+b\over 2}.$$ Multiply by $\sqrt{a}$ to obtain $${a+\sqrt{ab}\over 2}\leq \sqrt{a\left({a+b\over 2}\right)}.$$

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Oh I am sorry. I thought he was asking whether the arithmetic mean of a set of numbers was always less than or equal to the geometric mean of the numbers. I misread the question and will take down the answer –  Michael Chernick Jun 26 '12 at 18:25
no problem. we all make mistakes! –  Byron Schmuland Jun 26 '12 at 18:27