# Connection between arithmetic mean, geometric mean and sample variance

Let $x_1, \dots, x_n$ be positive real numbers. Arithmetic-geometric mean inequality tells us that:

$GM = \sqrt[n]{x_1 \dots x_n} \leq \frac{x_1 + \dots + x_n}{n} = AM$

and that equality occurs iff $x_1 = \dots = x_n$. This condition can be restated as $\sum_{k=1}^n (x_k - AM)^2 = 0$, i.e. sample variance of $x_1, \dots , x_n$ is zero. I'm curious: are there any inequalities connecting arithmetic mean, geometric mean and sample variance?

This is my point: Difference between $AM$ and $GM$ gets larger as $x_1, \dots, x_n$ get further away from each other, i.e. when their sample variance is big. Is there a way to account for the error in arithmetic-geometric mean inequality using sample variance?

• What you ask for might be related to the concept of majorization. Hardy, Littlewood and Polya had already developed the theory to a very advanced level. If you are willing to pursue the study, you should probably check the book "Inequalities: Theory of Majorization and Its Applications". – Troy Woo Aug 18 '14 at 10:33
• Maybe these can help: arxiv.org/pdf/1210.4417.pdf and arxiv.org/abs/1203.4454 and references therein. – pisoir Aug 18 '14 at 10:41

Power mean inequality can be used to get bounds the difference between various means. Not exactly using variances, but you have $$AM - GM \le \max_i x_i - \min_i x_i$$ Or if that's too large a bound, note $$AM - GM \le AM - HM$$
If you want variances involved, try manipulating $$\sqrt{\frac1n \sum x_i^2} - GM \ge AM-GM$$