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?

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
$$\sqrt{\frac1n \sum x_i^2} - GM \ge AM-GM$$
