Refering to this wikipedia page Unbiased estimation of standard deviation, it says that "it follows from Jensen's inequality that the square root of the sample variance is an underestimate".
I do know that for the concave square root function, Jensen's inequality says that the square root of the mean > mean of the square root.
So, how do we conclude that the square root of the sample variance underestimates population standard deviation?
Since we know from Jensen's inequality that square root of the mean > mean of the square root, does "square root of sample variance" somehow relate to "mean of the square root" while "population standard deviation" somehow relates to "square root of the mean"?
Added after joriki's response:
Given joriki's response about using a single sampling of data, I am now left with why $s=\sqrt{\frac{1}{N-1}\sum_{i=1}^N{(x_i-\overline{x})^2}}$ will underestimate pop std dev. In order to use Jensen's inequality (mean of the square root < square root of the mean). I need to somehow relate the expression for $s$ to "mean of square root". I do see the square root sign in the expression for $s$ but where is the "mean" of this square root quantity?