What does the numpy std documentation mean when it says it is always biased? The documentation for the numpy np.std() function states:

The average squared deviation is typically calculated as x.sum() / N,
where N = len(x). If, however, ddof is specified, the divisor N - ddof
is used instead. In standard statistical practice, ddof=1 provides an
unbiased estimator of the variance of the infinite population. ddof=0
provides a maximum likelihood estimate of the variance for normally
distributed variables. The standard deviation computed in this
function is the square root of the estimated variance, so even with
ddof=1, it will not be an unbiased estimate of the standard deviation
per se.

(emphasis added)
What does the final statement in bold mean?  Where does the estimate break down and become biased?
 A: While the estimate of the variance is unbiased, the estimate of the standard deviation is not.
An unbiased estimate $V_{est}$ is one such that the expected value of the estimate is the true value $V$ being estimated: $E[V_{est}] = V$.
But applying a nonlinear operator like the square or square root destroys this property. In general, $E[\sqrt{V_{est}}] \neq \sqrt{V}$ so $E[\sigma_{est}] \neq \sigma$. That is, the estimated std deviation $\sigma_{est}$ will be biased even though $V_{est}$ is not.
The distribution of V estimates is not symmetric -- since V must be positive, it will have a longer tail above the true variance and a shorter, fatter tail below the true variance.  When we take the square root of V, the new distribution's mode can be found by taking the square root of the location of the mode, but we can't say the same thing about the mean.
Other answers on earlier questions take this discussion further. For example, this answer points out that Jensen's inequality can prove in which direction the estimated standard deviation will be biased and this answer shows a plot of the bias for different sample sizes.
But my personal favorite is this answer which proves using simple expectation algebra that the estimated standard deviation will (in general) be less than the true standard deviation of the data.  This answer relies on the identity $\mathrm{Var}[S_n] = \mathrm{E}[S_n^2] - \mathrm{E}^2[S_n]$ which would have to also be explained to the uninitiated, but there is an elegance to it that ties it into basic statistical theory nicely.
