For non-negative data the sample mean is not smaller than its standard error. (1) Let $X_1, X_2, \dots, X_n$ be a random sample from a population
with non-negative values. Then show that $\bar X \ge S/\sqrt{n},$
where $S^2 = [\sum_{i=1}^n (X_i - \bar X)^2]/(n-1).$
I have not seen this inequality stated before. It is not difficult
to prove directly, but may be implied by some more general result I am
overlooking. 
(2) Also use this inequality to find a counterexample, showing that the sample mean and variance are not independent for exponential (or beta) data; perhaps use $n = 4$ for simplicity. [of course, $\bar X$ and $S$ are independent for normal data.]
 A: Answer to (1).
Note that $(n-1)S^2 = \sum X_i^2 - \frac{1}{n}(\sum X_i)^2$ follows from the definition of $S^2.$ 
Also, for nonnegative data,
$\sum X_i^2 \le (\sum X_i)^2$
because the RHS contains nonnegative cross-product terms that the LHS does not.
Then
$$n(n-1)S^2 = n\sum X_i^2 - (\sum X_i)^2 \le n(\sum X_i)^2- (\sum X_i)^2 = (n-1)n^2\bar X^2,$$
which implies (1).
A: For the counterexample: picking basically arbitrary numbers,  if $n=4$ and $S \geq 2$ then $\overline{X} \geq 1$. So do you have
$$P(S \geq 2 \wedge \overline{X} < 1)=P(S \geq 2)P(\overline{X} < 1)?$$
If not then you have your counterexample. Note that the left side is zero, so it's enough for both factors on the right side to be positive.
Qualitatively speaking, this argument says that $\overline{X}$ and $S$ are dependent because if $S$ is large then $\overline{X}$ must also be large.
A: Answer and graphs for (2).
The left panel of the figure shows sample SDs $S$ plotted against
sample means $\bar X$ for 5000 samples of size 5 from $Exp(1)$; $\bar X$ is highly correlated with $S$ (sample correlation $r = 0.77$) such
that $\bar X$ and $S$ tend to increase together. 
More specifically, the red dotted line shows $S/\sqrt{5} = \bar X$, so that no points lie above the the line. It is clear that $P(\bar X < 5\sqrt{5}) > 0$
and $P(S > 25) > 0,$ but $P(\bar X <5\sqrt{5},\; S > 25) = 0,$ so that
$\bar X$ and $S$ are not independent. (Also, see the Answer by @Ian for exponential data with $n = 4.$)

The right panel shows a similar plot for 5000 samples of size
5 from $Beta(.5, .5).$ Here, it is clear from symmetry that
$\bar X$ and $S$ are uncorrelated (sample $r = 0.007$ is 
consistent with population $\rho = 0.$) However, an argument based on
$\bar X \ge S/\sqrt{5}$ provides counterexamples (in the same way
as for the exponential distribution) to show that the sample mean
and SD are not independent.
Points representing five observations from $Beta(.5,.5)$ lie within the 5-D unit hypercube and tend to lie near its faces,
edges, and vertices. When this hypercube is
'squashed' into 2-D by the mean-SD transformation vertices
and some edges of the hypercube remain visible. The 5-D hypercube
has 32 corners, and there are 6 'horns' visible in the plot;
clockwise from lower left, 'multiplicities' in the mapping of
vertices to horns are 1, 5, 10, 10, 5, and 1. 
Points near the origin have relatively small $\bar X$ and points near the
vertex $(0,0,0,1,1)$ (and 19 others) have relatively large $S.$ By the 
inequality in (1), there can be no points with both 'small' mean and 'large' SD.
Note: Formal proofs notwithstanding, students are often skeptical
that $\bar X$ and $S$ are stochastically independent for normal data
even though not functionally independent. This intuitive skepticism
is not stupid because precisely normal data are rare in applications.
And it is only for normal data that $\bar X$ and $S$ are independent. It is intuitively clear why they are not independent
for exponential data; the plot in the right panel shows one
pattern that thwarts independence in symmetrical nonnormal
distributions.
