Maximum of the Variance Function for Given Set of Bounded Numbers Let $ \boldsymbol{x} $ be a vector of $n$ numbers in the range $ \left[0, c \right] $, where $ c $ is a positive real number.
What's is the maximum of the variance function of this $ n $ numbers?
Maximum in the meaning what spread of the number will maximize the variance?
What would be a tighter bound for other assumptions on the spread of the numbers.
The variance of the vector $ \boldsymbol{x} $ is given by:
$$ \operatorname{var} (\boldsymbol{x}) = \frac{1}{n} \sum_{i = 1}^{n} {\left( {x}_{i} - \overline{\mathbf{x}} \right )}^2 $$
Where the mean $\overline{\boldsymbol{x}}$ is given by:
$$ \overline{\boldsymbol{x}} = \frac{1}{n} \sum_{i = 1}^{n} {x}_{i} $$
 A: Since $x_i \leq c$,
$\displaystyle \sum_i x_i^2 = \sum_i x_i\cdot x_i \leq \sum_i c\cdot x_i = cn\bar{x}.$
Note also that $0 \leq \bar{x} \leq c$.  Then,
$$
\begin{align*}
n\cdot \text{var}(\mathbf{x}) &= \sum_i (x_i - \bar{x})^2= \sum_i x_i^2 - 2x_i\bar{x} + \bar{x}^2\\
&= \sum_i x_i^2 - 2\bar{x}\sum_i x_i + n\bar{x}^2= \sum_i x_i^2 - n\bar{x}^2\\
&\leq cn\bar{x} - n\bar{x}^2 = n\bar{x}(c-\bar{x})
\end{align*}
$$
and thus $$\text{var}(\mathbf{x}) \leq \bar{x}(c-\bar{x}) \leq \frac{c^2}{4}.$$
Added  note: (second edit)
The result $\text{var}(X) \leq \frac{c^2}{4}$ also applies to random variables
taking on values in $[0,c]$, and, as my first comment on the question says, putting half the mass at $0$ and the other half at $c$ gives the maximal variance of $c^2/4$.  For the vector $\mathbf x$, if $n$ is even, the maximal variance $c^2/4$ occurs when $n/2$ of the $x_i$ have value $0$ and the
rest have value $c$.
Someone else posted an answer -- it has since been deleted  -- which said the
same thing and added that if $n$ is odd, the variance is maximized when
$(n+1)/2$ of the $x_i$ have value $0$ and $(n-1)/2$ have value $c$, 
or vice versa.  This gives a variance of $(c^2/4)\cdot(n^2-1)/n^2$ which
is slightly smaller than $c^2/4$.  Putting the "extra" point at $c/2$
instead of at an endpoint gives a slightly smaller variance of 
$(c^2/4)\cdot(n-1)/n$, but both choices have variance approaching
$c^2/4$ asymptotically as $n \to \infty$.
A: An easier derivation can be done as follows (From [1]):
For any constant $c$, we have,
\begin{equation}
E[(X-c)^2] = E[X^2] - 2E[X]c + c^2
\end{equation}
The above quadratic is minimized when $c=E[X]$. It follows that,
\begin{equation}
\sigma^2 = E[(X-E[X])^2] \leq E[(X-c)^2], \text{for all } c
\end{equation}
By letting $c = (a+b)/2$, we obtain,
\begin{equation}
\sigma^2 \leq E\left[\left(X-\frac{a+b}{2}\right)^2\right] = E[(X-a)(X-b)] + \frac{(b-a)^2}{4} \leq \frac{(b-a)^2}{4}
\end{equation}
since for $x$ in $[a,b]$, $(x-a)(x-b)<0$
Further, the bound could be very conservative. However, in the absence of any other information about $X$, it can not be improved.
[1]  D. P. Bertsekas and J. N. Tsitsiklis, Introduction to Probability. Athena
Scientific, 2002.
