Although Qmechanics's answer is formally complete and correct, there is a more intuitive formulation of this identity that makes it self evident. Consider the operator B which is A minus its expectation value in some state.
$$B = A - \langle A\rangle $$
Then the expectation value of B is zero in the same state (obviously--- it has been shifted to make it so). The expected value of $B^2$ can be nonzero--- it is a measure of the spread in B in state $\psi$. It is positive, as you can see by the definition of matrix multiplication (or by "inserting the identity in a basis")
$$ \langle B^2 \rangle = \sum_i \langle |B|i\rangle\langle i|B\rangle $$
The last thing on the right is the sum of positive quatities of the form $c^*c$. If you now reexpress the expectation value of $B^2$ in terms of A,
$$ \langle B^2 \rangle = \langle (A-\langle A\rangle)^2\rangle = \langle A^2\rangle - 2 \langle A\langle A\rangle \rangle + \langle A\rangle^2 = \langle A^2\rangle - \langle A\rangle^2 $$
This manipulation justifies this thing.