Matrix inequality $A^2 \succeq A$ If $A$ symmetric positive semidefinite matrix is the following inequality true.
If $A \succeq I$ then 
\begin{align}
A^2
& \succeq  A
\end{align}
This is an equivalent of $a^2 \ge a$ is $a \ge 1$.

Yes, that relationship is true.  $A^2$, $A$, and $I$ can all be simultaneously diagonalized by an orthogonal matrix $E$.  If $\Lambda$ are the eigenvalues of $A$ then we know that
$$A^2 = E\Lambda^2 E^{-1} \,\,\,\, A = E\Lambda E^{-1} \,\,\,\, I = E I E^{-1}$$
The question of whether the ordering you specified is true is, by definition, the question of whether $A^2 - A$ is positive semi-definite.  From the above we see that the eigenvalues of $A$ are all greater than or equal to $1$. Furthermore, we see that
$$A^2 - A = E(\Lambda^2 - \Lambda)E^{-1} = E((\Lambda - I)\Lambda)E^{-1}$$
So $A^2 - A$ has eigenvalues that are greater than or equal to $0$, hence it is positive semi-definite.
