Producing lower bounds for $\text{trace}(A^2)$ for a positive semidefinite, symmetric matrix $A$ Are there any lower bounds on $\DeclareMathOperator{trace}{trace}$
\begin{align*}
\trace(A^2),
\end{align*}
where $A$ is positive semi-definite and symmetric?
I am aware of the inequality 
$$
\trace(A^2) \le\bigl(\trace(A)\bigr)^2
$$
Are there any inequalities in the other direction?
 A: For $A$ symmetric positive semi-definite, we have
$$\rho(A)^2\quad\leq\quad \sum_{k=1}^n \lambda_k^2 \quad =\quad\operatorname{trace}(A^2),$$
where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$ and $\rho(A)$ its spectral radius.
EDIT: In the book Matrix Analysis of A. Horn and C. Johnson, appears the bound 
$$ \operatorname{trace}(A^2) \geq \frac{\operatorname{trace}(A)^2}{\operatorname{rank}(A)}$$
(see exercise 13, p.175) which is in turn slightly better than $\frac{\operatorname{trace}(A)^2}{n}$ :).
A: $\langle X,Y\rangle=\operatorname{trace}(XY^T)$ is an inner product, therefore Cauchy-Schwarz inequality gives $\langle A,A\rangle \langle I,I\rangle \ge |\langle A,I\rangle|^2$, or $\operatorname{trace}(A^2)\ge\frac1n\operatorname{trace}(A)^2$ when $A$ is an $n\times n$ PSD matrix. This is slightly sharper than the $\rho(A)^2$ in one other answer.
A: Hints


*

*For any matrix $B$, $\text{tr } B$ is the sum of the eigenvalues of $B$, counting multiplicity.

*If $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $B$, counting multiplicity, the eigenvalues of $B^2$ are $\lambda_1^2, \ldots, \lambda_n^2$.

*Since $A$ is symmetric, its eigenvalues are real, and since it is positive semidefinite, they are nonnegative.

