TITLE: Matrix inequality $(A-B)^2 \leq c (A+B)^2$ ?
QUESTION [12 upvotes]: Let A and B be positive semidefinite matrices. It is not hard to see that $(A-B)^2 \leq 2A^2 + 2B^2$. In fact, $2A^2 + 2B^2 - (A-B)^2 = (A+B)^2$ is positive semidefinite. 
My question is: Is there a constant c (independent of A and B and the dimension) such that
$$(A-B)^2 \leq c (A+B)^2?$$
Thanks.

REPLY [3 votes]: A weaker version is also not true, that is, there is no constant $c$ such that 
$$\lambda_j(A-B)^2 \leq c \lambda_j(A+B)^2, ~~ j=1, 2,\ldots$$
where $\lambda_j(A)$ means the $j$-th largest eigenvalue of $A$.
Nevertheless, it is trivial to see that
$$trace(A-B)^2 \leq  trace(A+B)^2.$$