In the question, $A$ is a symmetric matrix, and $B$ is a positive semi-definite matrix. $A\preceq B$ means that $B-A$ is a positive semi-definite matrix. $\|\|_F$ means the Frobenius norm.
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This would be easy if we could conclude that $A^2 \preceq B^2$, but notably we cannot – Ben Grossmann Nov 02 '16 at 12:44
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The answer is yes. Suppose without loss of generality (via the spectral theorem) that $A$ is diagonal, so that its (real) diagonal entries are $\lambda_1,\dots,\lambda_n$, and its other entries are zero. (In other words: if the orthogonal $U$ is such that $U^TAU$ is diagonal, we now consider $U^TAU$ instead of $A$ and $U^TBU$ instead of $B$)
We have $-B \preceq A \preceq B$, which tells us that both $B + A$ and $B - A$ are positive semidefinite. Thus, the diagonal entries $(B+A)_{ii} = B_{ii} + \lambda_i$ and $(B - A)_{ii} = B_{ii} - \lambda_i$ are necessarily non-negative. Thus, $B_{ii} > |\lambda_i|$. Thus, we have $$ \|B\|_F^2 = \sum_{i,j} B_{ij}^2 \geq \sum_{i} B_{ii}^2 \geq \sum_{i}\lambda_i^2 = \|A\|_F^2 $$ as desired.
Ben Grossmann
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