How to go about solving this question on positive semi-definite matrices?

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When both $A$ and $B$ are PSD, $(B^T)^{1/2}A(B^T)^{1/2}$ is also PSD. Hence $$ \sum_{i,j}a_{ij}b_{ij}=\operatorname{tr}(AB^T)=\operatorname{tr}\left((B^T)^{1/2}A(B^T)^{1/2}\right)\ge0. $$ Conversely, if $A$ is not PSD, then there exists a vector $v$ such that $v^\ast Av$ is not real nonnegative. Thus $\operatorname{tr}(AB^T)=v^\ast Av$ is not real nonnegative for the PSD matrix $B=(vv^\ast)^T$.