TITLE: Does PSD imply on average diagonal dominant?
QUESTION [6 upvotes]: Suppose $A$ is a $N \times N$ positive semidefinite matrix.   This does not necessarily imply that $A$ is diagonally dominant.  But does it imply the following "average diagonal dominance" i.e.
$$\frac{1}{N} \sum_{i}A_{i,i}\ge \frac{2}{(N-1)N} \sum_{j=1}^{N}\sum_{i<j}A_{i,j}\text{ ?}$$

REPLY [2 votes]: Presumably you are talking about real matrices. If so, the inequality is true. We only need to consider only the case where the RHS is nonnegative, because the LHS is always nonnegative. So, it suffices to prove the inequality that
$$
\frac1N\sum_i a_{ii} \ge \frac2{N(N-1)}\sum_j\sum_{i<j}a_{ij}\tag{1}
$$
when the RHS is nonnegative. In turn, it suffices to prove $(1)$ in general, regardless of whether the RHS is nonnegative or not.
Rewrite $(1)$ as $\operatorname{trace}(A)\ge u^TAu$, where $u$ is the unit vector $\frac1{\sqrt{N}}(1,1,\ldots,1)^T$. Now the inequality obviously holds. In fact, when $A$ is PSD, we have $\operatorname{trace}(A)\ge \lambda_\max(A)\ge v^TAv$ for every unit vector $v$.