Take the obvious matrix interpretation: let $A\in\mathbb{R}^{n\times n}$ with $A_{ij}=a_{ij}$. In particular, $A$ is skew-symmetric: $A^T=-A$. Further, let $\boldsymbol{e}\in\mathbb{R}^n$ be the vector of all ones. Then the inequality is equivalent to \[ \frac1n \|A\boldsymbol{e}\|_2^2\le \frac12\|A\|_F^2. \]Using now the estimate $ \|A\boldsymbol{e}\|_2^2\le \|A\|_2^2 \|\boldsymbol{e}\|_2^2 = n\|A\|_2^2$, it suffices to prove $\|A\|_2^2\le \|A\|_F^2/2$. Recalling now the fact that the eigenvalues of a real skew-symmetric matrix are purely imaginary and partitioned into conjugate pairs (see e.g. Horn-Johnson), we get that \[ \|A\|_F^2 = \sum_{i\le n}|\lambda_i|^2 \ge 2|\lambda_1|^2 = 2\|A\|_2^2, \]where $|\lambda_1|\ge \cdots \ge |\lambda_n|$ are the eigenvalues of $A$ arranged in the order of magnitude. This completes the proof.

$n(RHS-LHS)=\sum\limits_{1 \le i < j < k \le n}(a_{i,j}+a_{j,k}+a_{k,i})^2\ge0$