Why is the largest element of symmetric, positive semidefinite matrix on the diagonal?
Suppose $A = [A_{ij}]$ is symmetric, and let $s$ be an arbitrary real number. If $x = se_{i} - e_{j}$, then $$ x^tAx = s^{2} A_{ii} - 2sA_{ij} + A_{jj}. \tag{1} $$
Assume, contrapositively, that the strictly largest entry is $A_{ij} = A_{ji}$ with $i \neq j$. Taking $s = 1$, i.e., $x = e_{i} - e_{j}$, (1) gives $$ x^tAx = A_{ii} - 2A_{ij} + A_{jj} = (A_{ii} - A_{ij}) + (A_{jj} - A_{ij}) < 0. $$
If $A_{ii} = 0$ but $A_{ij} \neq 0$ for some $j$, (1) gives $$ x^tAx = -2sA_{ij} + A_{jj}, $$ which is negative some $s$ of sufficiently large absolute value.