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For a real symmetric matrix, or more generally a Hermitian matrix, the finite-dimensional spectral theorem makes all eigenvalues real. Intersecting its Gershgorin discs with the real axis givesThe union preserves gaps that the single outer interval loses. If every $a_{ii}-r_i>0$, all eigenvalues are positive, so the matrix is a positive-definite matrix. This is the positive-diagonal strictly diagonally dominant matrix criterion. For the three-by-three example, $1.7\le \lambda \le 9.4$ gives a positive-definite matrix certificate and bounds its spectral condition number of a positive-definite matrix by $9.4/1.7\approx 5.529412$; the computed value is about $4.544364$.

Strictness matters. The symmetric matrix $\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$ has positive diagonal and is a diagonally dominant matrix, but its eigenvalues are $0$ and $2$: it is singular. For a general complex matrix, the condition $\mathrm{Re}{a}_{ii}>r_i$ only puts the eigenvalues in the open right half-plane. The Hermitian matrix assumption is what converts that statement into positive definiteness.