Prove that a symmetric matrix with a positive diagonal entry has at least one positive eigenvalue

If $A$ has all non-positive eigenvalues, then it is negative semidefinite so $x^t A x \le 0$ for all $x \in \mathbb R^n$. But this contradicts $e_i^t A e_i = a_{ii} > 0$. The contradiction implies that $A$ has at least one positive eigenvalue. You can check $e_i^t Ae_i = a_{ii}$ by just performing the necessary multiplication.

By contradiction assume that all the eigenvalues $\lambda_1,\ldots,\lambda_n$ of $A$ are non positive and by spectral theorem let $(v_1,\ldots,v_n)$ an orthonormal basis of eigenvectors then using the hint let $e_i=\alpha_1v_1+\cdots+\alpha_nv_n$ and then $$a_{ii}=e_i^tAe_i=\sum_{j=1}^n\lambda_j\alpha_j^2\le0$$ which is a contradiction.