Is there a fast way to prove a tridiagonal matrix is positive definite?

Notice $A$ can be rewritten as a sum of 5 matrices. $$A = \left[\begin{smallmatrix} 2 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 3 \end{smallmatrix}\right] + \left[\begin{smallmatrix} 2 & 2 & 0 & 0 & 0\\ 2 & 2 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 2 & 2 & 0 & 0\\ 0 & 2 & 2 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 2 & 2 & 0\\ 0 & 0 & 2 & 2 & 0\\ 0 & 0 & 0 & 0 & 0 \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 2 & 2\\ 0 & 0 & 0 & 2 & 2 \end{smallmatrix}\right] $$ The first matrix is diagonal with positive entries on diagonals, so it is positive definite. The remaining four matrices are clearly positive semi-definite. Being a sum of a positive definite matrix with a bunch of positive semi-definite matrices, $A$ itself is positive definite.

From Gershgorin’s disc theorem and the matrix being symmetric (real eigenvalues) you get that the eigenvalues are in the interval [1,9] and thus positive. Apparently it’s commonly known as Gershgorin’s circle theorem.

This particular matrix is symmetric diagonally dominant (SDD), meaning that the absolute values of each row's off-diagonal entries do not exceed the absolute value of the diagonal, ie. $$ \sum_{\substack{j\in[1,n] \\ i \neq j}} \lvert a_{i,j} \rvert \leq \lvert a_{ii} \rvert$$ Since the diagonals are positive, it is positive semidefinite, but it is actually positive definite because the inequality above is strict. This can be seen by splitting $A$ into $A = A' + \varepsilon I$ such that $A'$ remains SDD (and therefore PSD) and $\varepsilon > 0$. Multiplying on both sides by the same vector is always positive, since $\varepsilon I$ is positive definite.