= Solution
For the analogous positive off-diagonal entries in size $n$, take
$$
\boxed{d_1=1,\qquad d_i=\sqrt{\prod_{j=1}^{i-1}\frac{c_j}{b_j}}\quad(2\le i\le n),\qquad D=\operatorname{diag}(d_1,\ldots,d_n).}
$$
The same entrywise computation gives $(D^{-1}AD)_{ii}=a_i$ and $(D^{-1}AD)_{i,i+1}=(D^{-1}AD)_{i+1,i}=\sqrt{b_ic_i}$, with every other entry zero. Thus the result is a <symmetric matrix> with a <tridiagonal> pattern. Positivity of the diagonal entries $a_i$ is not needed for this <positive diagonal symmetrization of a tridiagonal matrix>; positivity of $b_i,c_i$ ensures the displayed positive, invertible $D$. In particular, the assertion is not an arbitrary-real-coefficient symmetrization claim when the off-diagonal signs differ.
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