= Solution
For $D=\operatorname{diag}(d_1,d_2,d_3)$, the upper entry of $D^{-1}AD$ in positions $i,i+1$ is $b_id_{i+1}/d_i$, and the corresponding lower entry is $c_id_i/d_{i+1}$. They are equal when $(d_{i+1}/d_i)^2=c_i/b_i$. All coefficients used here are positive, so choose
$$
\boxed{D=\operatorname{diag}\left(1,\sqrt{c_1/b_1},\sqrt{c_1c_2/(b_1b_2)}\right).}
$$
Direct <matrix multiplication> gives the <symmetric matrix>
$$
D^{-1}AD=\begin{pmatrix}
a_1&\sqrt{b_1c_1}&0\\
\sqrt{b_1c_1}&a_2&\sqrt{b_2c_2}\\
0&\sqrt{b_2c_2}&a_3
\end{pmatrix}.
$$
Its pattern remains <tridiagonal>. This is <positive diagonal symmetrization of a tridiagonal matrix>.
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