= Solution
Let $Ly_n=\lambda_ny_n$. By self-adjointness,
$$
\lambda_n\langle y_m,y_n\rangle_w=\langle y_m,Ly_n\rangle_w=\langle Ly_m,y_n\rangle_w=\lambda_m^*\langle y_m,y_n\rangle_w.
$$
Taking $m=n$ first shows that each <eigenvalue> is real: $\langle y_n,y_n\rangle_w>0$ for a nonzero <eigenfunction>, since $w>0$. Therefore
$$
(\lambda_n-\lambda_m)\langle y_m,y_n\rangle_w=0.
$$
For distinct eigenvalues this proves <orthogonality of Sturm-Liouville eigenfunctions>:
$$
\boxed{\int_a^b y_m^*(x)y_n(x)w(x)\,dx=0\quad(\lambda_m\ne\lambda_n).}
$$
The weight is essential: this need not be orthogonality under the unweighted integral.
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