= Solution
Use the weighted <inner product>
$$
\langle f,g\rangle=\int_a^bf(x)g(x)w(x)\,dx,\qquad h_n=\langle Q_n,Q_n\rangle>0.
$$
Here the interval is nondegenerate and the indicated integrals are finite, as required for the given <orthogonal polynomials>. Strict positivity follows because $w>0$ and a nonzero polynomial cannot vanish on an interval. Since each $Q_j$ is <monic> of degree $j$, $Q_0,\ldots,Q_{n+1}$ form a <basis> of the polynomials of degree at most $n+1$. Comparing leading coefficients, expand
$$
xQ_n=Q_{n+1}+\sum_{j=0}^nc_jQ_j.
$$
By <orthogonality>, $c_j=\langle xQ_n,Q_j\rangle/h_j$. For $j\leq n-2$, symmetry of multiplication by $x$ gives
$$
\langle xQ_n,Q_j\rangle=\langle Q_n,xQ_j\rangle=0,
$$
because $xQ_j$ has degree at most $n-1$ and is a <linear combination> of $Q_0,\ldots,Q_{n-1}$. Only the last two terms remain. Their coefficients are
$$
\boxed{a_n=\frac{\int_a^bxQ_n(x)^2w(x)\,dx}{\int_a^bQ_n(x)^2w(x)\,dx},\qquad
b_n=\frac{\int_a^bQ_n(x)Q_{n-1}(x)xw(x)\,dx}{h_{n-1}}=\frac{h_n}{h_{n-1}}>0\quad(n\geq1).}
$$
For the second equality, $xQ_{n-1}-Q_n$ has degree at most $n-1$, so its <inner product> with $Q_n$ vanishes. Rearranging proves the <three-term recurrence for monic orthogonal polynomials>:
$$
\boxed{Q_{n+1}=(x-a_n)Q_n-b_nQ_{n-1}.}
$$
For $n=0$ this reads $Q_1=x-a_0$ because $Q_{-1}=0$ and $Q_0=1$. The coefficient $b_0$ is therefore immaterial and is not determined by the polynomials. The usual convention is $b_0=0$; if a positive $b_0$ is desired, any positive value gives the identical recurrence. The strictly positive norm-ratio formula is the assertion for \b[$n\geq1$], not a division by the norm of $Q_{-1}$.
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