[proofplan] Since $p_{k+1}$ and $xp_k$ are both monic of degree $k+1$, their difference $r = xp_k - p_{k+1}$ lies in $P_k[x]$. Expanding $r$ in the orthogonal basis $\{p_0, \dots, p_k\}$ and using the symmetry $\langle xp_k, p_j \rangle = \langle p_k, xp_j \rangle$, only the $j = k$ and $j = k-1$ coefficients survive (since $xp_j \in P_{j+1}[x]$ and $\langle p_k, xp_j \rangle = 0$ when $j+1 < k$). This gives the three-term recurrence with explicit formulas for $\alpha_k$ and $\beta_k$. [/proofplan] [step:Express the remainder $xp_k - p_{k+1}$ in the orthogonal basis] Since $p_{k+1}$ and $xp_k$ are both monic of degree $k+1$, their difference $r(x) = xp_k(x) - p_{k+1}(x)$ lies in $P_k[x]$. Expanding $r$ in the orthogonal basis $\{p_0, \dots, p_k\}$: \begin{align*} r(x) &= \sum_{j=0}^{k} \frac{\langle r, p_j \rangle}{\langle p_j, p_j \rangle}\,p_j(x). \end{align*} [/step] [step:Show that only the $j = k$ and $j = k-1$ coefficients are nonzero] For each $j$, $\langle r, p_j \rangle = \langle xp_k, p_j \rangle - \langle p_{k+1}, p_j \rangle$. The second term vanishes for $j \le k$ by orthogonality of $p_{k+1}$ to $P_k[x]$. For the first term, the symmetry condition $\langle xp_k, p_j \rangle = \langle p_k, xp_j \rangle$ holds by hypothesis. Now $xp_j \in P_{j+1}[x]$, so $\langle p_k, xp_j \rangle = 0$ whenever $j + 1 < k$, i.e., $j \le k - 2$. Therefore the only nonzero coefficients are for $j = k-1$ and $j = k$: \begin{align*} r(x) &= \alpha_k\,p_k(x) + \beta_k\,p_{k-1}(x), \end{align*} where \begin{align*} \alpha_k &= \frac{\langle xp_k, p_k \rangle}{\langle p_k, p_k \rangle}, \qquad \beta_k = \frac{\langle xp_k, p_{k-1} \rangle}{\langle p_{k-1}, p_{k-1} \rangle} = \frac{\langle p_k, xp_{k-1} \rangle}{\langle p_{k-1}, p_{k-1} \rangle}. \end{align*} [/step] [step:Compute $\beta_k$ and derive the recurrence] Since $xp_{k-1}$ is monic of degree $k$, writing $xp_{k-1} = p_k + (\text{lower-order terms orthogonal to } p_k)$ gives $\langle p_k, xp_{k-1} \rangle = \langle p_k, p_k \rangle$. Hence $\beta_k = \langle p_k, p_k \rangle / \langle p_{k-1}, p_{k-1} \rangle > 0$. Rearranging $p_{k+1} = xp_k - r$ gives the stated recurrence: \begin{align*} p_{k+1}(x) &= (x - \alpha_k)\,p_k(x) - \beta_k\,p_{k-1}(x). \end{align*} [/step]