Since $p_{k+1}$ and $x p_k$ are both monic of degree $k+1$, their difference $r(x) = x p_k(x) - p_{k+1}(x)$ 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*} For each $j$, $\langle r, p_j \rangle = \langle x p_k, p_j \rangle - \langle p_{k+1}, p_j \rangle$. The second term vanishes for $j \leq k$ by orthogonality of $p_{k+1}$ to $P_k[x]$. For the first term, the symmetry condition $\langle x p_k, p_j \rangle = \langle p_k, x p_j \rangle$ holds by hypothesis. Now $x p_j \in P_{j+1}[x]$, so $\langle p_k, x p_j \rangle = 0$ whenever $j + 1 < k$, i.e. $j \leq k - 2$. Therefore the only non-zero 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 x p_k, p_k \rangle}{\langle p_k, p_k \rangle}, \qquad \beta_k = \frac{\langle x p_k, p_{k-1} \rangle}{\langle p_{k-1}, p_{k-1} \rangle} = \frac{\langle p_k, x p_{k-1} \rangle}{\langle p_{k-1}, p_{k-1} \rangle}. \end{align*} For $\beta_k$: since $x p_{k-1}$ is monic of degree $k$, $x p_{k-1} = p_k + (\text{lower-order terms orthogonal to } p_k)$, so $\langle p_k, x p_{k-1} \rangle = \langle p_k, p_k \rangle$. Hence $\beta_k = \langle p_k, p_k \rangle / \langle p_{k-1}, p_{k-1} \rangle > 0$. Rearranging $p_{k+1} = x p_k - r$ gives the stated recurrence. $\blacksquare$