WOLOG, let us assume $\alpha = 0$ and I'll only consider the case $n > 2$.
To shorten expressions, we will use a suffix
$0$ to denote the value of a polynomial evaluate at $0$.
Differentiate the ODE
$$ p(x)y'' + q(x)y' + r(x) y = 0,\tag{*1}$$
one more times and using the assumption $y_0 \stackrel{def}{=} y(0) = 0$, it is easy to show
$$\begin{cases}
y_0'' &= -\frac{q_0}{p_0} y_0'\\
y_0''' &= -\frac{1}{p_0}\left( (q_0 + p_0')\left(-\frac{q_0}{p_0}\right) + (r_0 + q_0')\right) y_0'
\end{cases}\tag{*2}$$
Let $\lambda_1, \lambda_2, \cdots, \lambda_{n-1}$ be the non-zero roots of $y(x)$.
Consider the polynomial
$$g(x) = \prod_{k=1}^{n-1} (x - \lambda_k) = x^{n-1} - \sigma_1 x^{n-2} + \cdots + (-1)^{n-2} \sigma_{n-2} x + (-1)^{n-1} \sigma_{n-1}$$
Up to a sign, the coefficients of $g(x)$ is the $k^{th}$ elementary symmetric polynomial for the list of $n-1$ numbers $\lambda_1,\ldots,\lambda_{n-1}$. i.e.
$$\sigma_k \stackrel{def}{=} \prod_{1\le i_1 < i_2 < \cdots < i_k \le n-1} \lambda_{i_1}\lambda_{i_2}\cdots\lambda_{i_k}$$
Let
$$S_k = \begin{cases}
1, & k = 0\\
\frac{\sigma_k}{\binom{n-1}{k}}, & 0 < k < n
\end{cases}
$$
be the corresponding elementary symmetry means. It is easy to check
$$\begin{cases}
S_{n-1} &= (-1)^{n-1} y_0'\\
S_{n-2} &= \frac{(-1)^{n-2}}{2(n-1)} y_0''\\
S_{n-3} &= \frac{(-1)^{n-3}}{3(n-1)(n-2)} y_0'''
\end{cases}\tag{*3}
$$
By assumption, all the $\lambda_i$ are real and distinct. By
Newton's inequalities, we have
$$S_{n-2}^2 \ge S_{n-1}S_{n-3}\tag{*4}$$
Substitute $(*2)$ and $(*3)$ into $(*4)$, this leads to
$$\left(\frac{-\frac{q_0}{p_0}y_0'}{2(n-1)}\right)^2
\ge
\frac{-\frac{1}{p_0}\left( (q_0 + p_0')\left(-\frac{q_0}{p_0}\right) + (r_0 + q_0')\right) (y_0')^2}{3(n-1)(n-2)}
$$
Notice $y_0', p_0 \ne 0$, we can cancel them in above expression and get
$$\begin{align}
& 3(n-2) q_0^2 \ge 4(n-1) \left( (q_0 + p_0') q_0 - (r_0 + q_0') p_0\right)\\
\iff &
4(n-1)\left((r_0 + q_0')p_0 - p_0' q_0\right) - (n+2) q_0^2 \ge 0
\end{align}
$$
