How to prove that the eigenvalues of this Sturm-Liouville problem are all positive? I have been trying to solve the following problem:

Let the Sturm-Liouville problem defined by $p(x)y''(x)+p'(x)y'(x)-q(x)y(x)+\lambda r(x)y(x)=0$ in the interval $[a,b]$, with the periodic conditions $y(a)=y(b)$, $y'(a)=y'(b)$ and $p(a)=p(b)$. Prove that if $p(x)$, $q(x)$ and $r(x)$ are positive defined, then the eigenvalues of the Sturm-Liouville operator are positive.

My approach:
This problem is equivalently written as:

$\frac{1}{r(x)}\frac{d}{dx}\left(p(x)\frac{dy}{dx}\right)-\frac{q(x)}{r(x)}y = -\lambda y$

What I have tried is to integrate both sides of the equality after reordering the terms as following:

$\int_a^b \frac{d}{dx} \left(p(x)\frac{dy}{dx}\right) dx = \int_a^b (q(x)-\lambda r(x))y dx$

The left hand side is equal to zero according to the Sturm-Liouville conditions. It remains:

$\int_a^b q(x)ydx = \lambda\int_a^br(x)ydx$

I am not quite sure at this point if the "positive defined" condition makes those two integrals positive, making $\lambda$ positive too. Is this reasoning correct?
 A: By $p, q, r$ positive defined, I will assume it is a typos that $p, q, r$ are positive definite. i.e. $p(x), q(x), r(x) > 0$ for $x \in [a,b]$.
Start from $$(py')' - qy + \lambda ry = 0 \iff \lambda ry = q y - (py')'$$
Multipy both sides by $y$ and integrate over $[a,b]$, one obtain${}^{\color{blue}{[1]}}$
$$\require{cancel}
\begin{align}\lambda \int_a^b ry^2 dx 
&= \int_a^2 (qy^2 - y (py')') dx
= \int_a^b (qy^2 + p(y')^2 - (p yy')') dx\\
&= \int_a^b (qy^2 + p (y')^2) dx - \color{red}{\cancelto{0}{\color{gray}{\left[ p yy'\right]_a^b}}}\\
&= \int_a^b (qy^2 + p (y')^2) dx\end{align}
$$
Since $y$ is non-zero and $p, q, r$ is positive over $[a,b]$, we have${}^{\color{blue}{[2]}}$
$$\int_a^b ry^2 dx > 0
\quad\text{ and }\quad
\int_a^b (qy^2 + p(y')^2)dx \ge \int_a^b qy^2 dx > 0$$
As a result,
$$\lambda = \frac{\int_a^b (qy^2 + p(y')^2)dx}{\int_a^b ry^2 dx} > 0$$
Notes


*

*$\color{blue}{[1]}$ Since $p(a) = p(b)$, $y(a) = y(b)$ and $y'(a) = y'(b)$, we have 
$$[pyy']_a^b = p(b)y(b)y'(b) - p(a)y(a)y'(a) = 0$$

*$\color{blue}{[2]}$ - In general, if you have two continuous function $f$, $y$ with $f$ positive definite and $y$ non-zero over $[a,b]$, we will have
$$\int_a^b f y^2 dx > 0$$
Let say's $y(c) = Y \ne 0$ for some $c \in (a,b)$. Let $F  = f(c) > 0$, Choose a $\epsilon$ so small so that $(c-\epsilon,c+\epsilon) \subset (a,b)$ and
for any $x \in (c-\epsilon,c+\epsilon)$, we have  $|y(x)| > \frac{Y}{2}$ 
and $f(x) > \frac{F}{2}$. This leads to
$$\begin{align}\int_a^b fy^2 dx &= \left(\int_a^{c-\epsilon} + \int_{c-\epsilon}^{c+\epsilon}+ \int_{c+\epsilon}^b\right) fy^2 dx
\ge \int_{c-\epsilon}^{c+\epsilon} fy^2 dx\\
&\ge \int_{c-\epsilon}^{c+\epsilon} \frac18 FY^2dx
= \frac14 FY^2\epsilon > 0\end{align}
$$ The analysis where $c = a$ or $b$ is similar and I won't repeat it here.
