Show that there aren't negative eigenvalues. I've been trying to solve this Sturm-Liouville theory problem.

Show that the problem:
$$\left\{\begin{matrix} y''+(x+\lambda)y = 0\\ y(0)=0\\y(1)=0\end{matrix}\right.$$
doesn't have nontrivial solutions if $\lambda<0$.

It actually asks to show that the operator $L[y]:=y''+xy$ just doesn't have negative eigenvalues. I've tried the trick of multiplying the equation by $y$, and integrating, but it doesn't seem to work.
An alternative approach or something will be thanked.
 A: What you want to be able to show is that, if $y$ is a solution of the equation,
$$
              0 \le \int_{0}^{1}(-y''(t)-ty(t))y(t)dt = \lambda \int_{0}^{1}y(t)^2dt.
$$
After integrating by parts, assuming $y(0)=y(1)=0$, you must show
$$
                 \int_{0}^{1}ty(t)^2dt \le \int_{0}^{1}y'(t)^2dt.
$$
This inequality holds more generally for a $C^2$ function with $y(0)=y(1)=0$ because the Cauchy-Schwarz inequality gives
\begin{align}
    \int_{0}^{1}ty(t)^2dt & = \int_{0}^{1}t\left(\int_{t}^{1}y'(s)ds\right)^{2}dt \\
   & \le \int_{0}^{1}t\int_{t}^{1}y'(s)^2ds\int_{t}^{1}1^2ds\,dt \\
   & = \int_{0}^{1}t(1-t)\int_{t}^{1}y'(s)^2ds \,dt \\
   & = \left.\left(\frac{t^2}{2}-\frac{t^3}{3}\right)\int_{t}^{1}y'(s)^2ds\right|_{t=0}^{1}+\int_{0}^{1}\left(\frac{t^2}{2}-\frac{t^3}{3}\right)y'(t)^2dt \\
   & = \int_{0}^{1}\left(\frac{t^2}{2}-\frac{t^3}{3}\right)y'(t)^2dt.
\end{align}
The function $f(t)=\frac{t^2}{2}-\frac{t^3}{3}$ satisfies $f(0)=0$ and $f'(t) > 0$ for $0 < t < 1$. Therefore, $f(t) \le f(1)=\frac{1}{6}$ for $0 \le t \le 1$. Hence,
$$
    \int_{0}^{1}ty(t)^2dt \le \frac{1}{6}\int_{0}^{1}y'(t)^2dt \\
    -\int_{0}^{1}ty(t)^2dt \ge -\frac{1}{6}\int_{0}^{1}y'(t)^2dt.
$$
Because of this, any real solution $y$ of the eigenvalue equation with eigenvalue $\lambda$ must satisfy
\begin{align}
  \lambda\int_{0}^{1}y(t)^2dt & = \int_{0}^{1}-y''(t)y(t)dt-ty(t)^2dt \\
   & = \left.-y'(t)y(t)\right|_{t=0}^{1}+\int_{0}^{1}y'(t)^2dt - \int_{0}^{1}ty(t)^2dt \\
   & = \int_{0}^{1}y'(t)^2dt - \int_{0}^{1}ty(t)^2dt \\
   & \ge \int_{0}^{1}y'(t)^2dt - \frac{1}{6}\int_{0}^{1}y'(t)^2dt \\
   & =\frac{5}{6}\int_{0}^{1}y'(t)^2dt \ge 0.
\end{align}
Therfore $\lambda \ge 0$, and $\lambda=0$ iff $y'\equiv 0$, which forces $y=0$ because $y(0)=0$ for such a solution. Hence, all eigenvalues are strictly positive.
