Consider the function $F \in C^{2}([0,L])$ which satisfies the eigenvalue problem $$F''(x) = -\lambda F(x)$$
and suppose that $F$ satisfies the following constraints on its boundary values $$F(L)F'(L) \leq F(0) F'(0)$$
Show that $\lambda \geq 0$.
Is there a straightforward way to approach this without having to deal with hyperbolic sines and cosines? My first thought is that if $\lambda < 0$, then $A\cosh(\sqrt{|\lambda|}) + B \sinh(\sqrt{|\lambda|})$ would be a solution for some $A,B$. But I don't know much about hyperbolic trig functions, so following this approach would lead me to pushing definitions that I have no intuition for.
Is there a straightforward way to prove this that doesn't rely on these things; if not, is there a straightforward way to make my approach work?
Hint: Notice that $$ 0 \leq F(0)F'(0) - F(L)F'(L) = -\int_0^L \frac{d}{dx}\big( F(x) F'(x)\big) \, dx = \int_0^L \big[ \lambda F(x)^2 - F'(x)^2\big] \,dx. $$