Suppose that $\phi$ solves $x'' + A \left( t \right) x = 0$, is positive on $\left(0, b \right)$, and $\phi \left( 0 \right) = \phi \left( b \right) = 0$. ($A(t)$ is continuous) Is it true that $$\int_{0}^{b} \left| A \left( t \right) \right| dt > \frac{4}{b}$$ Any help or hints appreciated.
One attempt: By the MVT for integrals, for some $0 < c < b$,
$$\int_0^b \left| A \left( t \right) \right| dt = b \left| A \left( c \right) \right| \geq b \inf_{0 \leq t \leq c} \left| A \left( t \right) \right| $$ However, I think this is too restrictive.
Another attempt: since $\int_0^b \left| A \left( t \right) \right| \phi \left( t \right) dt \leq M \int_0^b \left| A \left( t \right) \right| dt$, where $M = \sup \left| \phi \left( t \right) \right|$, $$\int_0^b \left| A \left( t \right) \right| dt \geq \frac{1}{M} \int_0^b \left| A \left( t \right) \right| \phi \left( t \right) dt = \frac{1}{M} \int_0^b \left| A \left( t \right) \phi \left( t \right) \right| dt = \frac{1}{M} \int_0^b \left| \phi'' \left( t \right) \right| dt \geq \frac{1}{M} \left| \int_0^b \phi'' \left( t \right) dt \right| = \frac{1}{M} \left| \phi' \left( b \right) - \phi' \left( 0 \right) \right| $$ or, again by the MVT for integrals, $$ = \frac{b}{M} \left| \phi'' \left( c \right) \right| $$ for some $0 < c < t$. Again, I'm not sure how to turn this into a solution.
FINAL EDIT: To anyone finding this result in the future, this is a famous result by Lyapunov. One may complete my second attempt as follows: Apply Rolle's theorem to obtain $\alpha$ and $\beta$, $s$ and $t$ such that $x'(t) = -M / \alpha$ and $x' (s) = M / \beta$, and $\beta + \alpha = b$. An application of the AGM inequality finishes the job.
Too long for a comment.
You might be able to get the result by playing around with energy inequalities. For example, if you multiply the equation by $\phi^{2p+1}$ (I'm using arbitrary $p$ to illustrate that you can multiply by any monotone $f(\phi)$ more generally), for any $p\ge 0$, and integrate from $0$ to $b$, you get $$ \int_0^b\phi(x)^{2p+1}\phi''(x)dx+\int_0^bA(t)\phi(x)^{2p+2}dx=0. $$ If you integrate by parts and apply $\phi(0)=\phi(b)=0$, you get the identity $$ \int_0^bA(t)\phi(x)^{2p+2}dx=(2p+1)\int_0^b\phi(x)^{2p}\phi'(x)^2dx. $$ The integrand on the right hand side is $[\phi(x)^p\phi'(x)]^2=\frac{1}{(p+1)^2}\psi'(x)^2$, where $\psi(x)=\phi(x)^{p+1}$. By Wirtinger's inequality, you get $$ \int_0^1A(t)\phi(x)^{2p+2}dx\ge \frac{\pi^2(2p+1)}{b^2(p+1)^2}\int_0^1 \phi(x)^{2p+2}dx. $$ This seems to imply $$ \max_{[0,b]}A(t)\ge \frac{\pi^2(2p+1)}{b^2(p+1)^2}. $$ So taking $p=0$, we see that $\max A\ge 2\pi^2/b^2$.