Let $g$ be a piecewise-smooth, zero-average function over $[0,1]$ such that $\min g^2 > 0$. I would like to show that
$$ \int_0^1 g\sqrt{1 - \frac{r}{g^2}} \int_0^1 \frac{1}{g\sqrt{1 - \frac{r}{g^2}}} \leq 1 $$
for all $r \in [-1,\min g^2[$. I don't know that this is true but I am persuaded that it is, based on numerical tests and on a couple of, admittedly trivial, particular cases (e.g., $g^2$ is constant).
If it helps: note that second integral is almost the derivative of the first with respect to $r$.
Also, one could rescale $g$ to get rid of $r$. Say $0<r<\min g^2$. Let $g_r=g/\sqrt{r}$ so that $\min g_r^2>1$. Then, we're after $$ \int_0^1 g_r\sqrt{1-1/g_r^2}\int_0^1 \frac{1}{g_r\sqrt{1-1/g_r^2}} \leq 1. $$ We should be able to do something similar for $r<0$.
I am aware of an inequality $E(X)E(1/X)\geq 1$ for $X$ positive. Here, $g$ has zero average meaning it can't be positive.
Motivation: I am trying to show that a certain linkage (something like a carpenter's ruler) grows in span. The problem boils down to the above inequality.
