Let $g$ be a piecewise smooth, zero average, function over $[0,1]$ such that $\min g^2>0$. I would like to show that
$$
\int\_0^1 g\sqrt{1-r/g^2}\int\_0^1 \frac{1}{g\sqrt{1-r/g^2}} \leq 1
$$
for all $r \in \mathopen[-1,\min g^2\mathclose[$. I don't know that this is true but I am persuaded that it is, based on intuition (see motivation below), 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. With that in mind, I've tried to apply something like Cauchy's arm lemma, to no avail.


The question is [cross-posted](https://math.stackexchange.com/questions/4587970/integral-inequality-prove-int-01-f-int-01-1-f-leq-1-for-some-function-f).


**For future generations:** Inequality disproven thanks to @fedja's answer. To construct a counter example consider the case where $g$ takes three values $g\_{1,2,3}$ over three intervals of lengths $p\_{1,2,3}$. Suppose $g\_{1,2}>0$ and $g\_3<0$. This case is enough by @Iosif's observation. Also, let $f=\sqrt{g^2-1}$. Now consider the limit $p\_1\to 0$, $g\_1\to 1$. To disprove the inequality, we would also like for $p\_1/f\_1$ to go to infinity. Thus, take $g\_1=1+p\_1^3$ for instance. Then, the inequality reads, to leading order, $(p\_2f\_2-p\_3f\_3)p\_1/f\_1<1$. Disproving the inequality amounts to finding $p\_{2,3}$ and $g\_{2,3}$ such that $p\_2+p\_3=1$ and $p\_2g\_2+p\_3g\_3=0$ and such that $p\_2f\_2-p\_3f\_3>0$. It suffices to take $g\_2>-g\_3>1$ and $p\_2=-g\_3/(g\_2-g\_3)$ and $p\_3=g\_2/(g\_2-g\_3)$. For instance, $g\_2=3$, $g\_3=-2$, $p\_1=0.004$ appear to do the trick.



Integral inequality: Prove $\int_0^1 f\int_0^1 1/f \leq 1$ for a certain function $f$
Matematika, shmatematika: any minimally decent CAS should immediately detect and tell the human operator that, for $r=1$, the inequality reads
$$
(\int\_X F-\int\_Y G)(\int\_X 1/F-\int\_Y 1/G)\le 1
$$
where $F=\sqrt{f^2-1}, G=\sqrt{g^2-1}$, $f,g> 1$, $\mu(X)+\mu(Y)=1$, and $\int\_X f=\int\_Y g$. Now it becomes obvious that the inequality can easily fail: Take any $f,g$ with the expression in the first parentheses not $0$. That first difference doesn't feel small moving of the distribution of $f$ or $g$ but the second one is extremely unstable: if you move a tiny portion of values of $f$ or $g$ sufficiently close to $1$ without changing the full integral, you can get arbitrarily large size and any sign you want.


The second part (negative r) is true and equally trivial:
Let $r=-1$ (scaling) and $F=\sqrt{f^2+1}, G=\sqrt{g^2+1}$. Then
$|F-f|,|G-g|\le 1$, so the first difference is at most $\mu(X)+\mu(Y)=1$ in absolute value (after subtracting integrals of $f$ and $g$ that compensate each other) and the second one is trivially bounded by the same quantity (since the denominators are at least $1$). So we get the bound $1$ for the absolute value of the product.


I would *love* to see an *intelligent* computer program for solving elementary inequalities and do not see any principal obstacles to creating it and even have some ideas about how this task could be possibly accomplished, but, alas, what is around is still nowhere close :-(.


