It's a follow-up of my previous question answered by user Apass.Jack nicely. See this post for motivation and background.
Conjecture :
Let $f$ be a positive increasing continuous function on $x\in[0,1]$ and $g$ be an integrable decreasing positive continuous function on $x\in(0,\infty)$. Let $$J=\left(\int_{0}^{1}\left(f\left(x\right)\right)^{2}dx\int_{0}^{1}\left(g\left(x\right)\right)^{2}dx-\left(\int_{0}^{1}f\left(x\right)g\left(x\right)dx\right)^{2}\right)\max_{x\in\left[0,1\right]}\left(g\left(x\right)\right)$$ and $$I=\left(\int_{0}^{1}\left(f\left(x\right)\right)^{2}dx\int_{0}^{1}\left(g\left(x\right)\right)^{2}dx-\left(\int_{0}^{1}f\left(x\right)g\left(x\right)dx\right)^{2}\right).$$ Then $$\left(\int_{0}^{I}g\left(t\right)dt\right)\leq J.$$
Now the usual counterexample doesn't works @Apass.jack.
Here is one example :
Choosing $g(x)=\arccos(x),f(x)=\arcsin(x)$ we have :
$$2\frac{\left(1/4(2(2+((1+\pi)(4-(\pi-3)\pi^{2}))^{\frac{1}{2}})+\pi((\pi-3)\pi\arccos(1/4(\pi-3)\pi^{2})-((1+\pi)(4-(\pi-3)\pi^{2}))^{\frac{1}{2}}))\right)}{\pi}< \left(\frac{1}{4}\left(\pi^{2}-8\right)\left(\pi-2\right)-\left(2-\frac{\pi}{2}\right)^{2}\right)$$
$$I=\int_{0}^{1}e^{-x^{2}}dx\int_{0}^{1}\left(1+\frac{1}{2}e^{-x}\right)dx-\left(\int_{0}^{1}\sqrt{e^{-x^{2}}\left(1+\frac{1}{2}e^{-x}\right)}dx\right)^{2},$$ $$\left|\int_{0}^{I}e^{-t^{2}}dt-I\right|<6\cdot10^{-7}$$
As I have no idea for a proof, here is my question.
How to (dis)prove my conjecture?
$$\max_{x \in [0, I]} g(x) \le\max_{x\in[0,\infty)}g(x)=\max_{x\in[0,1]}g(x)$$ where the last equality holds because $g(x)$ is decreasing on $[1,\infty)$.
Bunyakovsky integral inequality says $I\ge0$. As Martin R observed,
$$\int_0^I g(t)\,\text{d}t \le\int_0^I\max_{x \in [0, I]} g(x)\,\text{d}t = I \cdot \max_{x \in [0, I]} g(x) \le I \cdot \max_{x \in [0, 1]} g(x) = J.$$