Can we make the inequality $A-B \leq \frac{A^2}{4}$ strict? I have an interesting problem:
Given that $$
A=\frac{1}{2022}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2022}\right)$$
$$B=\frac{1}{2023}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2023}\right)
$$
I got an upper bound on $A-B$.The Lower bound is zero, that is $A-B>0$ is easy to prove.For the upper bound , here is my analysis:
$$\text{Let}\:H=\sum_{k=1}^{2022}\frac{1}{k}$$
So we have $$2022=\frac{H}{A} \implies 2023=\frac{H+A}{A}$$
So $$B=\frac{H+\frac{1}{2023}}{2023}=\frac{\left(H^2+A H+A\right) A}{(H+A)^2}$$
Finally we get a quadratic in $H$ as:
$$(A-B) H^2+\left(A^2-2 A B\right) H+A^2(1-B)=0$$ and since $H \in \mathbb{R}$ the discriminant should be non negative.
So we have
$$(A^2-2AB)^2-4(A-B)A^2(1-B)\geq 0$$
Simplifying we get
$$A-B \leq \frac{A^2}{4}$$
Now is it possible to make it strict?
EDIT: After deep thinking i came to the conclusion that inequality should be strict:
It is evident that $A<2, B<1$
Let us assume equality holds true, that is $$A-B=\frac{A^2}{4}$$
So we get
$$(A-2)^2=4(1-B) \implies A=2-2\sqrt{1-B}$$
Now i am pretty sure that $\sqrt{1-B}$ is irrational. So that makes $A$ irrational, which contradicts that $A$ is rational. Hence the inequality is strict.

If $\;A-B=\dfrac{A^2}4\;,\;$ then , by solving the quadratic equation in $H$, we get that $\;H=2-A\;,\;$ hence ,
$H=\dfrac{4044}{2023}<2<1+\dfrac12+\dfrac68<H_8<H_{2022}=H\;.$
So, $\;A-B\neq\dfrac{A^2}4\;,\;$ consequently ,$\;\;A-B<\dfrac{A^2}4\,.$
