A set system $\mathcal{A}\subseteq\mathcal{P}([n])$ has the property that for any distinct $A,B\in\mathcal{A}$ we have $|A-B|>1$ and $|B-A|>1$. By considering the shadow of $\mathcal{A}$ show that \begin{equation}\sum_{r=1}^n\frac{r|\mathcal{A}_r|}{{n}\choose{r-1}}\le 1,\end{equation} where $\mathcal{A}_r:=\{A\in\mathcal{A}:|A|=r\}$.
My idea is to try and modify the LYM inequality to get this result. The LYM inequality says that \begin{equation}\sum_{r=1}^n\frac{|\mathcal{A}_r|}{{n\choose r}}\le 1.\end{equation} So, observing that \begin{equation}\frac{r|\mathcal{A}_r|}{{n\choose{r-1}}}=\frac{r!(n-r+1)!|\mathcal{A}_r|}{n!},\end{equation} if we could somehow show that our condition on $|A-B|,|B-A|$ means we over/under count $n-r+1$ things at each step, we would be done. But I have no ide ahow to do this, nor how to use shadows to dedude the result.
The shadow of $\mathcal{A}_r$ are sets of size $r-1$ that are in one of the sets in $\mathcal{A}_r$ i.e., $$\delta \mathcal{A}_r = \{A\in \binom{[n]}{r-1}:\exists B\in \mathcal{A}_r\text{ and }A\subset B\}.$$ Consider then the union of all those, say $$\delta A = \bigcup _{r\in [n]}\delta \mathcal{A}_r,$$ If we show that $\delta A$ doesn't contain two sets s.t one is inside the other, we can use the LYM inequality on $\delta A$, we have that $$\sum _{r \in [n]}\frac{|\delta \mathcal{A}_r|}{\binom{n}{r-1}}\leq 1.$$
Consider $A,B\in \delta \mathcal{A}$ and suppose that $A\subset B$, so there are $x,y\in [n]$ such that $A\cup \{x\}$ and $B\cup \{y\}\in \mathcal{A}$, and so $|A\cup \{x\} \setminus (B\cup \{y\})|>1,$ but we have that $|A\setminus B|=0$ and so $|A\cup \{x\}\setminus B|\leq 1$ which is a contradiction.
The only thing left to show is that $|\delta \mathcal{A}_r|\geq r|\mathcal{A}_r|$. Consider the function $$\varphi : [r]\times \mathcal{A}_r \longrightarrow \delta \mathcal{A}_r,$$ given by $\varphi (i,A)=A\setminus \{ a_i \}$, where $a_i$ is the $i$-th element of $A$ ordered with the usual order of numbers. If $\varphi (i_1,A_2)=\varphi(i_2,A_2)$, then $|A_1|=|A_2|$ and if $A_1\neq A_2$, then $|A_1\setminus A_2|=1$ which is a contradicton, and so $A_1=A_2$ and clearly $i_1=i_2$, so $\varphi$ is injective and the inequality holds.