LYM Inequality equivalent versions I've seen the LYM inequality expressed as: Let $\mathcal{F} \subset \mathcal{P}(n)$ be an antichain.  Then
$$
\sum\limits_{F \in \mathcal{F}} \frac{1}{\binom{n}{|F|}} \leqslant 1.
$$
However, I know it as:
Let $\mathcal{F} \subset \mathcal{P}(n)$ be an antichain.  Then
$$
\sum\limits_{i=0}^n \frac{|\mathcal{F} \cap [n]^{(i)}|}{\binom{n}{i}} \leqslant 1,
$$
I'm struggling to see how these two are equivalent?  Any help appreciated, thanks!
 A: Let $[n]^{(i)}:=\{S\in P([n]) \, ; \, |S|=i\}$ where $P([n])$ is the power set of $[n]:=\{1,2,3,\dots ,n\}$. The LYM inequality is obtained by counting the maximal chains in the poset $(P([n]),\subseteq)$ (ordered by inclusion). A maximal chain is always of the type,
$$\{e_1\}\subset \{e_1,e_2\}\subset \dots \subset \{e_1,e_2,e_3,\dots ,e_n\}$$
Hence, there are $n!$ maximal chains. Let $\mathcal{F}$ be an antichain. 
A maximal chain can contain at most one member of $\mathcal{F}$. We can can count the number of maximal chains containing some $F\in \mathcal{F}$ in two different ways.


*

*Suppose $\mathcal{F}$ contains $f_i$ number of subsets of size $i$. A
particular subset of size $i$ sits in a maximal chain at exactly
$i$th place from left. This leaves us with $i!(n-i)!$ ways to form
the chain. Totally, there are $f_ii!(n-i)!$ chains containing subsets
of size $i$. Summing over $i$, we get, $$\sum_{i=1}^{n}f_ii!(n-i)!\le
   n!$$ Noticing that $f_i=|\mathcal{F}\cap [n]^{(i)}|$ and rewriting
the previous inequality we get,
$$\sum_{i=1}^{n}\frac{|\mathcal{F}\cap [n]^{(i)}|}{\binom{n}{i}}\le
   1$$

*Let $F\in \mathcal{F}$. Then, $|F|!(n-|F|)!$ maximal chains contain
$F$. Hence, $$\sum_{F\in \mathcal{F}}|F|!(n-|F|)!\le n!$$ We
rearrange this to get the familiar looking form of LYM inequality,
$$\sum_{F\in \mathcal{F}}\frac{1}{\binom{n}{|F|}}\le 1$$


Hence, both the inequalities are equivalent.
