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  "body": "\n## Statement\n\nLet $n \\ge 1$ be an integer and let $\\mathcal{F}$ be a [[family|0C21Q694AG]] of subsets of $[n] = \\set{1, \\ldots, n}$.\nSuppose that $\\mathcal{F}$ is an [[antichain|03HF498P1J]] in the [[power set|00E0YZCERH]] ordered by inclusion: no member of $\\mathcal{F}$ is a proper subset of another.\nThen\n$$\\sum_{A \\in \\mathcal{F}} \\frac{1}{\\binom{n}{|A|}} \\le 1.$$\n\n## Remarks\n\nA maximal chain $\\emptyset \\subset \\set{a_1} \\subset \\set{a_1, a_2} \\subset \\cdots \\subset [n]$ corresponds to an ordering $a_1, \\ldots, a_n$ of $[n]$, so there are $n!$ of them.\nA set $A$ with $|A| = k$ lies on exactly $k! (n - k)!$ maximal chains.\nA chain meets an antichain at most once, so $\\sum_{A \\in \\mathcal{F}} |A|! (n - |A|)! \\le n!$, and division by $n!$ gives the inequality.\n\nSince $\\binom{n}{k} \\le \\binom{n}{\\floor{n/2}}$ for all $k$, the inequality gives $|\\mathcal{F}| \\le \\binom{n}{\\floor{n/2}}$, which is [[Sperner's theorem|0J87V9K1GE]].\nEquality holds for the family of all $k$-subsets of $[n]$, for any $k$.\nWithout the antichain hypothesis the sum can exceed $1$: for $\\mathcal{F} = \\set{\\emptyset, [n]}$ it equals $2$.\n\n## Explanations\n\n### L1 human\n\nEach set of size $k$ gets the weight $1/\\binom{n}{k}$, so a full layer of sets of the same size has total weight $1$.\nThe inequality says that a family in which no set contains another can never have total weight more than one full layer.\nSo sets of rare sizes are expensive, and sets of the middle size are cheap.\n\n### L2 human\n\nThe LYM inequality is the strong form of [[Sperner's theorem|0J87V9K1GE]], and its chain-counting proof is a model double-counting argument.\nIt extends to posets with a suitable chain structure, called LYM posets.\nIt is also used, through the same idea of averaging over orderings, in Katona's proof of the [[Erdős–Ko–Rado theorem|FJ92Q6VTJH]].\nThe [[Kruskal–Katona theorem|G6WNYB0WMY]] gives the exact minimum size of the shadow of a family of sets of one size.\n",
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