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  "body": "\n## Proof\n\nWe show that a graph with the most edges among those with no large clique is a complete multipartite graph. Then we compare part sizes.\n\n*Setup.*\n\n1. Fix integers $n \\ge r \\ge 1$ and a set $V$ with $n$ elements. Call a simple [[graph|0JD5C82QMB]] on $V$ *admissible* if it has no [[clique|0G52BPNPKG]] with $r + 1$ vertices. Write $e(H)$ for the number of edges of a graph $H$, and $d_H$ and $N_H$ for [[degree|0T0F6CCS9W]] and [[neighbourhood|01T3TMGTTE]].\n2. Let $V = V_1 \\sqcup \\dots \\sqcup V_r$, where some parts may be empty, and let $n_i = \\abs{V_i}$. The complete $r$-partite graph on these parts joins two vertices exactly when they lie in different parts.\n3. This graph has $\\sum_{i \u003c j} n_i n_j$ edges. Since $n^2 = \\left(\\sum_i n_i\\right)^2 = \\sum_i n_i^2 + 2 \\sum_{i\u003cj} n_i n_j$, it has $\\frac{1}{2}\\left(n^2 - \\sum_i n_i^2\\right)$ edges.\n4. This graph is admissible. Two vertices in the same part are not adjacent, so a clique has at most one vertex in each part, hence at most $r$ vertices.\n\n*Cloning a vertex.*\n\n5. Let $H$ be a graph on $V$, and let $p \\ne q$ be nonadjacent vertices. Let $H'$ be obtained from $H$ by deleting all edges at $q$ and then joining $q$ to every vertex of $N_H(p)$. Note that $q \\notin N_H(p)$.\n6. Then $e(H') = e(H) - d_H(q) + d_H(p)$, and $q$ is still not adjacent to $p$.\n7. If $H$ is admissible, so is $H'$. Let $S$ be a clique of $H'$. If $q \\notin S$, then $S$ is a clique of $H$, because only edges at $q$ changed.\n8. If $q \\in S$, then $p \\notin S$. Each vertex of $S \\setminus \\set{q}$ is adjacent to $q$ in $H'$, hence to $p$ in $H$. So $(S \\setminus \\set{q}) \\cup \\set{p}$ is a clique of $H$ of the same size as $S$.\n9. In both cases $\\abs{S} \\le r$ when $H$ is admissible.\n\n*An extremal graph is complete multipartite.*\n\n10. There are finitely many graphs on $V$, and the empty graph is admissible. Let $M$ be the largest value of $e(H)$ over admissible $H$, and let $G^*$ be admissible with $e(G^*) = M$. Write $d = d_{G^*}$.\n11. Let $x, y, z$ be pairwise distinct vertices such that $y$ is adjacent to neither $x$ nor $z$. We claim that $x$ is not adjacent to $z$. Suppose instead that $x$ and $z$ are adjacent.\n12. If $d(y) \u003c d(x)$, cloning $x$ onto $y$ as in Step 5 gives an admissible graph with $M - d(y) + d(x) \u003e M$ edges. This contradicts the choice of $M$. So $d(y) \\ge d(x)$, and in the same way $d(y) \\ge d(z)$.\n13. Clone $y$ onto $x$: this gives an admissible graph $G_1$ with $M - d(x) + d(y)$ edges. In $G_1$, the vertex $z$ lost its neighbour $x$ and gained nothing, since $z \\notin N_{G^*}(y)$. So $d_{G_1}(z) = d(z) - 1$.\n14. In $G_1$ the vertex $y$ has the same neighbours as in $G^*$, and $y$, $z$ are still not adjacent. Clone $y$ onto $z$ in $G_1$. This gives an admissible graph $G_2$.\n15. By Step 6, $e(G_2) = M - d(x) + d(y) - (d(z) - 1) + d(y) = M + (d(y) - d(x)) + (d(y) - d(z)) + 1$.\n16. By Step 12, $e(G_2) \\ge M + 1$. This contradicts the choice of $M$, and proves the claim of Step 11.\n17. Define $u \\equiv v$ if $u = v$ or $u$ is not adjacent to $v$ in $G^*$. This relation is reflexive and symmetric. Let $u \\equiv v$ and $v \\equiv w$. If two of $u, v, w$ coincide, then $u \\equiv w$ at once. Otherwise $u \\equiv w$ by Step 11. So $\\equiv$ is an equivalence relation.\n18. Let $W_1, \\dots, W_k$ be its classes. Two distinct vertices in one class are not adjacent. Two vertices in different classes are distinct and not related, so they are adjacent. Thus $G^*$ is the complete multipartite graph with parts $W_1, \\dots, W_k$.\n19. One vertex from each class gives a clique with $k$ vertices. As $G^*$ is admissible, $k \\le r$. Add $r - k$ empty parts. Then $G^*$ is the complete $r$-partite graph of Step 2 for some part sizes $n_1, \\dots, n_r$.\n\n*Balancing the parts.*\n\n20. Suppose that $n_i \\ge n_j + 2$ for some $i, j$. Move one vertex from part $i$ to part $j$, and take the complete $r$-partite graph on the new parts. It is admissible by Step 4.\n21. The sum of squares changes by $(n_i - 1)^2 + (n_j + 1)^2 - n_i^2 - n_j^2 = 2(n_j - n_i) + 2 \\le -2$. By Step 3, the number of edges grows by at least $1$. This contradicts the choice of $M$.\n22. So the sizes $n_1, \\dots, n_r$ of $G^*$ differ pairwise by at most $1$. Let $a$ be the least of them, and let $t$ be the number of parts of size $a + 1$. Then $n = ra + t$ with $0 \\le t \\le r - 1$.\n23. Hence $a = \\floor{n/r}$ and $t = n - r\\floor{n/r}$. The same holds for the part sizes of the [[Turán graph|0VTCABADBE]] $T(n, r)$, whose $r$ parts also differ in size by at most $1$.\n24. So $G^*$ and $T(n, r)$ have the same multiset of part sizes. By Step 3 they have the same number of edges, and $M = e(T(n, r))$.\n\n*Conclusion.*\n\n25. Let $G$ be a simple graph with $n$ vertices and no clique with $r + 1$ vertices. Identify its vertex set with $V$. Then $G$ is admissible, so $e(G) \\le M = e(T(n, r))$.\n26. Let $m_1, \\dots, m_r$ be the part sizes of $T(n, r)$. Apply the [[Cauchy–Schwarz inequality|0RFWW56Y22]] in $\\R^r$ to $(m_1, \\dots, m_r)$ and $(1, \\dots, 1)$. This gives $n^2 = \\left(\\sum_i m_i\\right)^2 \\le r \\sum_i m_i^2$.\n27. By Step 3, $e(T(n, r)) = \\frac{1}{2}\\left(n^2 - \\sum_i m_i^2\\right) \\le \\frac{1}{2}\\left(n^2 - \\frac{n^2}{r}\\right) = \\left(1 - \\frac{1}{r}\\right) \\frac{n^2}{2}$.\n28. Therefore $G$ has at most $e(T(n, r))$ edges, and at most $\\left(1 - \\frac{1}{r}\\right) \\frac{n^2}{2}$ edges.\n\n## Idea\n\nIf two nonadjacent vertices have different degrees, replacing the poorer one by a copy of the richer one adds edges and creates no new clique. Pushing this to the extreme shows that a best graph is complete multipartite, with at most $r$ parts. Moving vertices from large parts to small parts adds edges, so the best parts are balanced, which is the Turán graph.\n",
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