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  "body": "\n## Proof\n\nWe double count the sum of the squares of the degrees and apply the Cauchy–Schwarz inequality.\n\n1. Let $G$ be a finite simple [[graph|0JD5C82QMB]] with vertex set $V$, $\\abs{V} = n$, and $m$ edges. Suppose that $G$ has no [[cycle|07CT9RDY7Q]] of length $3$. For $v \\in V$, write $d(v)$ for its [[degree|0T0F6CCS9W]] and $N(v)$ for its [[neighbourhood|01T3TMGTTE]].\n2. If $m = 0$, then $m = 0 \\le \\floor{n^2/4}$. From now on let $m \\ge 1$, so $n \\ge 2$.\n3. Let $xy$ be an edge. If a vertex $z$ were adjacent to both $x$ and $y$, then $z \\ne x, y$ and $x, y, z, x$ would be a cycle of length $3$. So $N(x) \\cap N(y) = \\varnothing$.\n4. Hence $d(x) + d(y) = \\abs{N(x)} + \\abs{N(y)} = \\abs{N(x) \\cup N(y)} \\le n$ for every edge $xy$.\n5. We [[double count|0F8YBHF2W3]] the sum $\\sum_{xy} \\big(d(x) + d(y)\\big)$ over all edges $xy$. A vertex $v$ is an end of exactly $d(v)$ edges, and each of them contributes $d(v)$ to the sum.\n6. So $\\sum_{v \\in V} d(v)^2 = \\sum_{xy} \\big(d(x) + d(y)\\big) \\le m n$, by Step 4.\n7. By the [[handshaking lemma|0XAP8WHNES]], $\\sum_{v \\in V} d(v) = 2m$.\n8. Apply the [[Cauchy–Schwarz inequality|0RFWW56Y22]] in $\\R^n$, with the standard inner product, to the vectors $(d(v))_{v \\in V}$ and $(1)_{v \\in V}$. This gives $\\left(\\sum_v d(v)\\right)^2 \\le n \\sum_v d(v)^2$.\n9. By Steps 6, 7 and 8, $4m^2 \\le n \\cdot m n = m n^2$.\n10. Divide by $4m \u003e 0$ to get $m \\le n^2/4$.\n11. Since $m$ is an integer, $m \\le \\floor{n^2/4}$.\n\n## Idea\n\nIn a triangle-free graph, the two ends of an edge have no common neighbour, so their degrees add up to at most $n$. Summing over edges bounds the sum of squared degrees by $mn$. Cauchy–Schwarz bounds the same sum from below by $4m^2/n$, and comparing the two gives $m \\le n^2/4$.\n",
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