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Prove that if a simple graph of order n has more than n^2/4 edges then it contains a triangle.

I know Martels theorem states the opposite condition for a triangle free graph but I'm not sure how to prove this condition.

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    $\begingroup$ Just to be sure, are you allowed to apply Martel's Theorem to prove what you want, or is the intent of the question to prove Martel's Theorem in the first place? $\endgroup$ – Casteels Nov 26 '14 at 17:15
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Let $G$ be a simple graph of order $n$. Then Mantel's Theorem states that:

If $G$ is triangle-free, then $G$ has at most $n^2/4$ edges.

But by taking the contrapositive of this implication, we get exactly what we want:

If $G$ has more than $n^2/4$ edges, then $G$ contains a triangle.

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Let $G$ be a graph of order $n$ and size $m$. Assume that $G$ is $K_3$-free. Then by Mantel's Theorem we know that $m\leq {n^2\over 4}$. However, this is a contradiction since we are given that $m> {n^2\over 4}$. Thus if $G$ has order $n$ and $m>{n^2\over 4}$, then $G$ contains a triangle.

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