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I need to show that if I have $G(V, E)$ that is simple and undirected, such that $|V|=n$ and $|E|=m$ and $m>\frac{n^2}{4}$ then there is a triangle in the graph.

I started with induction : base cause for $0,1,2$. assume this works for $n$ and try to prove for $n+1$.

Now if I have graph with $|v|=n+1$ such that $m>\frac{(n+1)^2}{4}$. I randomly choose a vertice and remove it. the maximum amount of edges it could have is $n$ ($1$ for each other vertice).

Now I'm stuck, cause I got that in the new graph after removing $1$ vertice I have $\frac{(n-1)^2}{4}$ edges, and that's not greater than $\frac{n^2}{4}$ so I can't use the induction assumption.

Where is my mistake? Can I choose the vertice a better way? *my solution: In the end I proved by removing vertex. I sperated for $2$ cases : there is vertex such that $0\le$deg($v$)$\le n/2$ , and then the new graph have $|E|> \frac{n^2}{4}+\frac{1}{4}$ edges , so the induction assumtion works .

If there is no vertex that fits the condition ,then each vertex is :$\frac{n}{2}\lt$deg($v$)$\le n$ then I choose one and delete it. After I delete, I stay with $n$ vertices, where each have at least $\frac{n}{2}-1$ edges (minus $1$ cause they maybe were connected to the vertex I deleted). Now I demand that $n(\frac{n}{2}-1)$ will be greater then $\frac{n^2}{4}$ , and this holds for all $n\gt4$. so I prove base for $3$ and $4$ and then my induction is proven. Am I right?

19021605
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user25778822
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  • Where are you getting the n^2/4+1/4 from? But ok, let's say you meant (n+1)^2/4, which would work. So the first case is good. But in the second case, you forget to divide by $2$: The number of edges is half the sum of the degrees. I think your argument works if all vertices of $G$ have degree $> n/2+1$, but there is a bit of space between this and $\leq n/2$ that I'm not sure you can easily fill. – darij grinberg Aug 19 '24 at 21:51

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Here is a hint: if $v$ is really connected to every other vertex as in the scenario you describe, then certainly it is connected to two vertices $x$ and $y$ that are connected to eachother. But then $v, x$ and $y$ form a triangle. So the only scenario we need to worry about is the case that $v$ has so few neighbors that no two of them share an edge...

Vincent
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This kind of induction works better if you remove an edge, not a vertex. See Theorem 2.4.6 in my An Introduction to Graph Theory (arXiv:2308.04512v1) for this proof in detail.

Why is removing a single vertex a bad idea? Restate the claim in the contrapositive: If $G$ has no triangles, then the inequality $\left|E\right| \leq n^2/4$ holds. In this form, the inequality can become an equality for every even $n$, but not for any odd $n$ (since $n^2/4$ is not an integer when $n$ is odd). If you remove a vertex, the even-$n$ case turns into the odd-$n$ case, so the equality case disappears. But it is hard to derive a stronger inequality from a weaker one, and certainly an inequality that becomes an equality is stronger than one that doesn't! (This reasoning doesn't completely rule out a proof along these lines -- after all, you can strengthen inequalities by multiplying them with something small, and you might be able to fully avoid using the induction hypothesis in some cases. But it shows that such a proof will certainly not be straightforward.)

The result you are proving is known as Mantel's theorem and has been discussed on m.se many times (see, e.g., Mantel's Theorem proof verification ).

darij grinberg
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