Problem 1202

Prove that when $n=2$, $n^{2}+1=5$, these 4 points $A, B, C, D$ are connected by 5 line segments, only $C_{4}^{2}-5=$ 1 pair of points are not connected, let's assume only $C$ and $D$ are not connected, thus there exist two triangles: $\triangle A B C$ and $\triangle A B D$, the conclusion holds.

Assume that when $n=k$, the conclusion holds. When $n=k+1$, we first prove that there is at least one triangle. Let $A B$ be a connected line segment, and let the number of line segments drawn from $A, B$ to the remaining $n-2=2 k$ points be $a$ and $b$ respectively.
(1) If $a+b \geqslant 2 k+1$, then among the remaining $2 k$ points, there must be a point $C$ that is connected to both $A$ and $B$, thus $\triangle A B C$ exists.
(2) If $a+b \leqslant 2 k$, then removing points $A, B$ and the line segments from $A, B$ (including $A B$), there are still $2 k$ points, and the number of line segments between them is at least $(k+1)^{2}+1-(2 k+1)=k^{2}+1$ line segments. Therefore, by the induction hypothesis, there must be a triangle.

Let $\triangle A B C$ be a triangle formed by the connected line segments, and let the number of line segments drawn from $A, B, C$ to the remaining $2 k-1$ points be $\alpha, \beta, \gamma$ respectively.
(1) If $\alpha+\beta+\gamma \geqslant 3 k-1$, then among the remaining $2 k-1$ points, at least $(3 k-1)-(2 k-1)=k$ points are connected to both endpoints of one of the line segments $A B, B C, C A$, forming at least $k$ triangles, plus $\triangle A B C$, there are at least $k+1$ triangles in total.
(2) If $\alpha+\beta+\gamma \leqslant 3 k-2$, i.e., $(\alpha+\beta)+(\beta+\gamma)+(\gamma+\alpha) \leqslant 6 k-4$, then at least one of $\alpha+\beta, \beta+\gamma, \gamma+\alpha$ is no more than $2 k-2$. Assume $\alpha+\beta \leqslant 2 k-2$, at this time, removing points $A, B$ and the line segments from $A, B$ (including the three sides of $\triangle A B C$), there are still $2 k$ points, and the number of line segments between them is at least $(k+1)^{2}+1-[(2 k-2)+3]=k^{2}+1$. Thus, by the induction hypothesis, there are at least $k$ triangles formed by the connected line segments, plus $\triangle A B C$, there are at least $k+1$ triangles in total. Therefore, when $n=k+1$, the conclusion holds, completing the induction proof.