A lower bound for the number of triangles that contain a particular edge

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If $u \leftrightarrow v$ in a graph $G$, prove that $uv$ belongs to at least $d(u) + d(v) − n(G)$ triangles in $G$, where $n(G)=$ number of vertices in $G$.

In this question, I tried a lot. I know we are supposed to start with $2$ vertices and check the vertices they are adjacent to, but I'm unable to write the entire solution. Please help.

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Working in the direction you mentioned, let the set of vertices adj. to $u$ be $A$ and to $v$ be $B$. The set of vertices adjacent to both $u$ & $v$ $=A\cap B$. Note that $$|A\cap B|=|A|+|B|-|A\cup B|$$ $$|A|=d(u), |B|=d(v) \quad and \quad|A\cup B|\leq n(G)$$ Also, $|A\cap B|=$the no. of triangles asked since they'll formed with the vertices common to $u$ and $v$. So, $$|A\cap B|=|A|+|B|-|A\cup B|\geq d(u)+d(v)-n(G)$$