Upper diameter bound of simple and connected graphs with $dm(G)\geq 3$

Question

Given a simple, undirected graph $G=(V,E)$. Let $dm(G)$ denote the diameter of $G$ and $\delta(G)$ denotes the min-degree of $G$.

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Claim: If $G$ is connected and $dm(G)\geq 3$ then $dm(G)\leq|V|-2\delta(G)+1$

Proof by contradiction (EDIT2: Proof incorrect!):

Assume $dm(G)>|V|-2\delta(G)+1$. Because $G$ is connected the min-degree is at least 1, that is $\delta(G)\geq 1$. Therefore we have: $dm(G) > |V|-2+1=|V|-1$.

Furthermore, the diameter $dm(G)$ is naturally bounded by $n-1$ which is the longest path in $G$, so $dm(G)\leq n-1$.

Hence we have: $|V|-1

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EDIT1: When $G$ is connected I cannot infer $\delta(G)\geq 1$ due to e.g. having 2 complete triangles - not connected. . Thus my proof is incorrect. I might need help with the proof.

Moreover, for the cases $dm(G)=2$ and $dm(G)=1$ there are counterexamples: $K_3$ and $C_4$.

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EDIT3:

Proof: We show the equivalent inequality $dm(G)+2\delta(G)-1 \leq |V|$. Let $N(x):=\{v\in V|xv\in E\}$ denote then neighbors of $x\in V$.

Let $a,b\in V$ be the vertices such that $dm(G) = d(a,b)$.

dm(G)-1 denotes the amount of vertices on the path $a-b$ without both end-vertices, i.e. the amount of vertices between $a$ and $b$. Furthermore $|N(a)|=\deg(a)\geq \delta(G)$ and $|N(b)| = \deg(b)\geq\delta(G)$.

We therefore have $dm(G)-1+2\delta(G) \leq dm(G)-1+|N(a)|-1+|N(b)|-1$. Because $dm(G)-1$ already counts exactly 2 vertices which are each neighbor of $a$ and $b$ the inequality holds. In addition $N(a)\cap N(b) = \emptyset$ because we have $dm(G)\geq 3$.

Now we can easily add the vertices $a$ and $b$ and get $dm(G)-1+|N(a)|-1+|N(b)|-1 \leq dm(G)-1+|N(a)|+|N(b)|$.

Can I now infer that the last term is lower or equal to $|V|$ due to the fact that I have counted exactly the minimum amount of vertices given $dm(G)\geq 3$ and considered the neighborhoods of the end-vertices $a$ and $b$?