Prove $G$ of order $n \geq 3$ is nonseparable if $\text{deg}\; u + \text{deg}\; v \geq n$ for every non-adjacent vertices $u,v \in V(G)$.

Question

Prove $G$ of order $n \geq 3$ is nonseparable if $\text{deg}\; u + \text{deg}\; v \geq n$ for every non-adjacent vertices $u,v \in V(G)$.

Here's my attempt: Suppose the contrary, that all $u-w$ paths cross $v$, (i.e it is a cut vertex). then removing $v$ will create two disjointed vertices sets $|G_1|=n_1,|G_2|=n_2$ s.t $u \in G_1$ and $w \in G_2$, and $\text{deg}\; u-1 + \text{deg}\; v-1 \geq n$.

from this point i'm not sure how to show a contradiction.

Answer 1

Using Ore's theorem One can prove such a graph contains a Hamiltonian cycle.
Since a graph with a Hamiltonian cycle is:

1. Connected
2. Does not contain cut vertex

It is non-separable.

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Edit: This was a huge overkill.

It is sufficient to prove:

Prop: Every pair of non-adjacent vertices have at least two common neighbors.

Answer 2

Looks like i finally figured this out: we would have $1 \leq w \leq n_i-1$, for $w\in G_{\{1,2\}}$ and $n_1 + n_2 = n-1$

Now by the hypothesis: $\text{deg } v + \text{deg } u \geq n \\ \implies n_1-1 + n_2-1 =n_1+n_2-2\geq n \\ n-1 \geq n+2$.

which is clearly a contradiction.

is this proof valid ?