This is known as Dirac's theorem. I don't know if an inductive proof is possible; the standard proof is extremal. You consider the longest sequence of friends and show by contradiction that it must be a Hamiltonian circuit. I can edit with more details if you want the full argument; it's quite nice.
Edit: the argument is as follows. It can be found, for example, in Bollobas' Modern Graph Theory (p. 106). First let's establish the graph-theoretic language. Let $G$ be a graph on $n$ vertices, $n \ge 3$. For each $v \in G$ let $d_v$ denote the number of neighbors of $v$. Then the claim is that if $d_v \ge \frac{n}{2}$ for each $v \in G$, then $G$ is Hamiltonian.
We claim that $G$ is connected. Indeed, if $v, w \in G$ then $v$ and $w$ each have $\frac{n}{2}$ neighbors out of a possible $n-1$, so they must share some neighbor. (This step works under the weaker assumption that $d_v + d_w \ge n$ for each $v \neq w$.)
Now suppose that $G$ is not Hamiltonian. Let $v_1, ... v_k$ be a path all of whose vertices are distinct with $k$ maximal. Then $v_1$ and $v_k$ have no neighbors outside the path (by maximality), so all of their neighbors must be one of $v_1, ... v_k$. In addition, $v_1$ cannot be connected to $v_k$; otherwise we would have a cycle of length $k$, and by connectivity this cycle would be connected to some $w$ not in the cycle, giving a path of length $k+1$, which contradicts maximality.
More generally, $v_1$ cannot be connected to a vertex $v_{i+1}$ with the property that $v_i$ is connected to $v_k$, since otherwise $v_{i+1} v_1 v_2 ... v_i v_k v_{k-1} ... v_{i+2}$ is a cycle of length $k$. It follows that the sets $\{ v_i : v_1 \sim v_i \}$ and $\{ v_{i+1} : v_i \sim v_k \}$ are disjoint subsets of $\{ v_2, ... v_k \}$, but by assumption their union must have size at least $n$ (this step also works under the weaker assumption that $d_v + d_w \ge n$ for each $v \neq w$); contradiction.
Take complete graph $K_8$ on $8$ vertices, whose edges are colored red, if two vertices $\{i,j\}$ are friends, and blue, if $\{i,j\}$ are enemies. Looking at one person $v$ in particular, who has $6$ friends within the group.
These 6 people he is friends with now make up a complete graph $K_6$ on $6$ vertices. As suggested in the comments, this graph is known to have either a red $K_3$ or a blue $K_3$, since $R(3,3)=6$. Therefore there is either a blue triangle, in which case we'd be done, or there is a red triangle. In the event that there is a red triangle, it joins with the vertex that wasn't friends with our original $v$. Therefore we are done.
Best Answer
Recall the handshake lemma: the number of odd vertices must be even. If $|G| = 9$, then take an even vertex $v$ (one must exist since otherwise there are $9$ odd vertices). $d(v)$ must be one of $0,2,4,6,8$.
If $d(v) \ge 6$, then take all the neighbors of $v$. Since $R(3,3) = 6$, then there must be an independent set of $3$ vertices (in which case we are done) or $3$ vertices which are mutually adjacent, so form a clique of order $4$ with $v$.
If $d(v) \le 4$, then take all $4$ non-neighbors of $v$. Either these form a clique of order $4$ or there are two non-adjacent vertices, which together with $v$ form an independent set of order $3$.
This proves that $R(3,4) \le 9$.