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.
Consider the diagram below.
Suppose we want to find the number of elements in the union of sets $A$, $B$, and $C$.
If we simply add the number of elements in sets $A$, $B$, and $C$, then we count each element that is in two sets twice, once for each set to which it belongs. We only want to count such elements once, so we subtract $|A \cap B| + |A \cap C| + |B \cap C|$ from $|A| + |B| + |C|$. However, if we do that we will not have counted the elements in $A \cap B \cap C$ at all since we added them three times, once for each set to which they belong, and subtracted them three times, once for each pair of sets to which they belong. Therefore, we need to add $|A \cap B \cap C|$ to the total. Doing so yields
$$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$$
Let $C$ be the set of employees who speak Chinese; let $M$ be the set of employees who speak Malay; let $T$ be the set of students who speak Tamil. We are given
\begin{align*}
|C \cup M \cup T| & = 36\\
|C| & = 24\\
|M| & = 20\\
|T| & = 7\\
|C \cap M| & = 8\\
|C \cap T| & = 3\\
|M \cap T| & = 6
\end{align*}
You need to find $|C \cap M \cap T|$ and $|C^C \cap M^C \cap T|$. Can you proceed?
Best Answer
Assuming that all participants at least did one session, I get that everything is determined uniquely:
$|A \cup B| = |A| + |B| - |A \cap B|$, where $A$ are the morning session participants, $B$ the afternoon ones. The union is 80. And we know |A| = 61, |B| =43, so $80 = 104 - |A \cap B|$, so 24 people did both sessions.
Now $A \setminus B = 61 - 24 = 37$ and $B \setminus A = 43 - 24 = 19$. So 37 only attended the morning sessions, and 19 only the afternoon ones.
I'm not sure how to arrive at your target answers, or your own. Could you expand on that?