TITLE: A conference uses $4$ main languages. Prove that there is a language that at least $\dfrac{3}{5}$ of the delegates know.
QUESTION [21 upvotes]: A conference uses $4$ main languages. Any two delegates always have a common language that they both know. Prove that there is a language that at least $\dfrac{3}{5}$ of the delegates know.

Source: Romania TST 2002

My attempt:
Suppose we have $n$ people and that $l_i$ people speak language $L_i$. 
We want to prove that $l_i$ is at least $3n/5$ for some $i\in\{1,2,3,4\}$.
Say it is not true. Then we have $l_i<3n/5=:m$ for all $i$.
Because of condition we have
$$ {n\choose 2} \leq \sum_{i=1}^4{l_i\choose 2} < 4\times {m\choose
2}$$ From here we get $11n>35$ and there is no contradiction if
$n>3$.

Update (1. 20. 2019): I am now trying to develop the idea that was mentioned by Servaes in the commentary. 
Lemma: Among any $4$ delegates there are $3$ that speaks the same languague.
Proof: If somebody speaks only one languague we are done. Say there is somenone that speaks exactly 2 languagues $L_1$ and $L_2$. Then by pigeonhole out of 3 there are two in the same "hole" and we are done again. 
Suppose that everyone speaks at least 3 languague. Then by double counting between languagues and 4 people we have $$4\cdot 3\leq 4\cdot l\implies l\geq 3$$
where $l$ is maximum cardinality of $L_i$ in this double counting and we are done. $\square $
Any idea how to finish now?

REPLY [3 votes]: Here's an alternative argument. Suppose that this is not the case, and the number of people is $n$. If everyone spoke at least three languages then by double counting some language would be spoken by at least $3n/4$. So there is someone who speaks exactly two languages, A and B, say. There is some set $X$ of more than $2n/5$ who don't speak $A$, and a set $Y$ of more than $2n/5$ not speaking $B$; since everyone has to have a language in common with the first person, $X$ and $Y$ are disjoint. 
From now on we'll only check whether all pairs one from $X$ and one from $Y$ have a common language. Within $X\cup Y$ there must be at some people who don't speak C, and some who don't speak D. We can't have someone who doesn't speak C in one set and someone who doesn't speak D in the other, so all these people are in the same set, say $X$. This means everyone in $Y$ speaks both C and D, and everyone in $X$ speaks either C or D (or both). Now one of C and D is spoken by at least half of $X$ and all of $Y$, which totals more than $3n/5$, contradiction.