7 fishermen caught exactly 100 fish and no two had caught the same number of fish. Then there are three who have together captured at least 50 fish. 
$7$ fishermen caught exactly $100$ fish and no two had caught the same number of fish. Prove that there are three fishermen who have captured together at least $50$ fish.


Try: Suppose $k$th fisher caught $r_k$ fishes and that we have
$$r_1<r_2<r_3<r_4<r_5<r_6<r_7$$
and let $r(ijk) := r_i+r_j+r_k$.
Now suppose $r(ijk)<49$ for all triples $\{i,j,k\}$.
Then we have $$r(123)<r(124)<r(125)<r(345)<r(367)<r(467)<r(567)\leq 49$$
so  $$300\leq 3(r_1+\cdots+r_7)\leq 49+48+47+46+45+44+43= 322$$
and no contradiction. Any idea how to resolve this?
Edit: Actually we have from $r(5,6,7)\leq 49$ that $r(4,6,7)\leq 48$ and $r(3,6,7)\leq 47$ and then $r(3,4,5)\leq r(3,6,7) - 4 \leq 43$ and $r(1,2,5)\leq r(3,4,5)-4\leq 39$ and $r(1,2,4)\leq 38$ and  $r(1,2,3)\leq 37$ so we have:
$$300\leq 49+48+47+43+39+38+37= 301$$
but again no contradiction.

If the maximum number of fish caught is $m$, then the total number of fish caught is no more than $m+(m-1)+...+(m-6)$. So there is one fisherman that caught at least 18 fish. Repeat this process for the second and third highest number of fish caught and you should be good.
I should add that this is a common proof technique in combinatorics and graph theory. To show that something with a certain property exists, choose the "extremal" such something, and prove that property holds for the extremal object. For instance, to show in a graph where each vertex has degree at least $d$ there is a path of length at least $d$, and one proof starts by simply showing a maximal path has length at least $d$.
