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.

Minimizing the problem:
$7$ dwarfs with tiny buckets - a bucket can not contain more than $7$ fish - go fishing.
As dwarfs have a strict hierarchy, no two of them may catch the same number of fish.
At evening they return with $1+2+3+4+5+6+7 = 28$ fish.  
Dwarfs are quite hungry, so the next day they use their larger $10+7$-fish-buckets,
and they return with $11+12+13+14+15+16+17 = 98$ fish.  
On their way home they find $2$ more fish. After a short battle they agree that those $2$ fish go to the dwarfs which were most successful, thus maintaining the hierarchy:  
$11+12+13+14+15+17+18 = 100$ fish  
The most fishy dwarfs caught (at least) $15+17+18 = 50$ fish 
