TITLE: 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.
QUESTION [63 upvotes]: $7$ 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.

REPLY [2 votes]: Let me present still another approach, which actually expands on @Vlad 's answer.
Let's arrange, like you did, the seven people according to the captured amounts
$$
\left\{ \matrix{
  r_{\,1}  < r_{\,2}  <  \cdots  < r_{\,7}  \hfill \cr 
  r_{\,1}  + r_{\,2}  +  \cdots  + r_{\,7}  = 100 \hfill \cr}  \right.
$$
Each fisherman so ranked shall have at least one fish more than the one preceding him.
This difference bias amounts to $0+1+\cdots+6= 21 fishes$.
Take out the bias from each of them, so that we can write
$$
\left\{ \matrix{
  r'_{\,1}  \le r'_{\,2}  \le  \cdots  \le r'_{\,7}  \hfill \cr 
  r'_{\,1}  + r'_{\,2}  +  \cdots  + r'_{\,7}  = 79 \hfill \cr}  \right.
$$
i.e.: they can have also the same unbiased amount of fishes, and some in the first positions even nothing.
Now let's associate to this the sequence of the progressive sums
$$
\left\{ \matrix{
  s'_{\,0}  = 0,\;\;s'_{\,j}  = \sum\limits_{1\, \le \,j\, \le \,7} {r'_{\,j} }  \hfill \cr 
  0 = s'_{\,0}  \le s'_{\,1}  \le  \cdots  \le s'_{\,7}  = 79 \hfill \cr}  \right.
$$
Thus clearly the graph of the progressive sums can reach at most the straight line
$(0,0),\;(7,79)$ or remain below that.
The straight line is reached when the $r'$ are all equal.
That means that 
$$
s'_{\,4}  \le \left\lfloor {{4 \over 7}79} \right\rfloor  = 45
$$
But $45$ is not divisible by $4$; the extra fish cannot be given to $r'_{\,4}$ because 
that will make it greater than  $r'_{\,5}$ and shall be assigned to the last group
which will have $2$ extra fishes besides the $11$ flatly assigned to all the seven.
Since $6$ of the biased fishes were deducted from the first four, then we conclude that
$$
s_{\,4}  \le 50
$$
which demonstrates the thesis.