# 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
$$r_1
and let $$r(ijk) := r_i+r_j+r_k$$.
Now suppose $$r(ijk)<49$$ for all triples $$\{i,j,k\}$$.
Then we have $$r(123)
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:

$$300\leq 49+48+47+43+39+38+37= 301$$

Supposing there is a counterexample, then the lowest four must add to at least $$51$$ (else the highest three add to $$50$$ or more).
Since $$14+13+12+11=50$$ the lowest four numbers would have to include one number at least equal to $$15$$ to get a total as big as $$51$$.
Then the greatest three numbers must be at least $$16+17+18=51$$, which is a contradiction to the assumption that there exists a counterexample.
The examples $$18+17+15+14+13+12+11=100$$ and $$19+16+15+14+13+12+11=100$$ show that the bound is tight.