$7$ 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.

**Answer**

Let’s work with the lowest four numbers instead of the other suggestions.

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.

**Attribution***Source : Link , Question Author : Aqua , Answer Author : Malady*