$4$ piles of stones in to $5$ piles I wanted to ask for verification of "proof" of the following. We are given $4$ piles of stones and we form $5$ piles of stones from those. The question asks to prove that there are at least $2$ stones in a smaller pile
My attempt:
Since we are forming more piles , this suggests that there is a pile with less amount of stones , since otherwise we would have more stones than what we started with. Thus there is a pile with at least $1$ stone . But this stone came from another pile , so there is a decrease there too.
It's possible when we formed the new piles , we tried to add and compensate for the decreases pile , but in any case one of the piles has less stones , since if they had the same or more stones there would be more stones which is not true.
The proof is quite formal in the book , that's why I asked here.
Thanks
Second attempt:
Say the pile containing not less than any of the other 3 piles has f stones . Now there must a pile in the 5 piles that has less than f stones .since if there is not there is at least 5f stones which is contradictory.Now exclude this pile containing less than f stones from the system and we have 4 piles now . Now this new 4 piles must have in total less than the original set since excluded pile has at least 1 stone. Now there must be a pile that has less than one of the previous piles , sinc if each one has at least the same number of stones , then those 4 piles would have more stones than they should have. Combining those two piles , they are the ones q was looking for.(i hope)
 A: Your solution has a gap in the beginning. Suppose the four piles are 1,5,5,5, and we put them into five piles 2,2,4,4,4, which pile has "less stone"?

The solution is as follows.

 When the stones are in the 4 piles, write a fraction on each of them: For each stone in a pile of size $n$, write $\frac1n$. Note that all the written numbers sum to 4.

 Now form 5 piles. Write a new fraction on each of the stone the same way. Note that all the newly written numbers now sum to 5.

 A bunch of numbers previously summed to 4, now they summed to 5. Can you see why two of them must have grown bigger?

Usually, doing a "microscopic" analysis like you did is very hard and easy to screw up. The best way is to do it "macroscopically".
A: Suppose you have $4$ piles of stones with $n$ stones in each and now you form $5$ piles from them. Then, the newer of the $5$ piles cannot have all $\ge n$ stones because then we would have a total $5n$ stones which is $>4n$. So, now one of the piles must have $1, 2, ... n-1$ stones. Clearly, this pile has $\ge2$ and $\le n$ stones if we exclude the $n=1$ case. If this pile has $n=1$ stone then, some other pile must have a number of stones $\le n$ (since otherwise, you would have $>4n$ stones) and that smaller pile has $<n$ and $>1$ stones (so atleast two condition is satidfied).
