Different number of apples are given to five children such that any $3$ receive more apples than the remaining $2$

Question

A friend asked me this:

A woman gave a different number of apples to each of her five children. Any three of her children together received more apples than the remaining two children. What is the least number of apples that she could have given?

I did some 'trial and error' and I think the answer is $5$. I checked this for the cases where the least number of apples were $1, 2, 3, 4$ and $5$ and in each of those cases I chose the number of apples given to the other children to be $(x+1)$, $(x+2)$, $(x+3)$, $(x+4)$ where $x$ the least number of apples given. The reason why I chose them is because if I had chosen bigger numbers, then I'll have less chance that the sum of the lowest $3$ numbers is more than the sum of the remaining $2$ numbers (the highest).

As you can see that my line of reasoning is in no way rigorous. So my question is whether there is a way to mathematically (and without trial-and-error) solve this?

Answer 1

Let $a_1<a_2<\ldots <a_5$ be the numbers of apples given in ascending order. Then there are numbers $x_i\geq0$ $\ (1\leq i\leq 4)$ such that $$\eqalign{a_2 &=a_1+1+x_1\cr a_3&=a_2+1+x_2=a_1+2+x_1+x_2\cr a_4&=a_3+1+x_3=a_1+3 +x_1+x_2+x_3\cr a_5&=a_4+1+x_4=a_1+4+x_1+x_2+x_3+x_4\cr}\quad.$$ The condition $a_1+a_2+a_3\geq a_4+a_5+1$ leads to $$a_1\geq x_2+2x_3+x_4+5\ ,$$ which shows that $a_1=5$ is the minimum possible value. We can realize it with arbitrary $x_1\geq0$ and $x_2=x_3=x_4=0$. The minimal total number of apples is then 35.