How would I show if I choose a set $10$ unique numbers from $0$ to $14$, there exists two numbers in the set such that their sum is greater than the largest number in the set?
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If you add any number of the set other than $0$ to the largest, the sum will be larger than the largest. Since you have picked three numbers, at most one is $0$ and you can use the largest and second largest.
I suspect the real problem is to show that if you choose $10$ unique numbers from $0$ to $14$, there are two besides the largest that sum to more than the largest. The worst case is that the largest is $14$ and the rest are (what)? And so (what)? This is more believable as the problem as $10$ is the smallest number that works.
If you take the two largest numbers, they will be both different from 0 hence their sum will be larger than the largest number.
Not a formal proof but my idea is:
Consider the worst case: Try to get the case that there does not exist such 2 numbers whose sum is greater than the biggest number of your set of 10 elements. That is; pick $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $14$. I tried to choose the set that we will pick two numbers and their sum as small as possible and our greatest one as great as possible. Now, your greatest is 14 and even in this case you can find $8$ $+$ $7$ $=$ 15 which is greater than $14$.
(You should have meant that we choose that 2 numbers that none of them is the biggest one in your 10 set)