Pigeon-hole with the sum of 3 numbers

Question

In any set consisting of exactly 7 different numbers chosen from the first 9 positive whole numbers, there are always 3 different numbers whose sum is 15. Is this true or false?

There's a follow-up question that asks if the same is true when we choose only 6 different numbers. In class, we showed the follow-up question was false by using the counter-example 1,9,6,7,3,4, and apparently, the question I've listed above (when we use 7 numbers) is true.

I'm not sure how to go about proving the first question (We're working on pigeon-hole in class, so I assume we'll have to somehow show that 6 of the numbers we choose will not produce 15 when added to two others from the group but when a 7th is chosen they will), and for the second, I was wondering if there was a more rigorous way to prove it as opposed to just randomly looking for a counter-example.

Thanks.

Answer 1

The answer to the first question is affirmative: $(8,3,4),(9,1,5),(2,6,7)$

Answer 2

HINT: Consider this $3\times 3$ magic square:

$$\begin{array}{c|c|c} \color{brown}8&\color{brown}1&6\\ \hline \color{brown}3&5&\color{brown}7\\ \hline 4&\color{brown}9&\color{brown}2 \end{array}$$

The six brown cells are significant, as is the square itself.