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I was working on this question and came up with this answer: Take any prime number $p$ that’s greater than $\frac{n}{2}$.

Let $a=p$ and $b=n-p$, then b is necessarily smaller than a because $p>\frac{n}{2}$ and $gcd(a,b)=1$, because $p$ is prime.

Then we found $a$ and $b$ with $a+b=n$.

Does this count as a prove or is there a smarter way? Got an exam on Monday and appreciate any help!

J3ck_Budl7y
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    Any prime less than $n$, but not dividing $n$, would be good enough as $p+(n-p)$ gets the job done. So you just need to prove that the product of the primes $<n$ is greater than $n$. And you need something like that, because if the product of all those primes did divide $n$ then no expression of the form you want would be possible. Not sure if that's easier that Bertrand's Postulate or not. – lulu Feb 23 '23 at 11:00
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    (Non standard) Induction works, and arguably that's much easier (or at least less machinery) than Bertrand's postulate or Lulu's suggestion. We even can state that one of the numbers is either 2, 3, or 6. $\quad$ Also depends on what your exam is on, and how much mathematical knowledge you have. – Calvin Lin Feb 23 '23 at 16:06

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