bulgarian Mathematical olympiad

[jamescollis]

It's also the case that the arithmetic mean is greater than the harmonic mean. This gives k^2<=9 almost immediately

[rafael]

I think that for k=3, must be proven that there is not more solutions that (3, 3, 3). It is not obvious.

[riadsouissi]

Or without using AM-GM

x1+..xk=9 implies xi in [1, 9] for positive integers

1/x1+..1/xk=1 implies
xi cannot be 1 (1/1=1 => k=1 but 1 not = 9)
xi cannot be 5 or 7, there has to be at least another xi with 5 or 7 as factors to produce a sum of 1

So only valid xi are 2, 3, 4, 6, 8
9-xi needs to be in that list as well, but 9-xi is 7, 6, 5, 3, 1 out of which 7, 5, 1 are not permitted
So only valid xi are 3, 6

With these 2 numbers, 9 can be written as

3+3+3 which verifies 1/3+1/3+1/3=1

3+6 which does not verify 1/3+1/6=1