Maximize the value of $(a,b) + (b,c) + (a,c)$ given is $a + b + c = 5n$; $a,b,c,n$ are positive integers
$(a,b)$ stands for the gcd of $a$ and $b$
Maximize the value of $(a.b) + (b,c) + (a,c)$
So I found that since $a \geq(a,x)$, that $5n \geq (a.b) + (b,c) + (a,c)$
And for $n=0 $ mod 3 , we can set $a=b=c$, and find the maximum. I've been trying to do the cases $n$=1 and 2 mod 3 separately, but I haven't made any progress with those cases.
 A: We are going to explore how large the fraction $\frac{\gcd(a,b) + \gcd(b,c) + \gcd(c,a)}{a+b+c}$ can be, where $a,b,c$ are any positive integers such that all of them are not equal. (We forget about the sum being $5n$).
Case $1: a=b$.
Notice $\gcd(a,b) + \gcd(b,c) + \gcd(c,a) = a + \gcd(a,c) + \gcd(c,a)$.
If $a$ is a multiple of $c$ then we get $kc + 2c = (k+2)c$ which when compared to $2a+c = (2k+1)c$ gives a ratio of at most $\frac{4}{5}$.
If $c$ is a multiple of $a$ then we get $a + 2a = 3a$ which when compared to $2a+c = a(2+k)$ gives a ratio of at most $\frac{3}{5}$.
If neither number is a multiple of another then $a + \gcd(a,c) + \gcd(c,a) \leq a + a/2 + c/2 = \frac{3a}{2} + c/2 $ which when compared to $2a+c$ gives a ratio of at most $\frac{3}{4}$

If no number is equal to another without loss of generality $a<b<c$ and now notice:
$\gcd(a,b) + \gcd(b,c) + \gcd(c,a) \leq b/2 + c/2 + a < \frac{2}{3}(a+b+c)$
So the best ratio possible is $\frac{4}{5}$ with equality if and only if $a=b=2c$, in other words if the solution is of form $(n,2n,2n)$
