Sum of two co-prime integers I need some help in a proof:
Prove that for any integer $n>6$ can be written as a sum of two co-prime integers $a,b$ s.t. $\gcd(a,b)=1$.
I tried to go around with "Dirichlet's theorem on arithmetic progressions" but didn't had any luck to come to an actual proof.
I mainly used arithmetic progression of $4$, $(4n,4n+1,4n+2,4n+3)$, but got not much, only to the extent of specific examples and even than sometimes $a,b$ weren't always co-prime (and $n$ was also playing a role so it wasn't $a+b$ it was $an+b$).
I would appriciate it a lot if someone could give a hand here.

Later:  the numbers between $1$ and $n-1$ that are relatively prime to $n$ itself come in pairs that add up to $n$ and are relatively prime to each other as well. If $n=5$ or $n \geq 7$ both such numbers can be chosen strictly larger than $1.$
Original:
A different emphasis: if Euler's totient $\phi(n) \geq 3,$ then there is some integer $a$ with $\gcd(a,n) = 1$ and $1 < a < n-1.$ If we then name $b = n-a,$ we find that $\gcd(a,b) = 1$ as well, since a prime $p$ that divides both $a,n-a$ also divides $n,$ and this contradicts $\gcd(a,n) = 1.$
So, when is $\phi(n) \geq 3 \; ? \; \;$ If $n$ is divisible by any prime $q \geq 5,$ then $\phi(n)$ is a multiple of $\phi(q) = q-1,$ and that is at least $4.$
Next, if $n = 2^c \; 3^d \; . \;$ When $d=0$ we find $\phi(n) = 2^{c-1}$ is at least $3$ when $c \geq 3,$ leaving $2,4$ out. When $c=0$ we find $\phi(n) = 2 \cdot 3^{d-1}$ is at least $3$ when $d \geq 2,$ leaving $3$ out. When $c,d \geq 1,$ we find $\phi(n) = 2^c \cdot 3^{d-1}$ is at least $3$ when either $c \geq 2$ or $d \geq 2,$ so this leaves out $6.$
Put it together, for $n=5$ or $n \geq 7,$ there is some $a$ with $1 < a < n-1$ and $\gcd(a,n) = 1.$ 
