This very confusing. First you say "let a and b be two numbers greater or equal than 1 from the set of real numbers R"
But then you say "there is always a and b such that equation is true: n=a+b+1". You cannot select a and b first (which is what your first sentence means) and then say "there exist a and b such that".
I think what you mean is "If n is any real number larger than 3 then there exist real numbers a and b, larger than 1, such that n= a+ b+ 1".
That's obviously true but I can see no reason to use "reductio ad absurdum" (proof by contradiction). If n> 3, then n- 1> 2. Let a= b= (n-1)/2> 1. Then n-1= a+ b so n= a+ b+ 1.
(Of course, there exist an infnite number of choices for a and b. To show "there exists" you only need to demonstrate one.)