Use the formal definition of Big-Oh to prove that if f (n) and g(n) are nonnegative functions such that f (n) = O(g(n)), f (n) + g(n) = Ω(g(n)).
By the definition of Big-Oh:
If f(n) and g(n) are non-negative functions such that f(n) = O(g(n)) there must be positive constants c and n_0 such that 0 ≤ f(n) ≤ c(g(n)) for all n ≥ n_0.
My approach so far: If 0 ≤ f(n) ≤ c(g(n)) and f(n) and g(n) are non-negative functions we know c(g(n) ≤ c( f(n) + g(n) ) we can rewrite the original inequality as:
0 ≤ f(n) ≤ c( (f(n) + g(n) )
0 ≤ (1/c)f(n) ≤ f(n) + g(n)
But I've reached a dead end here.
Starting from $$0 \le f(n) \le \text{c}g(n)~~\text{for all $n\ge n_0$}$$ we have $$f(n) \ge 0, ~\text{c}g(n) \ge 0 \implies \text{c}g(n) \le \text{c}(f(n)+g(n))~~\text{for all $n\ge n_0$}$$ and as $c$ is a positive constant, we can divide by $c$ to form $$f(n)+g(n)\ge 1*g(n)~~\text{for all $n\ge n_0$}$$