Prove or disprove ：f(n)+g(n) =Θ(min(f(n),g(n)))

We will prove that f(n)g(n) ≠ Θ(min(f(n),g(n))) by providing a counterexample.

Consider f(n) = n and g(n) = n^2. Then, min(f(n), g(n)) = n and f(n)g(n) = n^3.

To prove that f(n)g(n) ≠ Θ(min(f(n),g(n))), we need to show that for any positive constants c1 and c2, there exist a value n such that:

c1 * min(f(n), g(n)) > f(n)g(n) > c2 * min(f(n), g(n))

Let's assume that such constants exist. Then, we can write:

c1 * min(f(n), g(n)) > f(n)g(n)
c2 * min(f(n), g(n)) < f(n)g(n)

Substituting the values of f(n) and g(n), we get:

c1 * n > n^3
c2 * n < n^3

Dividing both sides by n^3, we get:

c1/n^2 > 1
c2/n^2 < 1

Both of these inequalities cannot be true simultaneously because as n approaches infinity, the left-hand side of the first inequality approaches zero while the left-hand side of the second inequality approaches infinity. Therefore, we have a contradiction, and f(n)g(n) ≠ Θ(min(f(n),g(n))).