f(x) = x^2 + 3x + 2

For a real number m, f(m) + f(m + 1)

=> m^2 + 3m + 2 + (m + 1)^2 +3(m + 1) + 2

=> m^2 + 3m + 2 + m^2 + 1 + 2m + 3m + 3 + 2

=> 2m^2 + 8m + 8

=> 2(m^2 + 4m + 4)

=> 2*(m + 2)^2

As the square of any real number is >= 0, 2*(m + 2)^2 >= 0

**This proves that f(m) + f(m + 1) >= 0 for any real m**

We'll calculate f(m) + f(m+1):

f(m) + f(m+1) = m^2 + 3m + 2 + (m+1)^2 + 3(m+1) + 2

We'll expand the square and we'll remove the brackets:

f(m) + f(m+1) = m^2 + 3m + 2 + m^2 + 2m + 1 + 3m + 3 + 2

We'll combine like terms:

f(m) + f(m+1) = 2m^2 + 8m + 8

We'll factorize by 2:

f(m) + f(m+1) = 2(m^2 + 4m + 4)

We notice that within brackets is a perfect square:

f(m) + f(m+1) = 2(m + 2)^2

**Since the square (m + 2)^2 is always positive, except for m = -2, when it is cancelling, the sum of functions f(m) + f(m+1) >=0, for any real value of m.**