Verify if f(m)+f(m+1)>=0, if m is a real number?f(x)=x^2+3x+2

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.