f(a) + f(a+1) >= 0 if f(x) = x^2 + 3x +2

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

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

We need to prove that:

f(a) + f(a+1) => 0

Let us substitute:

= a^2 + 3a + 2 + (a+1)^2 + 3(a+1) + 2

Expand Brackets:

==> a^2 + 3a + 2 + a^2 + 2a + 1 + 3a + 3 + 2

Group similars:

==> 2a^2 + 8a + 8

Factor 2:

==> 2(x^2 + 4a + 4)

==> 2(x+2)^2

It is obvious that (x+2)^2 => 0

==> 2(x^+2)^2 => 0

First, we'll calculate the values f(a) and f(a+1).

f(a = a^2 + 3a + 2

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

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

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

f(a+1) = a^2 + 5a + 6

Now, we'll put f(a) and f(a+1) in the inequality which has to be verified:

a^2 + 3a + 2 + a^2 + 5a + 6>=0

We'll combine like terms and we'll get:

2a^2 + 8a + 8>=0

We'll factorize:

2(a^2 + 4a + 4)>=0

We'll divide by 2:

(a^2 + 4a + 4)>=0

But (a^2 + 4a + 4) is the result of square expanding (a+2)^2

**So, (a+2)^2>=0 for any value of a!**

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

Therefore f(a)+f(a+1) = (a^2+3a+2)+(a+1)^2+3(a+1)+2

=a^2+3a+2+a^2+2a+1+3a+3+2

= 2a^2+8a+8

= 2(a^2+4a+4)

= 2(a+2)^2 > 0 as a2 times square is always positive.