You have $\displaystyle \displaystyle \frac{3}{a + 1} \geq 3$.
First, note that $\displaystyle \displaystyle a + 1 \neq 0 \implies a \neq -1$.
In order to solve for $\displaystyle \displaystyle a$, you need to consider two cases, since multiplying or dividing by a negative number reverses the inequality sign...
Case 1: $\displaystyle \displaystyle a + 1 < 0 \implies a < -1$.
Then $\displaystyle \displaystyle \frac{3}{a + 1} \geq 3$
$\displaystyle \displaystyle 3 \leq 3(a + 1)$
$\displaystyle \displaystyle 1 \leq a + 1$
$\displaystyle \displaystyle 0 \leq a$.
So in this case, you are told $\displaystyle \displaystyle a < -1$ and $\displaystyle \displaystyle a \geq 0$. Since there are not any values of $\displaystyle \displaystyle a$ that satisfy both conditions, there are not any solutions in this case.
Case 2: $\displaystyle \displaystyle a + 1 > 0 \implies a > -1$.
$\displaystyle \displaystyle \frac{3}{a + 1} \geq 3$
$\displaystyle \displaystyle 3 \geq 3(a + 1)$
$\displaystyle \displaystyle 1 \geq a + 1$
$\displaystyle \displaystyle 0 \geq a$.
So in this case, the conditions are $\displaystyle \displaystyle a >-1$ and $\displaystyle \displaystyle a \leq 0$.
Therefore, the solution in this case is $\displaystyle \displaystyle -1 < a \leq 0$.
So the solution of your inequality is $\displaystyle \displaystyle -1 < a \leq 0$.