$a \leq 8$

Briefly, the solution is omitted. Therefore, the range of values for $a$ is $\boxed{a \leq 8}$.

Proposition: There exists $x \in [1, 2]$ such that $x^2 + 2x + a \geq 0$.

The negation of the original proposition is: For all $x \in [1, 2]$, $x^2 + 2x + a < 0$ (1).

For (1), it can be transformed into: $x^2 + 2x < -a$ holds true for all $x$ in $[1, 2]$.

Since $y = x^2 + 2x$ is monotonically increasing in $[1, 2]$,

Therefore, $x^2 + 2x \leq 8 < -a$.

Therefore, $a < -8$.

Based on the relationship between a proposition and its contrapositive,

The range of $a$ for the original proposition is the complement of $a < -8$, which is $a \geq -8$.

Hence, the answer is: $\boxed{a \geq -8}$.

The range of $a$ is $a \leq 8$.

Therefore, the range of values for $a$ is $\boxed{a \leq 8}$.