If $f(x) = x^3 + ax^2 + bx + c$ attains it’s local minima at certain negative real number then

$a^2 − 3b > 0, a > 0, b > 0$

$f'(x)x = 3x^2 + 2ax + b = 3(x − x_1)(x − x_2)$ where $x_1 < x_2$.

Clearly $f'(x) < 0 ∀ x ∈ (x_1, x_2)$ and $f'(x) > 0 ∀ x ∈ (−∞, x_1) ∪ (x_2, ∞)$.

Thus $x = x_1$ is the point of local maxima and $x = x_2$ is the point of local minima.

Thus bigger root of $f'(x) = 0$ must be negative. Hence $a^2 − 3b > 0, a > 0, b > 0$.