I am having trouble solving this.
I know that the vertex is $\left(-\frac{b}{2a}, p(-\frac{b}{2a})\right)$, where $p(x) = ax^2+bx+c$, which is $(-\frac{b}{2a},c)$.
After that I am lost, how to show that $c \ge -\frac{1}{4a}$?
check around and I forgot that the vertex lines on the line y=x.