If $ax^2 + bx + c = 0$ has no real roots, then its discriminant is negative: $b^2 - 4ac < 0$, or $c > (b^2)/(4a)$. Hence, $a > 0$ in order to ensure zero real roots, or else we have a contradiction otherwise. Let $c > -(a + b)$ (i) such that: $-(a + b) = \frac{b^2}{4a}$; or $-4a^2 - 4ab - b^2 = 0$; or $-(2a + b)^2 = 0$; or $b = -2a$ (ii). Substitution of (ii) into (i) yields $c > a$. Since $a > 0$ in order for no real roots to occur in $ax^2 + bx + c = 0$, we have shown that $c > 0$ holds.