Parabola and line proof

Question

Given are three non-zero numbers $a, b, c \in \mathbb{R}$. The parabola with equation $y=ax^2+bx+c$ lies above the line with equation $y=cx$.

Prove that the parabola with equation $y=cx^2-bx+a$ lies above the line with equation $y=cx-b$.

Answer 1

The parabola with equation $y=ax^2+bx+c$ lies above the line with equation $y=cx$. So

$ax^2 + bx + c = cx \quad \Rightarrow \quad ax^2 + (b - c)x + c = 0 \quad \Rightarrow \quad \Delta = (b - c)^2 - 4ac$.

On the other hand, the parabola with equation $y=cx^2-bx+a$ lies above the line with equation $y=cx - b$. So

$cx^2 - bx + a = cx - b \quad \Rightarrow \quad cx^2 - (b + c)x + a + b = 0 \quad \Rightarrow$

$\Delta = (b + c)^2 - 4(a + b)c = b^2 + 2bc + c^2 - 4ac - 4bc \quad \Rightarrow \quad \Delta = (b - c)^2 - 4ac$

Answer 2

Hint : "The parabola with equation $y=ax^2+bx+c$ lies above the line with equation $y=cx$" means that the discriminant of the equation $ax^2+(b-c)x+c=0$ is smaller than $0$. Now compute this discriminant and that of the equation $cx^2-(b+c)x+(a+b)=0$ and compare!