Parabola and line proof 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 $y=cx-b$.

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