## Task A-2.1.

Let $a, b$ and $c$ be real numbers such that $a+2b+c>0$ and $a-2b+c>0$ and $a > c$.
## First Solution.

Consider the quadratic function

$$
f(x)=a x^{2}+2 b x+c
$$

From the conditions, we have

$$
\begin{aligned}
a+2 b+c>0 & \Longrightarrow f(1)>0, \quad \text { i.e., } \\
a-2 b+c>0 & \Longrightarrow f(-1)>0
\end{aligned}
$$

Since \( f(1) > 0 \) and \( f(-1) > 0 \), we conclude that the function \( f \) must have a root between -1 and 1. Since \( f \) is a quadratic function, it follows that \( f \) has two (distinct) real roots. Therefore, its discriminant must be positive:

$$
4 b^{2}-4 a c>0 \Longrightarrow b^{2}>a c
$$

which is what we needed to show.