1. Reformulation of the Problem:
Let $a, b, c$ be three real numbers. Prove that:
- (a) If $a \geq 0$, $c \geq 0$, and $4ac \geq b^2$, then for any pair of vectors $\overrightarrow{f}$ and $\overrightarrow{g}$, we have $a \cdot \overrightarrow{f}^2 + b \cdot \overrightarrow{f} \cdot \overrightarrow{g} + c \cdot \overrightarrow{g}^2 \geq 0$.
- (b) If for every pair of vectors $\overrightarrow{f}$ and $\overrightarrow{g}$, we have $a \cdot \overrightarrow{f}^2 + b \cdot \overrightarrow{f} \cdot \overrightarrow{g} + c \cdot \overrightarrow{g}^2 \geq 0$, then $a \geq 0$, $c \geq 0$, and $4ac \geq b^2$.

2. Proof of (a):
- Given $a \geq 0$ and $c \geq 0$, their square roots $\sqrt{a}$ and $\sqrt{c}$ are real numbers.
- Since $4ac \geq b^2$, we have $\sqrt{4ac} \geq \sqrt{b^2}$, which implies $2\sqrt{ac} \geq |b|$.
- This yields $2\sqrt{ac} \geq b$ and $2\sqrt{ac} \geq -b$, or equivalently, $2\sqrt{ac} - b \geq 0$ and $2\sqrt{ac} + b \geq 0$.

3. **Case 1: $\overrightarrow{f} \cdot \overrightarrow{g} \geq 0$:**
$$a \cdot \overrightarrow{f}^2 + b \cdot \overrightarrow{f} \cdot \overrightarrow{g} + c \cdot \overrightarrow{g}^2 = \left(a \cdot \overrightarrow{f}^2 - 2\sqrt{ac} \cdot \overrightarrow{f} \cdot \overrightarrow{g} + c \cdot \overrightarrow{g}^2\right) + \left(2\sqrt{ac} + b\right) \cdot \overrightarrow{f} \cdot \overrightarrow{g}$$
$$= \left(\sqrt{a} \cdot \overrightarrow{f} - \sqrt{c} \cdot \overrightarrow{g}\right)^2 + \left(2\sqrt{ac} + b\right) \cdot \overrightarrow{f} \cdot \overrightarrow{g} \geq 0$$
Since $\left(\sqrt{a} \cdot \overrightarrow{f} - \sqrt{c} \cdot \overrightarrow{g}\right)^2 \geq 0$ and $\left(2\sqrt{ac} + b\right) \cdot \overrightarrow{f} \cdot \overrightarrow{g} \geq 0$.

4. **Case 2: $\overrightarrow{f} \cdot \overrightarrow{g} \leq 0$:**
$$a \cdot \overrightarrow{f}^2 + b \cdot \overrightarrow{f} \cdot \overrightarrow{g} + c \cdot \overrightarrow{g}^2 = \left(a \cdot \overrightarrow{f}^2 + 2\sqrt{ac} \cdot \overrightarrow{f} \cdot \overrightarrow{g} + c \cdot \overrightarrow{g}^2\right) - \left(2\sqrt{ac} - b\right) \cdot \overrightarrow{f} \cdot \overrightarrow{g}$$
$$= \left(\sqrt{a} \cdot \overrightarrow{f} + \sqrt{c} \cdot \overrightarrow{g}\right)^2 - \left(2\sqrt{ac} - b\right) \cdot \overrightarrow{f} \cdot \overrightarrow{g} \geq 0$$
Since $\left(\sqrt{a} \cdot \overrightarrow{f} + \sqrt{c} \cdot \overrightarrow{g}\right)^2 \geq 0$ and $-\left(2\sqrt{ac} - b\right) \cdot \overrightarrow{f} \cdot \overrightarrow{g} \geq 0$.

5. Proof of (b):
- Let $\overrightarrow{f} = \overrightarrow{0}$ and $\overrightarrow{g}$ be a vector of length 1. Then:
$$a \cdot \overrightarrow{0}^2 + b \cdot \overrightarrow{0} \cdot \overrightarrow{g} + c \cdot \overrightarrow{g}^2 = c \cdot 1^2 = c \geq 0$$
Similarly, by choosing $\overrightarrow{g} = \overrightarrow{0}$ and $\overrightarrow{f}$ of length 1, we get $a \geq 0$.

6. **Case when $a = 0$:**
- The inequality becomes $b \cdot \overrightarrow{f} \cdot \overrightarrow{g} + c \cdot \overrightarrow{g}^2 \geq 0$.
- If $b \neq 0$, for fixed $\overrightarrow{g}$ and varying $\overrightarrow{f}$, the term $b \cdot \overrightarrow{f} \cdot \overrightarrow{g}$ can be negative, contradicting the inequality. Hence, $b = 0$ and $4ac \geq b^2$ trivially holds.

7. **Case when $c = 0$:**
- Similar argument as above shows $b = 0$ and $4ac \geq b^2$ trivially holds.

8. **Case when $a > 0$ and $c > 0$:**
- Let $\overrightarrow{f}$ be a vector of length 1 and $\overrightarrow{g} = \frac{\sqrt{a}}{\sqrt{c}} \cdot \overrightarrow{f}$. Then:
$$a \cdot \overrightarrow{f}^2 + b \cdot \overrightarrow{f} \cdot \overrightarrow{g} + c \cdot \overrightarrow{g}^2 = \left(\sqrt{a} \cdot \overrightarrow{f} - \sqrt{c} \cdot \overrightarrow{g}\right)^2 + \left(2\sqrt{ac} + b\right) \cdot \overrightarrow{f} \cdot \overrightarrow{g}$$
$$= \left(2\sqrt{ac} + b\right) \cdot \frac{\sqrt{a}}{\sqrt{c}} \geq 0$$
This implies $2\sqrt{ac} + b \geq 0$. Similarly, using $\overrightarrow{g} = -\frac{\sqrt{a}}{\sqrt{c}} \cdot \overrightarrow{f}$, we get $2\sqrt{ac} \geq b$.

9. Conclusion:
- Combining $2\sqrt{ac} \geq b$ and $2\sqrt{ac} \geq -b$, we get $2\sqrt{ac} \geq |b|$, which implies $4ac \geq b^2$.

$\blacksquare$