Given that $A(x,y), B(x,y),$ and $C(x,y)$ are homogeneous real-coefficient polynomials of $x$ and $y$ with degrees 2, 3, and 4 respectively, and knowing that there exists a real-coefficient polynomial $R(x,y)$ such that $B(x,y)^2 - 4A(x,y)C(x,y) = -R(x,y)^2$, we need to prove that there exist two polynomials $F(x,y,z)$ and $G(x,y,z)$ such that
$$F(x,y,z)^2 + G(x,y,z)^2 = A(x,y)z^2 + B(x,y)z + C(x,y)$$
if for any real numbers $x, y, z$, $A(x,y)z^2 + B(x,y)z + C(x,y) \geq 0$.

### Proof:
Assume that $A(x,y)z^2 + B(x,y)z + C(x,y) \geq 0$ for all $x, y, z \in \mathbb{R}$. This implies that $A(x,y) \geq 0$ for all $x, y \in \mathbb{R}$.

We will consider two cases based on the nature of $A(x,y)$:

#### Case 1: $A(x,y)$ is a perfect square.
Let $A(x,y) = N(x,y)^2$. For any roots $(x,y)$ of $N$, we must have $B(x,y) = 0$. Hence, $N$ divides both $B$ and $R$, i.e., $N | B$ and $N | R$. Therefore, we can write:
$$A(x,y)z^2 + B(x,y)z + C(x,y) = \left( N(x,y)z + \frac{B(x,y)}{2N(x,y)} \right)^2 + \left( \frac{R(x,y)}{2N(x,y)} \right)^2.$$

#### Case 2: $A(x,y)$ is not a perfect square.
Since $A(x,y)$ is irreducible over $\mathbb{R}[x,y]$:

##### Subcase 2.1: $B(x,y)$ has a common non-constant factor with $A(x,y)$.
This implies that $A$ divides $B$ (since $A$ is irreducible over $\mathbb{R}[x,y]$), which further implies that $A$ divides $R$ and $C$. Let $B = KA$, $R = MA$, and $C = LA$. Since $A = S^2 + T^2$ for some polynomials $S, T \in \mathbb{R}[x,y]$, we have:
$$A(x,y)z^2 + B(x,y)z + C(x,y) = A(x,y)(z^2 + Kz + L) = (S^2 + T^2)\left( (z + \frac{K}{2})^2 + (\frac{M}{2})^2 \right).$$
Thus,
$$A(x,y)z^2 + B(x,y)z + C(x,y) = \left( Sz + \frac{SK + TM}{2} \right)^2 + \left( Tz + \frac{TK - SM}{2} \right)^2.$$

##### Subcase 2.2: $B(x,y)$ does not have a common non-constant factor with $A(x,y)$.
This implies that $R$ does not have a common non-constant factor with $A$ either, thus $R \not\equiv 0$. Let $G_1$ and $F_1$ be homogeneous real-coefficient polynomials in $x$ and $y$ of degree 1 such that $A$ divides $B - G_1x^2$ and $A$ divides $R - F_1x^2$. It follows that $A$ divides $G_1R - F_1B$ and $A$ divides $F_1R + G_1B$. Let:
$$F_2 = \frac{F_1B - G_1R}{2A}, \quad G_2 = \frac{F_1R + G_1B}{2A}.$$
Consider the polynomial:
$$(F_1z + F_2)^2 + (G_1z + G_2)^2.$$
By algebraic manipulation, we find that the roots of the polynomial are $z = \frac{-B \pm Ri}{2A}$. Hence, we have:
$$(F_1z + F_2)^2 + (G_1z + G_2)^2 = k(Az^2 + Bz + C)$$
for some positive real number $k$. Finally, let:
$$F = \frac{F_1z + F_2}{\sqrt{k}}, \quad G = \frac{G_1z + G_2}{\sqrt{k}}.$$

Thus, we have shown that there exist polynomials $F(x,y,z)$ and $G(x,y,z)$ such that:
$$F(x,y,z)^2 + G(x,y,z)^2 = A(x,y)z^2 + B(x,y)z + C(x,y).$$

The answer is: $\boxed{F(x,y,z)^2}$ + G(x,y,z)^2 = A(x,y)z^2 + B(x,y)z + C(x,y)}.