Three, (50 points) Try to determine all positive constants $a, b$, such that the inequality $x y+y z+z x \geqslant a\left(y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2}\right)+$ $b x y z$ holds for non-negative numbers $x, y, z$ satisfying $x+y+z=1$.


Three, Solution: All solutions $(a, b)=(a, 9-a)$, where $0<a \leqslant 4$. Taking $(x, y, z)=\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$ and $(x, y, z)=\left(\frac{1}{2}, \frac{1}{2}, 0\right)$, we get $a+b \leqslant 9$ and $0<a \leqslant 4$. Below, we prove: $x y+y z+z x \geqslant a\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+(9-a) x y z(0<a \leqslant 4)$ for non-negative real numbers $x, y, z$ satisfying $x+y+z=1$.
It suffices to prove the homogeneous form for $x, y, z$:
$$
(x+y+z)^{2}(x y+y z+z x) \geqslant a\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+(9-a) x y z(x+y+z)
$$

for non-negative real numbers $x, y, z$ satisfying $x+y+z=1$.
Let $P=(x+y+z)^{2}(x y+y z+z x)$, $Q=x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}$, $R=(x+y+z) x y z$.
(※) The left side of the equation $=2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+5(x+y+z) x y z+\left(x^{3} y+x y^{3}\right)+\left(y^{3} z+y z^{3}\right)+\left(z^{3} x\right.$ $+z x^{3}$ )
$$
\begin{array}{l}
\geqslant 2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+5(x+y+z) x y z+2 x^{2} y^{2}+2 y^{2} z^{2}+2 z^{2} x^{2} \\
=a\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+(4-a)\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+5(x+y+z) x y z .
\end{array}
$$

By the Cauchy-Schwarz inequality,
$\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)\left(x^{2} z^{2}+x^{2} y^{2}+y^{2} z^{2}\right) \geqslant(x y \cdot x z+y z \cdot x y+z x \cdot y z)^{2}$. Since $x, y, z$ are all non-negative real numbers,
$\therefore x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \geqslant x y \cdot x z+y z \cdot x y+z x \cdot y z=(x+y+z) x y z$. Combining with $4-a \geqslant 0$,
$\therefore\left(\right.$ the left side of (※) $\geqslant a\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+(4-a)(x+y+z) x y z+5(x+y+z) x y z$
$\geqslant a\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+(9-a)(x+y+z) x y z$. Hence ( $\%$ ) is proved.
In summary, all solutions are $(a, b)=(a, 9-a)(0<a \leqslant 4)$.