That expression isn't even cyclic!

By AM-GM, we have

$\begin{aligned} \sqrt{(xy+z)(xz+y)} &\leq \dfrac{(xy+z)+(xz+y)}{2}\\ &= \dfrac {(x+1)(y+z)}{2}\\ &\leq \dfrac{1}{2} \left ( \dfrac{(x+1)+(y+z)}{2} \right )^2\\ &= \dfrac {1}{2} \times 2^2\\ &= 2\\ \implies (xy+z)(xz+y) &\leq 4 \end{aligned}$

Thus, the maximum value of $(xy+z)(xz+y) = 4$ , attained at $x+1=y+z$ , or $x=1$ , $y+z=2$ .

Therefore, all equality cases are $x=1$ , $y \in [0,2]$ , $z=2-y$ .

As $x,y,z$ are non-negative reals then we can apply A.M.-G.M. inequality

$(xy+z) + (xz+y) \geq 2 \sqrt{(xy+z)(xz+y)} \\ \implies (x+1)(y+z) \geq 2 \sqrt{(xy+z)(xz+y)} \\ \implies (x+1)(3-x) \geq 2 \sqrt{(xy+z)(xz+y)} \\ \implies 3+2x-x^2 \geq 2 \sqrt{(xy+z)(xz+y)} \\ \implies 4-(x-1)^2 \geq 2 \sqrt{(xy+z)(xz+y)}$

The maximum value can be attained if $x-1=0 \implies x=1$ and $y+z=3-1=2$

So, $2 \sqrt{s} \leq 4 \implies s \leq 4$

$(xy+z)(xz+y)$ attains maximum at $x=1$ and for all non-negative reals $y,z$ where $y+z=2$

If $x + y + z = 3$ , then $xyz \geq 1$ . So plug in $xy = \frac{1}{z}$ and $xz = \frac{1}{y}$ . Thence the expression above can be written as $(\frac{1}{z} + z)(\frac{1}{y} + y)$ . Now by applying AM-GM inequality, we have $z + \frac{1}{z} \leq 2\sqrt{z \times \frac{1}{z}}$ and $y + \frac{1}{y} \leq 2\sqrt{y \times \frac{1}{y}}$ . Therefore $(xy + z)(xz + y) \leq 4$ when $x = y = z = 1$ .