Let $f(x,y,z)$ denote the difference between the left- and right-hand sides of your inequality. We have to show that $f(x,y,z)\ge0$ if $x,y,z\ge0$.

The minimum of $f(x,y,z)$ in $z\ge0$ is attained at $z_*:=\max(0,x+y-xy)$. If $x+y\le xy$ then $z_*=0$, whereas $f(x,y,0)=1+(x-y)^2>0$. So, without loss of generality (wlog) $x+y\ge xy$, and it remains to show that $$g(x,y):=f(x,y,x+y-xy)\ge0$$ if $$x+y\ge xy.$$ We have $g(x,y)=1-xy(x-2)(y-2)$, and so, $g(x,y)\ge1\ge0$ if $x\le2\le y$ or $y\le2\le x$. So, wlog either $x,y\le2$ or $x,y\ge2$.

Note that $$g(x,y)=1+xy[4(x+y)-xy-4]\ge1+xy[4xy-xy-4]=3(xy)^2-4xy+1=(3xy-1)(xy-1)>0$$ if $x,y\ge2$. So, wlog $x,y\le2$.

So, wlog $(x,y)$ is one of the critical points of $g$ in $(0,\infty)^2$, which are $(1,1)$ and $(2,2)$. We have $g(1,1)=0$ and $g(2,2)=1>0$.

It follows that the minimum of $f$ is $0$, and it is attained only at $(1,1)$.

Let $f(x,y,z)$ denote the difference between the left- and right-hand sides of your inequality. We have to show that $f(x,y,z)\ge0$ if $x,y,z\ge0$.

The minimum of $f(x,y,z)$ in $z\ge0$ is attained at $z_*:=\max(0,x+y-xy)$. If $x+y\le xy$ then $z_*=0$, whereas $f(x,y,0)=1+(x-y)^2>0$. So, without loss of generality (wlog) $x+y\ge xy$, and it remains to show that $$g(x,y):=f(x,y,x+y-xy)\ge0$$ if $$x+y\ge xy.$$ We have $g(x,y)=1-xy(x-2)(y-2)$, and so, $g(x,y)\ge1\ge0$ if $x\le2\le y$ or $y\le2\le x$. So, wlog either $x,y\le2$ or $x,y\ge2$.

Note that $$g(x,y)=1+xy[4(x+y)-xy-4]\ge1+xy[4xy-xy-4]=3(xy)^2-4xy+1=(3xy-1)(xy-1)>0$$ if $x,y\ge2$.

So, wlog $x,y\le2$. Then $0\le x(2-x)\le1$, $0\le y(2-y)\le1$, and hence $$g(x,y)=1-[x(2-x)]\,[y(2-y)]\ge0.$$

It follows that the minimum of $f$ is $0$, and it is attained only at $(1,1)$.