Elementary inhomogeneous inequality for three non-negative reals

Write $x = 1-X$, $y=1-Y$, $z=1-Z$. Then the inequality reduces to $$2XYZ \leq X^2 + Y^2 + Z^2$$ for $X, Y, Z \leq 1$. If $X, Y, Z < 0$ then the inequality is trivial, since LHS < 0. Otherwise suppose $X \in [0, 1]$ wlog. Then $$\text{RHS} - \text{LHS} = X^2 + (1-X^2)Y^2 + (Z-XY)^2 \geq 0.$$ Equality holds iff $X = Y = Z = 0$.

Denote $x^2=a^3,y^2=b^3,z^2=c^3$. By AM-GM we have $1+2xyz=1+(abc)^{3/2}+(abc)^{3/2}\geqslant 3\sqrt[3]{1\cdot (abc)^{3/2}\cdot (abc)^{3/2}}=3abc$, so LHS is not less then $$a^3+b^3+c^3+3abc\geqslant ab(a+b)+bc(b+c)+ac(a+c)\\ \geqslant 2(ab)^{3/2}+2(bc)^{3/2}+2(ca)^{3/2}=2(xy+yz+zx),$$ the first inequality is Schur, the second is three AM-GM's.

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 $(1,1)$.