Elementary inhomogeneous inequality for three non-negative reals I need the following estimate for something I am working on, but I don't immediately see how to establish it.
For $x, y, z \in \mathbb{R}_{\ge 0}$, show that
$$2xyz + x^2 + y^2 + z^2 + 1 \ge 2(xy + yz + zx),$$
and I suspect the only point of equality is (1,1,1).
It feels like the sort of thing that ought to have a simple Olympiad-style proof using standard inequalities, but I haven't had any luck thus far; of course, I'll take any proof that I can get.
Also, if this inequality is reminiscent of any others, I would be grateful for references!
 A: 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.
A: 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 $(1,1)$.
A: Another way.
Since $$\prod_{cyc}((x-1)(y-1))=\prod_{cyc}(x-y)^2\geq0,$$ we can assume that
$$(x-1)(y-1)\geq0,$$ which gives $$z(x-1)(y-1)\geq0$$ or
$$xyz\geq xz+yz-z.$$
Id est, it's enough to prove that:
$$2xz+2yz-2z+x^2+y^2+z^2+1\geq2(xy+xz+yz)$$ or
$$(x-y)^2+(z-1)^2\geq0$$ and we are done!
A: 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 $X = Y = Z = 0$.
A: Another way.
The first step is a homogenization as Fedor Petrov.
We need to prove that:
$$x^2+y^2+z^2+3\sqrt[3]{x^2y^2z^2}\geq2(xy+xz+yz).$$
Now, for $xyz=0$ our inequality is obvious.
let $xyz>0$, $x=e^{\frac{a}{2}}$,$y=e^{\frac{b}{2}}$ and $z=e^{\frac{c}{2}}.$
Thus, we need to prove that
$$\sum_{cyc}e^a+3e^{\frac{a+b+c}{3}}\geq2\sum_{cyc}e^{\frac{a+b}{2}},$$ which is the T.Popoviciu's inequality for the convex function $f(x)=e^x$.
About Popoviciu see here: https://en.wikipedia.org/wiki/Popoviciu%27s_inequality
Also, the inequality $$x^2+y^2+z^2+3\sqrt[3]{x^2y^2z^2}\geq2(xy+xz+yz)$$ we can get for $n=3$ from the following F.Shleifer's inequality.

Let $x_i\geq0$. Prove that:
$$(n-1)\sum_{i=1}^nx_i^2+n\sqrt[n]{\prod_{i=1}^nx_i^2}\geq\left(\sum_{i=1}^nx_i\right)^2.$$

