Olympiad Inequality AM-GM (easy) Prove that
$(1 + x + y)^2 + (1 + y + z)^2 + (1 + z + x)^2 ≤ 3(x + y + z)^2$, with equality if and only if $x = y = z = 1$ ($xyz \ge 1$) ($x,y,z$ positive reals)
This simplifies to 
$x^2 + y^2 + z^2 + 4(xy + yz + zx) - 4(x + y + z) \geq 3$
I am looking for a proof using AM-GM preferably.

$$x^2 + y^2 + z^2 + 4(xy + yz + zx) - 4(x + y + z) =(x+y+z-2)^2-4+2(xy + yz + zx)$$
Now, by AM-GM:
$$x+y+z\ge3\sqrt[3]{xyz}\ge3\tag{1}$$
$$xy+yz+zx\ge3\sqrt[3]{x^2y^2z^2}\ge3\tag{2}$$
which pretty much settles the matter:
$$\underbrace{(x+y+z-2)^2}_{\ge1}-4+2(\;\underbrace{xy + yz + zx}_{\ge3}\;)\ge1-4+6=3$$
