I don't quite see why you are interested in this particular square root, $\sqrt{xy+yz+zx}$. In general, adjoining an element to a ring may or may not affect irreducibility of a given element. For example, $13$ is irreducible in $\mathbb{Z}$ factors as $(3+2i)(3-2i)$ in $\mathbb{Z}[i]$, remains irreducible in $\mathbb{Z}[\sqrt{-5}]$, factors as $(4+\sqrt3)(4-\sqrt3)$ in $\mathbb{Z}[\sqrt3]$, remains irreducible in $\mathbb{Z}[\sqrt7]$ et cetera.
Anyway, you can prove irreducibility of $x^2+y^2+z^2$ for example as follows.
$\mathbb{C}[x,y,z]=\mathbb{C}[u,v,z]$ with $u=x+iy$, $v=x-iy$. Your polynomial then looks like $x^2+y^2+z^2=uv+z^2$. If this were not irreducible, it would be a product of two linear polynomials. As this polynomial is homogeneous, so are the presumed factors. So we need to rule out the possibility
$$
uv+z^2=(au+bv+cz)(a'u+b'v+c'z)
$$
for some constants $a,b,c,a',b',c'$. As $aa'=0$ one of those constants is zero, w.l.o.g. $a'=0$, $a\neq0$. Similarly from $bb'=0$ we see that one of those also needs to be zero. Clearly we must assume $b=0, b'\neq0$. This leaves us
$$
uv+z^2=(au+cz)(b'v+c'z)
$$
with $a,c,b',c'$ all non-zero ($cc'=1$). This forces non-zero coefficients to terms $vz$ and $uz$, so no factorization is possible.