$x^2 +y^2 + z^2$ is irreducible in $\mathbb C [x,y,z]$

Also, this is part of a more general pattern, using Gauss' Lemma and Eisenstein criterion: for positive integers $\ell,m,n$, $x^\ell+y^m+z^n$ is irreducible over any field $k$ of characteristic not dividing the exponents. Prove this via Gauss/Eisenstein by first noting that $k[x,y,z]$, $k(z)[x,y]$, and such are UFDs. By Gauss, irreducibility in $k[z,y,z]$ is equivalent to that in $k(z)[x,y]$. By Eisenstein, the polynomial is irreducible in $k(z)[y][x]$ if $y^m+z^n$ has some prime factor in $k(z)[y]$ that does not occur twice. Since the char of the field does not divide the exponents, this is easily so. (For $m=n$ and $k$ alg closed it is easy to write down factors $y+z$, etc.)

There is a more geometrical way to prove that $x^2+y^2+z^2$ is irreducible in $\mathbb{C}[x,y,z]$. Notice that, since the polynomial is homogeneous, the equation $x^2+y^2+z^2 = 0$ describes a curve $\mathcal{C}$ in $\mathbb{P}^2(\mathbb{C})$.

If the polynomial is reducible, $\mathcal{C}$ has two irreducible components (counted with multiplicity, i.e. $\mathcal{C}$ is made of two possibly coincident straight lines). By Bézout's theorem, these components must intersect in at least one point, which must be a singular point of $\mathcal{C}$. Then it suffices to check that $\mathcal{C}$ has no singular points to conclude that the polynomial is irreducible, and this is trivial.

This argument can also be applied for the more general case of $x^n+y^n+z^n$ (which correspond to the so-called Fermat curves).

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.