= Solution
Set $s=x+y+z$ and $Q=x^2+y^2+z^2$. The expression in the equation is $4(Q+4)-(s+2)^2$. Apply the <Cauchy-Schwarz inequality> in $\mathbb R^4$ to
$$
\mathbf v=(x,y,z,2),\qquad\mathbf e=(1,1,1,1).
$$
Their <dot product> is $s+2$, and $|\mathbf e|^2=4$. Thus the expression is nonnegative, and it is zero precisely when equality holds in <Cauchy-Schwarz>. That requires $\mathbf v=\lambda\mathbf e$. The fourth coordinate forces $\lambda=2$, and the other three coordinates then give
$$
\boxed{x=y=z=2.}
$$
Substitution gives $3(4+4+4+4)-2(4+4+4)-4(2+2+2)=48-24-24=0$, so this unique possible choice really is a solution.
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