Replace $x,y,z$ with $\cos A,\cos B,\cos C$ where $A+B+C=\pi$ and you receive the well known $\cos A+\cos B+\cos C\le\frac{3}{2}$.

Before you do that, you need show $x,y,z \leq 1$. Surely, this comes easily from writing the condition as $(z+xy)^2 = (x^2-1)(y^2-1)$; if both $x,y >1$, then $(z+xy)^2 < x^2y^2$ leads to $z<0$, absurd (and similarly cyclical), but it needs be said and done. Notice that, in the absence of $x,y,z\geq 0$, we have $\sup 2(x+y+z) = +\infty$.

mavropnevma wrote: Before you do that, you need show $x,y,z \leq 1$. Well if say $x>1$ then $x^2>1=x^2+y^2+z^2+2xyz$, no?

Yes, but you needed to say it! Now you've said it.

mavropnevma wrote: Yes, but you needed to say it! Now you've said it. Fair enough!

momo1729 wrote: Let $x$, $y$, and $z$ be three real positive numbers such that $x^{2}+y^{2}+z^{2}+2xyz=1$. Prove that $2(x+y+z)\leq 3$. We set $x=ab, y=bc, z=ca$ then the condition will be $a^2b^2+b^2c^2+c^2a^2+2a^2b^2c^2=1 \iff \frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}=1$ And we have to prove that $ab+bc+ca \le \frac{3}{2}$ By C-S $1=\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1} \ge \frac{(a+b+c)^2}{a^2+b^2+c^2+3}$ $\iff ab+bc+ca \le \frac{3}{2}$

Mavropnevma....

https://artofproblemsolving.com/community/c6h1099628p4948146