6. Find the minimum value of the expression $x^{2}+y^{2}+3 z^{2}+2 x y$ given that $x y+z=1$ and $x, y, z>0$.
Answer: $\frac{8}{3}$. Solution. Fix $z$ and find the minimum of $|x+y|$ given $x y=1-z$. It is achieved at $x=y=\sqrt{1-z}$. Substituting these values into the expression $(x+y)^{2}+3 z^{2}$, we get $4-4 z+3 z^{2}$. This is a quadratic function, which reaches a minimum of $\frac{8}{3}$ at $z=\frac{2}{3}$.