17. $[\mathbf{1 0}]$ Given positive real numbers $x, y$, and $z$ that satisfy the following system of equations:
$$
\begin{aligned}
x^{2}+y^{2}+x y=1 \\
y^{2}+z^{2}+y z=4, \\
z^{2}+x^{2}+z x=5,
\end{aligned}
$$
find $x+y+z$.
Answer: $\sqrt{5+2 \sqrt{3}}$
Let $O$ denote the origin. Construct vectors $O A, O B$, and $O C$ as follows: The lengths of $O A, O B$, and $O C$ are $x, y$, and $z$, respectively, and the angle between any two vectors is $120^{\circ}$. By the Law of Cosines, we have $A B=1, B C=2$, and $A C=\sqrt{5}$. Thus $A B C$ is a right triangle, and $O$ is its Fermat point. The area of triangle $A B C$ is equal to $\frac{1}{2} \cdot 1 \cdot 2=1$. But the area is also equal to the sum of the areas of triangles $A O B, B O C$, and $C O A$, which is equal to $\frac{1}{2} \cdot x y \cdot \frac{\sqrt{3}}{2}+\frac{1}{2} \cdot y z \cdot \frac{\sqrt{3}}{2}+\frac{1}{2} \cdot z x \cdot \frac{\sqrt{3}}{2}$. We thus obtain $x y+y z+z x=\frac{4}{\sqrt{3}}$. Adding the three equations given in the problem and subtracting both sides by $x y+y z+z x$, we obtain $x^{2}+y^{2}+z^{2}=5-\frac{2}{\sqrt{3}}$. Therefore $(x+y+z)^{2}=\left(x^{2}+y^{2}+z^{2}\right)+2(x y+y z+z x)=$ $5+2 \sqrt{3}$