1. Determine the minimum value of the expression $\frac{x^{2}+y^{2}+z^{2}}{x y+y z}$, where $x>0, y>0, \quad z>0$.
Solution. One

$$
x^{2}+\frac{1}{2} y^{2} \geq 2 \sqrt{\frac{1}{2} x^{2} y^{2}}=x y \sqrt{2} \text { and } z^{2}+\frac{1}{2} y^{2} \geq 2 \sqrt{\frac{1}{2} z^{2} y^{2}}=z y \sqrt{2}
$$

we obtain

$$
\frac{x^{2}+y^{2}+z^{2}}{x y+y z}=\frac{x^{2}+\frac{1}{2} y^{2}+\frac{1}{2} y^{2}+z^{2}}{x y+y z} \geq \frac{x y \sqrt{2}+y z \sqrt{2}}{x y+y z}=\sqrt{2} .
$$

Since for $x=z=1, y=\sqrt{2}$ we have

$$
\frac{1^{2}+(\sqrt{2})^{2}+1^{2}}{1 \cdot \sqrt{2}+1 \cdot \sqrt{2}}=\sqrt{2}
$$

it follows that the minimum value of the expression $\frac{x^{2}+y^{2}+z^{2}}{x y+y z}$ is $\sqrt{2}$.