4.2.  * Let $x, y, z$ be 3 non-zero real numbers, find the maximum value of $\frac{xy + 2yz}{x^2 + y^2 + z^2}$.
To find the maximum value of $\frac{x y+2 y z}{x^{2}+y^{2}+z^{2}}$, it is sufficient to find the smallest constant $k$ such that $x y+2 y z \leqslant k\left(x^{2}+y^{2}+z^{2}\right)$. Since $x^{2}+\alpha y^{2} \geqslant 2 \sqrt{\alpha} x y, (1-\alpha) y^{2}+z^{2} \geqslant 2 \sqrt{1-\alpha} y z$, we have
$$
x^{2}+y^{2}+z^{2} \geqslant 2 \sqrt{\alpha} x y+2 \sqrt{1-\alpha} y z.
$$

Let $2 \sqrt{\alpha}-\sqrt{1-\alpha}$, then $4 \alpha=1-\alpha$, so $\alpha=\frac{1}{5}$. Since $x^{2}+\frac{1}{5} y^{2} \geqslant \frac{2}{\sqrt{5}} x y, \frac{4}{5} y^{2}+$ $z^{2} \geqslant \frac{4}{\sqrt{5}} y z$, we have $x^{2}+y^{2}+z^{2} \geqslant \frac{2}{\sqrt{5}}(x y+2 y z)$, which means $\frac{x y+2 y z}{x^{2}+y^{2}+z^{2}} \leqslant \frac{\sqrt{5}}{2}$. When $x=1$, $y=\sqrt{5}, z=2$, the equality holds. Therefore, the maximum value of $\frac{x y+2 y z}{x^{2}+y^{2}+z^{2}}$ is $\frac{\sqrt{5}}{2}$.