4. Given that $a, b, c$ are not all zero, then the maximum value of $\frac{a b+2 b c}{a^{2}+b^{2}+c^{2}}$ is
4. $\frac{\sqrt{5}}{2}$ Detailed Explanation: Let $x, y \in \mathbf{R}_{+}$, then $x^{2} a^{2}+b^{2} \geqslant 2 x a b, y^{2} b^{2}+c^{2} \geqslant 2 y c b$, thus $\frac{x}{2} a^{2}+\frac{b^{2}}{2 x} \geqslant a b, y b^{2}+\frac{1}{y} c^{2} \geqslant 2 b c$. Let $\frac{x}{2}=\frac{1}{2 x}+y=\frac{1}{y} \Rightarrow x=\sqrt{5}, y=\frac{2 \sqrt{5}}{5}$, substituting into the above equations gives $\frac{\sqrt{5}}{2}\left(a^{2}+b^{2}+c^{2}\right) \geqslant a b+2 b c, \therefore \frac{a b+2 b c}{a^{2}+b^{2}+c^{2}} \leqslant \frac{\sqrt{5}}{2}$, equality is achieved when $\sqrt{5} a=b=\frac{\sqrt{5}}{2} c$.