Prove that if $y=a b+2 a c, \alpha$ is a positive constant to be determined, then
$$\begin{aligned} \alpha y & =\alpha a(b+2 c) \leqslant\left[\frac{\alpha a+(b+2 c)}{2}\right]^{2} \\ & =\left[\frac{1}{2}(\alpha a+b)+c\right]^{2} \\ & =\left[\frac{1}{2} \sqrt{(\alpha a+1 \cdot b)^{2}}+c\right]^{2} \\ & \leqslant\left[\frac{1}{2} \sqrt{\left(a^{2}+1^{2}\right)\left(a^{2}+b^{2}\right)}+c\right]^{2} \end{aligned}$$

Given $\sqrt{a^{2}+b^{2}}+c=1$, we can set $\alpha=\sqrt{3}$,
then $\sqrt{3} y \leqslant\left(\sqrt{a^{2}+b^{2}}+c\right)^{2}=1$,
which implies $y=a b+2 a c \leqslant \frac{1}{\sqrt{3}}$.
Note that the equality in (2) holds if and only if
$$\left\{\begin{array} { l } { \alpha a = b + c , } \\ { a = \alpha b , } \\ { \alpha = \sqrt { 3 } , } \\ { \sqrt { a ^ { 2 } + b ^ { 2 } } + c = 1 , } \end{array} ( a , b , c > 0 ) \quad \text { i.e., } \left\{\begin{array}{l} b=c=\frac{1}{3}, \\ a=\frac{1}{\sqrt{3}} . \end{array}\right.\right.$$