(2) (2006 - Chongqing) If $a, b, c>0$ and $a(a+b+c)+b c=4-2 \sqrt{3}$, then the minimum value of $2 a+b+$ $c$ is ( ).
(A) $\sqrt{3}-1$
(B) $\sqrt{3}+1$
(C) $2 \sqrt{3}+2$
(D) $2 \sqrt{3}-2$
Solve $(2 a+b+c)^{2}=4 a^{2}+4 a(b+c)+(b+c)^{2}=4(4-2 \sqrt{3}-b c)+(b+c)^{2}=$ $4(4-2 \sqrt{3})+(b-c)^{2} \geqslant 16-8 \sqrt{3}$, where the equality holds when $b=c$, so, when $b=c$ and $a+$ $b=\sqrt{3}-1$, $2 a+b+c$ achieves the minimum value $2 \sqrt{3}-2$, the answer is D.