Inequality -- 1

Completing the square rewrites the main identity as $(2a + b + c)^2 \; =\; 16 - 8\sqrt{3} + (b-c)^2$ so that $(2a + b + c)^2 \; \ge \; 16 - 8\sqrt{3} \; = \; (2\sqrt{3}-2)^2$ with equality when $b=c$ . Thus we deduce that $2a+b+c \ge \boxed{2\sqrt{3}-2}$

Tweak the original equation a bit $a(a+b+c)+bc = 4-2\sqrt 3$ --> we get $\\$ $(a+b)(a+c) = 4-2\sqrt 3$ $\\$ Thus $2a + b + c = (a+b)(a+c) \geq 2\sqrt {(a+b)(a+c)} = 2\sqrt {4-2\sqrt 3} = 2(\sqrt 3 - 1)\quad$ [By AM-GM Inequality] $\\$ $\\$ $\\$ $.\\$ Explicit steps for simplifying square root inside square root: $\\$ Suppose $\\\sqrt {4-2\sqrt 3} =\sqrt x + y\\$ $4-2\sqrt3 = x + y^2 +2y\sqrt x \\ 4-2\sqrt 3 = y^2 + x + 2y\sqrt x \\$ We get $x = 3, y = -1$ .