prove that $\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}+(\sqrt{3}-1)(|a-1|+|b-1|+|c-1|)\ge 3\sqrt{3}$ if $a+b+c=3$ $a,b,c\in[0,2]$
observation

by triangle inequality $|a|+|b|\ge |a+b| $

$|a-1|+|b-1|+|c-1|\ge |a+b+c-3|$
but $a+b+c=3$
hence 
$\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}+(\sqrt{3}-1)(|a-1|+|b-1|+|c-1|)\ge \sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2} +0$
hence our question is
$\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}\ge 3\sqrt{3}$
wlog we try to prove

$\sqrt{4-a^2}\ge\sqrt{3}$

$4-a^2\ge 3 \implies a^2-1\le0 $
$a\in \left [-1,1\right ]$ but $a\in [0,2]$ so $a\in [0,1]$
this would mean the inequality works for $a,b,c\in [0,1]$
$a=b=c=1$
but i need to prove it for  $a,b,c\in[0,2]$
for all $  a+b+c=3$

it is evident that $6=a^2+b^2+c^2+2(ab+bc+ac)$

by GM-HM

$\sqrt{ab}\ge \frac{ab}{a+b}$

so
$\sqrt{4-a^2}=\sqrt{(4-a)(4+a)}\ge\dfrac{16-a^2}{8}$ 
$\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2} \ge 6-\frac{a^2+b^2+c^2}{8}\ge^{AM-GM} 6-3\dfrac{\sqrt[3]{a^2b^2c^2}}{8}$
 A: Here is a simple solution.
First I would like to emphasize that $a,b,c$ must be non-negative for the inequality to hold. Otherwise a counter-example is $(a,b,c)=(-1,2,2)$.
Thus, we need to prove the following inequality for $0\le a,b,c\le 2$:
$$\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}+(\sqrt{3}-1)(|a-1|+|b-1|+|c-1|)\ge 3\sqrt{3}.$$
It suffices to show that the following inequality holds for all $a\in [0,2]$:
\begin{equation}
\sqrt{4-a^2} + (\sqrt{3}-1)|a-1| \ge \sqrt{3} + 1 -a \tag{1}
\end{equation}
(Then similarly we get two other inequalities for $b,c$ and taking the sum of these 3 inequalities we get the result.)
Proof for $(1)$:


*

*If $1\le a\le 2$: $(1)$ becomes
\begin{align}
&\sqrt{4-a^2} + (\sqrt{3}-1)(a-1) \ge \sqrt{3} + 1 -a  \\
\Longleftrightarrow &\sqrt{4-a^2} \ge \sqrt{3}(2-a)\\
 \Longleftrightarrow &4(2-a)(a-1)\ge 0,
\end{align} which is clearly true.

*If $0\le a< 1$: $(1)$ becomes
\begin{align}
&\sqrt{4-a^2} + (\sqrt{3}-1)(1-a) \ge \sqrt{3} + 1 -a  \\
\Longleftrightarrow &\sqrt{4-a^2} \ge 2-(2-\sqrt{3})a.
\end{align} 
Note that the right hand side of the last inequality is positive, squaring the both sides and taking the difference we get the equivalent inequality $4(2-\sqrt{3})a(1-a)\ge 0$, which is clearly true.


Therefore, $(1)$ is proved and the conclusion follows. Equality holds if and only if $(a,b,c)=(1,1,1)$ or a permutation of $(0,1,2)$.
Remark. According to the proof, we see that the inequality also holds for the weaker condition $a+b+c\le 3$.
