How find this maximum and minimum of$|x+1|+|x-1|+\sqrt{4-x^2}$ 
show that
$$2+\sqrt{3}\le|x+1|+|x-1|+\sqrt{4-x^2}\le2\sqrt{5}$$

This problem have nice methods? Thank you

my ugly methods,

since $-2\le x\le 2$,and $f(x)=|x-1|+|x+1|+\sqrt{4-x^2}\Longrightarrow f(x)=f(-x)$
so we only find $x\in [0,2]$ $f(x)_{\max},f(x)_{\min}$
so when
(1):

$$0\le x\le 1\Longrightarrow f(x)=2+\sqrt{4-x^2}\le 4$$
(2):  when $1\le x\le 2$, then
$$f(x)=2x+\sqrt{4-x^2}\Longrightarrow f'(x)=0\Longrightarrow x=\dfrac{4}{\sqrt{5}}$$
so $\cdots\cdots$

My question This problem have nice methods?Thank you

If $x=a+ib,$ where $a,b$ are real
$\implies 4-x^2=4-(a+ib)^2=4+b^2-a^2-2ab i$
This needs to be real and $\ge0$
For real value of $4-x^2, ab=0$
Case $1:$  If $b=0,$
$4-a^2\ge0\iff a^2\le 4\iff -2\le a\le 2$
We know for real $x,$ $|x|=\begin{cases} x &\mbox{if } x\ge0 \\
-x & \mbox{if } x<0 \end{cases}$
$$|x+1|=|a+1|=\begin{cases} a+1 &\mbox{if } a+1\ge0\iff a\ge-1 \\
-(a+1) & \mbox{if } a<-1 \end{cases}$$
and 
$$|x-1|=|a-1|=\begin{cases} a-1 &\mbox{if } a\ge1 \\
-(a-1)=1-a & \mbox{if } a<1 \end{cases}$$
Check region by region $[-2,-1)[-1,1],(1,2]$
Case $2:$  If $a=0,$
$4+b^2\ge0$ which is true for real $b$
$4-x^2=4-(ib)^2=4+b^2$
$|x+1|=|1+ib|=\sqrt{1+b^2}$
$|x-1|=|-1+ib|=\sqrt{1+b^2}$
$|x+1|+|x-1|+\sqrt{4-x^2}=2\sqrt{1+b^2}+\sqrt{4+b^2}\ge 1+2=3$
Observe that it has no finite maximum value 
