<s>[INST] 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
 [/INST] The function is defined only for $|x| \le 2$, and is even. Hence let $x = 2 \sin A$, for $A \in [0, \frac{\pi}{2}]$. Then: 
$F = |x+1| + |x-1| + \sqrt{4-x^2} = 2 \sin A + 1 + |2\sin A - 1|+ 2\cos A$
For $A \in [0, \frac{\pi}{6}]$, $F = 2 + 2 \cos A $ has the obvious range $[2 + 2 \cos \frac{\pi}{6}, 2 + 2 \cos 0] = [2+\sqrt{3}, 4]$
and for $A \in [\frac{\pi}{6}, \frac{\pi}{2}]$, $F = 4\sin A + 2 \cos A  = 2\sqrt{5} \cos (A - \alpha)$ where $\tan \alpha = 2$. Noting $\alpha \in [\frac{\pi}{6}, \frac{\pi}{2}]$ we now have a range $[2\sqrt{5} \cos(\frac{\pi}{6} - \alpha), 2\sqrt{5} \cos 0] = [2 + \sqrt{3}, 2\sqrt{5}]$
Thus $2 + \sqrt{3} \le F \le 2 \sqrt{5}$
</s>