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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

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  • $\begingroup$ what is the domain of $x$? $\endgroup$ – lab bhattacharjee Sep 15 '13 at 10:59
  • $\begingroup$ it only such that $$4-x^2\ge 0$$ $\endgroup$ – china math Sep 15 '13 at 11:11
  • $\begingroup$ do you allow $4-x^2=4-(ib)^2=4+b^2>0$ $\endgroup$ – lab bhattacharjee Sep 15 '13 at 11:14
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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 $2 + \sqrt{3} \le F \le 2 \sqrt{5}$

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The function is piecewise differentiable. You should decompose it in its differentiable intervals and for every interval find extrema, if any, PLUS values at endpoints.

In $[-2, -1]$, $f(x)=-2x+\sqrt{4-x^2}$.

In $[-1,+1]$, $f(x)=2+\sqrt{4-x^2}$.

In $[+1,+2]$, $f(x)=2x+\sqrt{4-x^2}$.

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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

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  • $\begingroup$ Thank you,But $x$ is real numbers $\endgroup$ – china math Sep 15 '13 at 11:37
  • $\begingroup$ @chinamath, you said "it only such that $$4−x^2\ge0$$" in response to my question "what is the domain of $x$?" $\endgroup$ – lab bhattacharjee Sep 15 '13 at 11:39

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