Let x, y, z in [0, 1]. Then the maximum value | vedclass

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(A) Let $P = \sqrt{|x-y|} + \sqrt{|y-z|} + \sqrt{|z-x|}$.Without loss of generality,assume $0 \leq x \leq y \leq z \leq 1$.Then $P = \sqrt{y-x} + \sqr

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$1 + \sqrt{2}$

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(A) Let $P = \sqrt{|x-y|} + \sqrt{|y-z|} + \sqrt{|z-x|}$.Without loss of generality,assume $0 \leq x \leq y \leq z \leq 1$.Then $P = \sqrt{y-x} + \sqrt{z-y} + \sqrt{z-x}$.To maximize $P$,we set $x=0$ and $z=1$,which gives $P = \sqrt{y} + \sqrt{1-y} + 1$.Let $f(y) = \sqrt{y} + \sqrt{1-y} + 1$ for $y \in [0, 1]$.Using the substitution $y = \sin^2 	heta$,where $	heta \in [0, \pi/2]$,we get $f(	heta) = \sin 	heta + \cos 	heta + 1$.We know that $\sin 	heta + \cos 	heta = \sqrt{2} \sin(	heta + \pi/4)$,which has a maximum value of $\sqrt{2}$ at $	heta = \pi/4$.Thus,the maximum value of $P$ is $\sqrt{2} + 1$.

Let $x, y, z \in [0, 1]$. Then the maximum value of $\sqrt{|x-y|} + \sqrt{|y-z|} + \sqrt{|z-x|}$ is

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