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Problem Of The Week #444 November 23rd, 2020

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Here is this week's POTW:

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

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Hi MHB, I have been feeling so sick for the past few days (due to chicken pox) and even now, I just couldn't sit but have to lie down most of the time.

Therefore any activities involving me in this forum will be delayed until I feel much better.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Hi MHB! I am back, even though I am not fully recovered yet, but, I am back. (Nod)

Unfortunately, no one answered last two week's POTW.(Sadface) You can read the suggested solution of other as follows:
We may assume $0\le x \le y \le z \le 1$. Then we let

$M=\sqrt{y-x}+\sqrt{z-y}+\sqrt{z-x}$

Since $\sqrt{y-x}+\sqrt{z-y}\le \sqrt{2[(y-x)+(z-y)]}=\sqrt{2(z-x)}$, we have

$M\le \sqrt{2(z-x)}+\sqrt{z-x}=(\sqrt{2}+1)\sqrt{z-x}\le \sqrt{2}+1$

The equality holds if and only if $y-x=z-y,\,x=0, z=1,\,y=\dfrac{1}{2}$.

Hence, $\sqrt{|x-y|}+\sqrt{|y-z|}+\sqrt{|z-x|}\le \sqrt{2}+1$.
 
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