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

Now the given answer is $2\sqrt{2}$ but I am not able to obtain the corresponding values of $x,y,z$. Though when I take $|x-y|=|y-z|=\frac{1}{2}$ and $|x-z|=1$, I do obtain $1+\sqrt{2}$. Can there be a proper method to get to the answer.

Maverick
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1 Answers1

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WLOG fix $x\geq y\geq z$. Then our expression $E$ is $$E=\sqrt{x-y}+\sqrt{y-z}+\sqrt{x-z}$$ Now, note that as $x$ increases, so does $E$ and so for the maximum value of $E$ we should have $x=1$. Similarly, as $z$ increases, $E$ decreases so we want $z=0$. Then, the expression reduces to $$E=1+\sqrt{y}+\sqrt{1-y}$$ which I hope you can take from here.

Hope this helps. :)

ultralegend5385
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  • When I take the derivative (in my head), I find the extremum is $y=\frac 12$, which yields $E=1+\sqrt 2$, not $2 \sqrt 2$. – Robert Shore May 25 '22 at 03:05
  • @RobertShore: And that's correct. The OP probably misread the answer. It is the question 3 in the paper and the answer key rightly shows the answer, option A which is $1+\sqrt2$. – ultralegend5385 May 25 '22 at 03:46
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    +1! If it was my answer, I would also add the final hint to use $y=\sin^2 \alpha$ in my answer at the end. – User May 25 '22 at 05:44
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    @User: Why couldn't I think of that? That's so cool. The way I would prove it originally was using the QM-AM inequality. – ultralegend5385 May 25 '22 at 05:57