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Consider three real numbers a,b,c such that $a^2+b^2+c^2 =1$. What will be the maximum value of the expression $|a-b| + |b-c| + |c-a|$ ?

I tried two approaches. One I used the fact that $a^2+b^2+c^2 =1 = \frac{1}{2} (|a-b|^2 + |b-c|^2 + |c-a|^2) + ab + bc + ca$. Original plan was to somehow eliminate ab + bc + ca , but unable to do so - got stuck in a series of repeating expressions

Another variant was to try and assert that $a \geq b \geq c$ without loss of generality. Then $(|a-b| + |b-c| + |c-a|)^2 \leq 4(a-c)^2$ , but I am unable to use the given fact about $a^2 + b^2 + c^2$.

I realize that I am missing something very trivial here as this seems a fairly simple problem - so any help would be greatly appreciated.

Ashwin Krishnan
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1 Answers1

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Let's assume that $a\ge b\ge c$, then $|a-b|+|b-c|+|c-a|=2(a-c)$

And $(a-c)^2=a^2+c^2-2ac$

Because $a^2+c^2\le a^2+c^2+b^2\le1$ and $-2ac\le a^2+c^2\le a^2+c^2+b^2\le1$

We got $(a-c)^2=a^2+c^2-2ac\le 2$

So $(a-c)\le \sqrt 2$

So the answer is $2\sqrt 2$

Equality holds at $(a,b,c)=(\sqrt \frac12,-\sqrt \frac12,0)$

oyyko
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