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If $a^2+b^2+c^2=5$ and $a,b,c \in \mathbb{R},$ find the maximum

value of $(a-b)^2+(b-c)^2+(c-a)^2$.

My Try: $(a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2)-2(ab+bc+ac)$

$$=10-2(ab+bc+ac)$$

Now this implies $ab+bc+ac\le 5$. So

$$(a-b)^2+(b-c)^2+(c-a)^2 = 10-2(ab+bc+ac) \geq 20.$$

Could some help me to find max of $(a-b)^2+(b-c)^2+(c-a)^2$? Thanks.

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    $\begingroup$ Do you know how to use Lagrange multipliers? $\endgroup$ – GSofer Apr 17 at 13:13
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    $\begingroup$ What is $\sum(a-b)^2$ meant to be? Without further explanation this is nonsense. You should also edit your exercise. It is not clear what you are supposed to do. $\endgroup$ – amsmath Apr 17 at 13:14
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    $\begingroup$ @amsmath It's pretty clear from the context what this means. $\endgroup$ – Servaes Apr 17 at 13:19
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    $\begingroup$ I think OP meant $\sum (a-b)^2 = (a-b)^2+(b-c)^2+(c-a)^2$ $\endgroup$ – J. W. Tanner Apr 17 at 13:20
  • $\begingroup$ @Servaes Nope, it isn't. $\endgroup$ – amsmath Apr 17 at 13:24
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$$\sum_{cyc}(a-b)^2=3(a^2+b^2+c^2)-(a+b+c)^2\leq15.$$ The equality occurs for $a+b+c=0,$ which says that we got a maximal value.

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    $\begingroup$ whats wrong with cauchy Inequality. $(a^2+b^2+c^2)(b^2+c^2+a^2)\geq (ab+bc+ca)^2$ means $-5 \leq (ab+bc+ca)\leq 5$ $\endgroup$ – DXT Apr 17 at 13:28
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    $\begingroup$ @DXT It's not wrong. It just does not help. $ab+ac+bc\leq5$ gives $\sum\limits_{cyc}(a-b)^2\geq0,$ which is trivial without using $ab+ac+bc\leq5$. But $ab+ac+bc\geq-5$ gives $\sum\limits_{cyc}(a-b)^2\leq20,$ which does not give a maximal value because the equality does not occur. $\endgroup$ – Michael Rozenberg Apr 17 at 13:32
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Here's a more visual approach; for a point $(a,b,c)\in\Bbb{R}^3$ the condition that $$a^2+b^2+c^2=5,$$ is equivalent to being on the sphere of radius $\sqrt{5}$ centered at the origin, and $$(a-b)^2+(b-c)^2+(c-a)^2=||(a,b,c)-(b,c,a)||^2,$$ where the point $(b,c,a)\in\Bbb{R}^3$ is obtained from $(a,b,c)$ by a rotation of one third of a full turn around the line spanned by $(1,1,1)$. Then clearly the distance is maximal precisely when $(a,b,c)$ is on the equator w.r.t. the axis of rotation, and as the following picture shows; enter image description here by elementary geometry the distance between $(a,b,c)$ and $(b,c,a)$ is $\sqrt{3}$ times the radius, which is $\sqrt{5}$. Hence the desired maximum is $(\sqrt{3}\times\sqrt{5})^2=15$.

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