# maximum value of $\sum (a-b)^2$

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.

• Do you know how to use Lagrange multipliers? – GSofer Apr 17 at 13:13
• 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. – amsmath Apr 17 at 13:14
• @amsmath It's pretty clear from the context what this means. – Servaes Apr 17 at 13:19
• I think OP meant $\sum (a-b)^2 = (a-b)^2+(b-c)^2+(c-a)^2$ – J. W. Tanner Apr 17 at 13:20
• @Servaes Nope, it isn't. – amsmath Apr 17 at 13:24

$$\sum_{cyc}(a-b)^2=3(a^2+b^2+c^2)-(a+b+c)^2\leq15.$$ The equality occurs for $$a+b+c=0,$$ which says that we got a maximal value.
• whats wrong with cauchy Inequality. $(a^2+b^2+c^2)(b^2+c^2+a^2)\geq (ab+bc+ca)^2$ means $-5 \leq (ab+bc+ca)\leq 5$ – DXT Apr 17 at 13:28
• @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. – Michael Rozenberg Apr 17 at 13:32
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; 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 $$(\sqrt{3}\times\sqrt{5})^2=15$$.