<s>[INST] Maximum value of a vector If a,b,c are unit vectors , then we have to know $|a-b|^2$$+|b-c|^2$$+|c-a|^2$ does not exceed . 
I tried it a lot , but not able to get how to start it . 
 [/INST] Using $||\mathbf{a}||^2=\langle\mathbf{a},\mathbf{a}\rangle$ and the linearity properties of the inner product (if you don't want this you may just expand on components) you may prove:
$||\mathbf{a}+\mathbf{b}+\mathbf{c}||^2+||\mathbf{a}-\mathbf{b}||^2+||\mathbf{b}-\mathbf{c}||^2+||\mathbf{c}-\mathbf{a}||^2=3(||\mathbf{a}||^2+||\mathbf{b}||^2+||\mathbf{c}||^2)=9$
since you said that $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are on the unit circle.
So:
$||\mathbf{a}-\mathbf{b}||^2+||\mathbf{b}-\mathbf{c}||^2+||\mathbf{c}-\mathbf{a}||^2=9-||\mathbf{a}+\mathbf{b}+\mathbf{c}||^2\le9$
Equality is attained when $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$, that is when $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ form an equilateral triangle on the unit circle.
EDIT: to prove the last statement at OP's request:
If $\theta$ is the angle between two of the vectors (say $\mathbf{a}$ and $\mathbf{b}$) then:
$\cos\theta=\frac{\langle\mathbf{a},\mathbf{b}\rangle}{||\mathbf{a}||\cdot||\mathbf{b}||}=\langle\mathbf{a},\mathbf{b}\rangle$
since $||\mathbf{a}||=||\mathbf{b}||=1$
From $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$ we get $\mathbf{a}+\mathbf{b}=-\mathbf{c}$ so $||\mathbf{a}+\mathbf{b}||=||\mathbf{c}||=1$
Now:
$1=||\mathbf{a}+\mathbf{b}||^2=||\mathbf{a}||^2+||\mathbf{b}||^2+2\langle\mathbf{a},\mathbf{b}\rangle=2+2\langle\mathbf{a},\mathbf{b}\rangle$
$\cos\theta=\langle\mathbf{a},\mathbf{b}\rangle=-\frac{1}{2}$
so
$\theta=\frac{2\pi}{3}=120^{\circ}$
The angle between any two vectors is $120^{\circ}$, so the vectors form an equilateral triangle.
</s>