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 . 
 A: 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.
A: Hint:
I don't know from your question if you are doing 2D vectors or 3D vectors (or more). I will assume 2D (the reasoning doesn't change for higher dimensions.
Geometrically $||a-b||$ is the length between the 'points' $a$ and $b$ on the Cartesian plane.
So $||a-b||^2$ is the square of the distance between the 'points'.
So you want to arrange three points on the plane so that the square of the distance between each pair of points is a maximum.
See if you can find an arrangement such that the square of the distances between all the points is 9. Then prove that any other arrangement is smaller.
