TITLE: Prove that the maximum value of $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ is $9$
QUESTION [5 upvotes]: If $\vec{a},\vec{b},\vec{c}$ are unit vectors,then prove that the maximum value of $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ is $9$

$|\vec{a}-\vec{b}|^2=2-2\cos\theta$,where $\theta $ is the angle between $\vec{a}$ and $\vec{b}$.
$|\vec{b}-\vec{c}|^2=2-2\cos\phi$,where $\phi $ is the angle between $\vec{b}$ and $\vec{c}$.
$|\vec{c}-\vec{a}|^2=2-2\cos\psi$,where $\psi $ is the angle between $\vec{c}$ and $\vec{a}$.
So,$|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=6-2(\cos\theta+\cos\phi+\cos\psi)$
But then i am stuck.Please help me.Thanks.

REPLY [6 votes]: Given $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1\;,$ and Let $$K=|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$$
Now Using $$|\vec{a}-\vec{b}|^2 = |\vec{a}|^2+|\vec{b}|^2+2\left(\vec{a}\cdot \vec{b}\right) = 2+2\left(a\cdot \vec{b}\right)$$
So we get $$K = 2-2\left(\vec{a}\cdot \vec{b}\right)+2-2\left(\vec{b}\cdot \vec{c}\right)+2-2\left(\vec{c}\cdot \vec{a}\right)$$
So we get $$K=6-2\left(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}\right)...............(\star)$$
Now Using $$\left|\vec{a}+\vec{b}+\vec{c}\right|^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2\left(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}\right)$$
$$\implies\left|\vec{a}+\vec{b}+\vec{c}\right|^2 = 3+2\left(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}\right)\geq 0$$
$$\implies 2\left(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}\right)\geq -3$$
$$\implies -2\left(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}\right)\leq 3$$
$$\implies K= 6-2\left(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}\right)\leq 6+3=9$$
above we get from equation $(\star.)$