Prove that the maximum value of $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ is $9$ 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.
 A: 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 $(\star.)$
A: One thing we know is that $\theta+\phi+\psi=2\pi$.
Hence $\pi-\theta,\pi- \phi$ and $\pi-\psi$ form a triangle as their sum is $3\pi-2\pi=\pi$.
I have proved in this post Trigonometric inequality on triangle that
$\displaystyle \cos(\pi-\theta)\cos(\pi-\phi)\cos(\pi-\psi)\leq{1\over8}$
Hence $\displaystyle \cos(\theta)\cos(\phi)\cos(\psi)\geq -{1\over8}$
Hence $\displaystyle \cos(\theta)+\cos(\phi)+\cos(\psi)\geq 3\cdot{^3\sqrt{-{1\over8}}}=-{3\over2}$
A: Just adding another approach to a well-known problem here.
We know that $|\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2\space\vec{a}\cdot \vec{b}$ . 
So our expression is now $2\left(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2\right)-2\left(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}\right)$
Also we know that $(\vec{a}+\vec{b}+\vec{c})^2\ge 0\implies |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2\ge-2\left(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}\right)$
Now clearly the maximum of the expression is $3\left(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2\right)$
A: Draw a simple picture you can observe $\psi=2\pi-(\phi+\theta)$, so $\cos(\theta)+\cos(\phi)+\cos(\psi)=\cos(\theta)+\cos(\phi)+\cos(\phi+\theta)$. You can use calculus techniques to find out minimum and maximum of this function. Alternatively,
Denote $\cos(\frac{\theta+\phi}{2})=x$ and $\cos(\frac{\theta-\phi}{2})=y$
$$\cos(\theta)+\cos(\phi)+\cos(\phi+\theta)\\=2\cos(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2})+2\cos^2(\frac{\theta+\phi}{2})-1\\=2xy+2x^2-1\\=2(x+\frac{y}{2})^2-1-(\frac{y}{\sqrt{2}})^2$$. It has minimum when $x+\frac{y}{2}=0, y=\pm1$. To see these values can be achieved, take $\theta=\phi=\frac{2\pi}{3}$.
That makes $$\cos(\theta)+\cos(\phi)+\cos(\phi+\theta)$$ has minimum value $\frac{-3}{2}$ and your original expression has maximum value $9$.
