Least value of unit vector $|a+b|^2+|b+c|^2+|c+a|^2$ If $a, b, c$ are unit vectors, then least value of $|a+b|^2+|b+c|^2+|c+a|^2$ will be equal to 
(1) 1
(2) 3
(3) 9
(4) 12  
If am using the concept $a=c=-b$, I am getting the answer 4, but not matching with options provided
 A: $$|a+b|^2+|b+c|^2+|c+a|^2  = 3 + |a+b+c|^2 \geq 3$$
A: Let $a,b,c$ be te three cube roots of $1$ or $1, e^{\frac{2\pi i}{3}},e^{\frac{-2\pi i}{3}}$.  Each sum has magnitude $1$, so $(2)3$ is the answer.
A: Writing in terms of inner product, you have that
\begin{align*}
|a+b|^2+|b+c|^2+|c+a|^2
&= 2(|a|^2 + |b|^2 + |c|^2) + 2 (a \cdot b + b \cdot c + c \cdot a) \\
&= 6 + 2 (a \cdot b + b \cdot c + c \cdot a)
\end{align*}
Thus, your problem amounts to minimizing the sum of inner products above. Considering that $a, b$ are fixed, notice that the choice of $c$ that minimizes the expression is taking $c$ at the opposite direction of $a + b$, since
$$
a \cdot b + b \cdot c + c \cdot a = a \cdot b + (a+b) \cdot c
$$
and we have
$$
|(a+b)\cdot c| = |a+b| \cos \theta
$$
with $\theta$ being the angle between $a+b$ and $c$. Thus, a true minimizer should satisfy that $a+b$ has the opposite direction of $c$ and the same will happen when you permute $a, b, c$. This implies that we want $a, b, c$ as vertices of an equilateral triangle. This means
$$
|a+b|^2+|b+c|^2+|c+a|^2 = 6 + 2 \cdot 3 \cdot \cos \frac{2\pi}{3} = 3.
$$
A: Let $a,b,c$ be three unit vectors then $$(a+b+c)^2\ge 0 \implies a^2+b^2+c^2+2(a,b+b.c+c.a) \ge 0$$ $$ \implies 2(a.b+b.c+c.a) \ge -3 ~~~(1)$$
Then $$|a+b|^2+|b+c|^2+|c+a|^2=2(a^2+b^2+c^2)+2(a.b+b.c+c.a) \ge 6-3=3.$$
