# 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

$$|a+b|^2+|b+c|^2+|c+a|^2 = 3 + |a+b+c|^2 \geq 3$$
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 $$|a+b|^2+|b+c|^2+|c+a|^2=2(a^2+b^2+c^2)+2(a.b+b.c+c.a) \ge 6-3=3.$$
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.
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 $$|a+b|^2+|b+c|^2+|c+a|^2 = 6 + 2 \cdot 3 \cdot \cos \frac{2\pi}{3} = 3.$$