What are the maximum and minimum values of $\langle u, v\rangle + \langle v, w\rangle + \langle w, u\rangle$?

Question

Let $V$ be an inner product space over $\mathbb{R}$. Suppose that $u$, $v$ and $w $ are three unit vectors in the $xy$-plane. What are the maximum and minimum values that $$\langle u, v\rangle + \langle v, w\rangle + \langle w, u\rangle$$ can attain, and under what conditions?

My attempt:

I can say that the maximum value is $\langle u, v\rangle + \langle v, w\rangle + \langle w, u\rangle= \langle 1, 1\rangle + \langle 1, 1\rangle + \langle 1, 1\rangle= 3$.

The minimum value is $\langle u, v\rangle + \langle v, w\rangle + \langle w, u\rangle= \langle 0, 0\rangle + \langle 0, 0\rangle + \langle 0, 0\rangle= 0$.

Please tell me if my answers is correct or not, and help me.

Thanks in advance.

Answer 1

We have that

$$\langle u, v\rangle + \langle v, w\rangle + \langle w, u\rangle =\cos \alpha + \cos \beta + \cos \gamma$$

with the condition $\alpha + \beta + \gamma=2\pi \implies \frac \alpha 2 + \frac \beta 2 + \frac \gamma 2=\pi$ and therefore indicating with $A=\frac \alpha 2$, $B=\frac \beta 2$, $C= \frac \gamma 2$

$$\cos \alpha + \cos \beta + \cos \gamma=3-2(\sin^2 A + \sin^2B + \sin^2 C)$$

which reaches its maximum value of $3$ when $\sin A=\sin B=\sin C=0$ and since

$$\sin^2 A + \sin^2B + \sin^2 C\le \frac 9 4$$

(refer to here and Show that $\sin^2(x)+\sin^2(y)+\sin^2(z) \le9/4$ where $x, y, z$ are angles of a triangle)

we have that its minimum value is

$$3-2(\sin^2 A + \sin^2B + \sin^2 C)\ge 3-\frac92=-\frac 32$$

therefore

$$-\frac 3 2\le \langle u, v\rangle + \langle v, w\rangle + \langle w, u\rangle\le 3$$

Answer 2

Consider the identity \begin{align} |u+v+w|^2&=|u|^2+|v|^2+|w|^2+2(\langle u,v\rangle+\langle u,w\rangle+\langle v,w\rangle)=\\ &=3+2(\langle u,v\rangle+\langle u,w\rangle+\langle v,w\rangle). \end{align} Then you need to optimize $$ \langle u,v\rangle+\langle u,w\rangle+\langle v,w\rangle=\frac{|u+v+w|^2-3}{2}, $$ which is equivalent to optimizing $|u+v+w|$. The latter is quite obvious: the largest value when all vectors are parallel, the smallest when $u+v+w=0$.

Answer 3

Let $u= (x,y)$, $v= (a,b)$ and $w= (c,d)$, then we have

$$\langle u, v\rangle + \langle v, w\rangle + \langle w, u\rangle =ax+by+cx+cy+ac+bd =: E$$

Since $$|ax+by| \leq \sqrt{(a^2+b^2)(x^2+y^2)} = 1$$ we have by triangle inequality $$ |E|\leq 3$$

Answer 4

Hint $$ -\|u \| \|v \| \leq \langle u , v \rangle \leq \| u \| \| v\|$$ by Cauchy Schwarz, with equality when the vectors are proportional.

From here the maximum is easy to attain.

The minimum is trickier, as you cannot get the lower equality for all three vectors at the same time. It is easy to find vectors for which the sum is $-1$, but the min is smaller than that.

If the three vectors are in the same plane, and the inner product is the dot product, you can express the dot product of the vectors in terms of the angles between vectors and find the min of the sum of $\cos$ of the angles.