For three vectors in any inner product space, prove the following inequality.

Question

Let $x, y, z$ be three unit vectors in a real inner product space. Assume that we are given the inner products $\langle x, y \rangle = a \in \mathbb{R}$ and $\langle y, z \rangle = b \in \mathbb{R}$. How do you obtain a bound on the value of $\langle z, x \rangle$, in particular, prove the following? $$ \langle z, x \rangle \geq ab - \sqrt{1 - a^2} \sqrt{1 - b^2} $$

My Approach

I was able to prove this inequality for the vector space $\mathbb{R}^n$ geometrically. I assumed that the angle between $x$ and $y$ is $\theta$ and the angle between $y$ and $z$ is $\phi$.

Thus $a = \cos{\theta}, b = \cos{\phi}$. The inner product of $x$ and $z$ is minimized when the angle between them is the largest and this happens when all three vectors are coplanar and $x$ and $z$ are on opposite sides of $y$ $$\therefore \min \langle z, x \rangle = \cos{(\theta + \phi)} = \cos{\theta} \cos{\phi} - \sin{\theta} \sin{\phi} = ab - \sqrt{1 - a^2} \sqrt{1 - b^2}$$

I however want to approach this question from a purely algebraic perspective without involving trigonometric or visualization, using purely the results from the algebra of inner product spaces. Thanks in advance!

Answer 1

The vectors $u:=x-ay,v:=z-by$ are orthogonal to $y$ and their respective norms are $\sqrt{1-a^2},\sqrt{1-b^2}.$ Therefore, $$\langle z, x \rangle=\langle v+by,u+ay\rangle=ab+\langle v,u\rangle\ge ab-\sqrt{1-a^2}\sqrt{1-b^2}.$$

Answer 2

As an alternative, for a not very effective method, let $y=Ax+Bz+w$ with $w$ orthogonal to $x$ and $z$, then

• $\langle x, y \rangle=A+B\langle x, z \rangle=a$
• $\langle y, z \rangle=A\langle x, z \rangle+B=b$

from which we obtain

• $A\left(1-\langle x, z \rangle^2\right)=a-b\langle x, z \rangle$

• $B\left(1-\langle x, z \rangle^2\right)=b-a\langle x, z \rangle$

and by

• $A^2+B^2+2AB\langle x, z \rangle=\langle y, y \rangle-\langle w, w \rangle=1-\langle w, w \rangle\le 1$

we obtain

$$(a-b\langle x, z \rangle)^2+(b-a\langle x, z \rangle)^2+2\langle x, z \rangle(a-b\langle x, z \rangle)(b-a\langle x, z \rangle)\le (1-\langle x, z \rangle^2)^2$$

which leads to

$$\langle x, z \rangle\ge ab- \sqrt{1-a^2}\sqrt{1-b^2}$$