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!
Best Answer
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, $$\langle z, x \rangle=\langle v+by,u+ay\rangle=ab+\langle v,u\rangle\ge ab-\sqrt{1-a^2}\sqrt{1-b^2}.$$