Prove the statement is true given that $x,y$ are members of an inner product space $V$ over a field $\mathbb{F}$
I don't think my proof of this statement is correct:
$\begin{aligned} \frac{1}{|| x|| \cdot|| y||} | \langle x, y \rangle | &= \frac{1}{|| x||} \cdot \frac{1}{||y||} | \langle x, y \rangle | \newline &= \frac{1}{||y||} | \langle \frac{x}{||x||}, y\rangle | \newline &= \frac{1}{||y||} | \overline{ \langle y, \frac{x}{||x||} \rangle} | \newline &= |\overline{\langle \frac{y}{||y||}, \frac{x}{||x||} \rangle} | \newline &= | \langle \frac{ x}{|| x||}, \frac{ y}{|| y||} \rangle | \end{aligned}$
Is this the right way to prove this?
As pointed out by another user in a comment, your proof is correct but the fourth equality is a bit unclear. It is probably simpler to prove the statement as follows. Let $\alpha=\frac{1}{\|x\|}$ and $\beta=\frac{1}{\|y\|}$. If we adopt the convention that the inner product is linear in the first argument and conjugate-linear in the second argument, then $$ \alpha\beta\,|\langle x, y\rangle| =|\alpha||\beta||\langle x, y\rangle| =|\alpha\overline{\beta}\langle x, y\rangle| =|\alpha\langle x, \beta y\rangle| =|\langle \alpha x, \beta y\rangle|. $$