If $A$ and $B$ are vectors in $\mathbb C^n$, the exercise is to prove that:
$$-2 \le \frac{A\cdot B + \overline{A\cdot B}}{|A||B|} \le 2.$$
I tried using the Cauchy-Schwarz inequality: $|A\cdot B| \le |A||B|$, but it only work with the modulus.
I guess it is possible to convert this identity to real vectors, stating that $\mathbb C^n$ is isometric to $\mathbb R^{2n}$, but I am unable to formalise and prove this.
How about: $$(A\cdot B+\overline{A\cdot B})^2 = (A\cdot B)^2+2(A\cdot B)(\overline{A\cdot B})+(\overline{A\cdot B})^2\leq $$ $$\leq 2|A|^2|B|^2+2\Re((A\cdot B)^2)\leq 2|A|^2|B|^2+2|A\cdot B|^2\leq 4|A|^2|B|^2.$$ Here, if $z = A\cdot B = a+bi,$ then: $$z^2+\overline{z}^2 = (a+bi)^2+(a-bi)^2 = 2a^2-2b^2\leq 2a^2+2b^2=2|z|^2.$$