Proving $\sup\limits_{\theta,\psi}\{||e^{i\theta}x+e^{i\psi}y||^2:\theta,\psi\in \Bbb R\}=||x||^2+||y||^2+2 Re<x,y>$ where $x,y \in \mathbb{C}^n$

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Q. Let $x,y \in \mathbb{C}^n$. Consider $f(x,y)=\sup\limits_{\theta,\psi}\{\lVert e^{i\theta}x+e^{i\psi}y\rVert^2\colon \theta,\psi\in \Bbb R\}.$ Which of the following is/are correct?

  1. $f(x,y)\leq\lVert x\rVert^2+\lVert y\rVert^2+2|\langle x,y\rangle|$

  2. $f(x,y)= \lVert x\rVert^2+\lVert y\rVert^2+2 Re\langle x,y\rangle$

  3. $f(x,y)=\lVert x\rVert^2+\lVert y\rVert^2+2\lvert\langle x,y\rangle\vert$

  4. $f(x,y)>\lVert x\rVert^2+\lVert y\rVert^2+2\lvert\langle x, y\rangle\rvert$

I can easily reject the option $(4)$ by simply assuming $y=0 \in \Bbb C^n$, How can we deal others by connecting the quantity $\lVert x+y\rVert^2$ with $f(x,y)$?

How to continue with $\lVert e^{i\theta}x+e^{i\psi}y\rVert^2=\langle e^{i\theta}x+e^{i\psi}y, \overline{e^{i\theta}x+e^{i\psi}y}\rangle$ ?

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By expanding the inner product we get $$\|e^{i\theta}x+e^{i\psi}y\|^2 = \|e^{i\theta}x\|^2 + \|e^{i\psi}y\|^2 + 2\operatorname{Re} \langle e^{i\theta}x,e^{i\psi}y\rangle = \|x\|^2+\|y\|^2 + 2 \operatorname{Re}e^{i(\theta-\psi)}\langle x,y\rangle.$$ Now this can be majorized by $$\|x\|^2+\|y\|^2 + 2 |e^{i(\theta-\psi)}\langle x,y\rangle| \le \|x\|^2+\|y\|^2 + 2 |\langle x,y\rangle|.$$ Moreover, writing the complex number $\langle x,y\rangle$ in polar form, we have $$\langle x,y \rangle = |\langle x,y\rangle| e^{i\eta}, \quad \text{ for some } \eta \in \Bbb{R}$$ so by setting $\theta = 0, \psi = \eta$, we have

\begin{align} \|x\|^2+\|y\|^2 + 2 \operatorname{Re}e^{i(\theta-\psi)}\langle x,y\rangle &= \|x\|^2+\|y\|^2 + 2 \operatorname{Re}e^{-i\eta}\langle x,y\rangle \\ &= \|x\|^2+\|y\|^2 + 2 \operatorname{Re}|\langle x,y\rangle|\\ &= \|x\|^2+\|y\|^2 + 2|\langle x,y\rangle|. \end{align} We can conclude that $f(x,y) = \|x\|^2+\|y\|^2 + 2|\langle x,y\rangle|$ so $1$ and $3$ are correct.

To see that $2$ isn't correct in general, pick any two vectors $x,y \in \Bbb{C}^n$ such that $\operatorname{Re} \langle x,y\rangle < |\langle x,y\rangle|$.