Gram-Schmidt applied to an inequality (Exercise 4.12 of CSMC) I am stuck on the first part of Exercise 4.12 of The Cauchy-Schwarz Master Class. The exercise itself is below.
$\newcommand{\ang}[1]{{\left\langle #1 \right\rangle}}$

Show that if $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are elements of a (real or complex) inner product space $V$ and if $\Vert \mathbf{x} \Vert = \Vert \mathbf{y} \Vert = \Vert \mathbf{z} \Vert = 1$, then one has the inequality
\begin{equation} \label{eq:finale-1}
|\ang{\mathbf{x}, \mathbf{x}} \ang{\mathbf{y}, \mathbf{z}} - \ang{\mathbf{x}, \mathbf{y}} \ang{\mathbf{x}, \mathbf{z}}|^2 \leq \{\ang{\mathbf{x}, \mathbf{x}}^2 - |\ang{\mathbf{x}, \mathbf{y}}|^2\} \{\ang{\mathbf{x},\mathbf{x}}^2 - |\ang{\mathbf{x},\mathbf{z}} |^2\}
\end{equation}

The exercise encourages the use of Gram-Schmidt, so I used it to write $\mathbf{e}_1 = \mathbf{x}$ and
$$\mathbf{y} = y_1 \mathbf{x} + y_2\mathbf{e}_2,$$
and
$$\mathbf{z} = z_1 \mathbf{x} + z_2 \mathbf{e}_2 + z_3\mathbf{e}_3,$$
for some complex $y_1,y_2,z_1,z_2,z_3$ and orthonormal set $\{\mathbf{e}_1,\mathbf{e}_2, \mathbf{e}_3\}$ (although I replaced $\mathbf{e}_1$ in the above with $\mathbf{x}$ because they are equal.)
So now, $\ang{\mathbf{x}, \mathbf{x}} = 1$, $\ang{\mathbf{y},\mathbf{z}} = y_1 \bar{z_1} + y_2 \bar{z_2}$, $\ang{\mathbf{x}, \mathbf{y}} = \bar{y_1}$, and $\ang{\mathbf{x}, \mathbf{z}} = \bar{z_1}$. Substituting this into the inequality gives the equivalent inequality
$$|y_1\bar{z_1} + y_2 \bar{z_2} - \bar{y_1}\bar{z_1}|^2 \leq (1 - |\bar{y_1}|^2) (1 - |\bar{z_1}|^2).$$
I am stuck here. I can rearrange the right hand side (since all the norms are 1) so that we have
$$|y_1\bar{z_1} + y_2 \bar{z_2} - \bar{y_1}\bar{z_1}|^2 \leq |y_2|^2 |z_1|^2 + |y_2|^2 |z_3|^2.$$
The answer key has something similar to the above, except with $\ang{\mathbf{x}, \mathbf{y}} = y_1$, so that it simplifies nicely. But, I don't think this is true (at least for a complex inner product $\ang{\cdot, \cdot}$). I have a feeling I'm wrong in thinking this, but I'm not sure.
I'd be very grateful for some help!
 A: I$\newcommand{\ang}[1]{{\left\langle #1 \right\rangle}}$ found a counterexample. I'm almost positive now that the author intended to write
$$
|\ang{\mathbf{x}, \mathbf{x}} \ang{\mathbf{y}, \mathbf{z}} - \ang{\mathbf{y}, \mathbf{x}} \ang{\mathbf{x}, \mathbf{z}}|^2 \leq \{\ang{\mathbf{x}, \mathbf{x}}^2 - |\ang{\mathbf{x}, \mathbf{y}}|^2\} \{\ang{\mathbf{x},\mathbf{x}}^2 - |\ang{\mathbf{x},\mathbf{z}} |^2\}
$$
instead of
\begin{equation}
|\ang{\mathbf{x}, \mathbf{x}} \ang{\mathbf{y}, \mathbf{z}} - \ang{\mathbf{x}, \mathbf{y}} \ang{\mathbf{x}, \mathbf{z}}|^2 \leq \{\ang{\mathbf{x}, \mathbf{x}}^2 - |\ang{\mathbf{x}, \mathbf{y}}|^2\} \{\ang{\mathbf{x},\mathbf{x}}^2 - |\ang{\mathbf{x},\mathbf{z}} |^2\}
\end{equation}
(Basically, the $\ang{\mathbf{x}, \mathbf{y}}$ in the left hand side should be $\ang{\mathbf{y}, \mathbf{x}}$.) It would make sense in the answer key if this were the case (that is, using my notation above, $\ang{\mathbf{y}, \mathbf{x}} = y_1$).

The counterexample is this: Consider a complex inner product space $(\ang{\cdot, \cdot}, \mathbb{C}^3)$, where $\ang{\cdot, \cdot}$ is defined by $\ang{\mathbf{u}, \mathbf{v}} = \mathbf{u} \cdot \bar{\mathbf{v}}$, where $\cdot$ is the typical dot product.
This satisfies all the requirements for a complex inner product space according to here, except I take $\ang{c \mathbf{u}, \mathbf{v}} = c \ang{\mathbf{u}, \mathbf{v}}$, hence why the conjugate sign is on $\mathbf{v}$ instead of $\mathbf{u}$.
If you take $\mathbf{x} = (1,0,0)$, $\mathbf{y} = (i\sqrt{0.5}, \sqrt{0.5}, 0)$, and $\mathbf{z} = (\sqrt{0.9}, 0, \sqrt{0.1})$, then
$$\Vert \mathbf{x} \Vert = \Vert \mathbf{y} \Vert = \Vert \mathbf{z} \Vert = 1,$$
and
$$\ang{\mathbf{x}, \mathbf{y}} = -i\sqrt{0.5}, \ \ang{\mathbf{y}, \mathbf{z}} = i\sqrt{0.45}, \ \ang{\mathbf{x}, \mathbf{z}} = \sqrt{0.9}.$$
The left hand side of the inequality evaluates to
$$|1 \cdot i\sqrt{0.45} - (-i\sqrt{0.5}) \cdot \sqrt{0.9}|^2 = |2i\sqrt{0.45}|^2 = 1.8,$$
and the right hand side of the inequality evaluates to
$$\{1 - |-i\sqrt{0.5}|^2\} \{1 - |\sqrt{0.9}|^2\} = \{1-0.5\}\{1-0.9\}=0.05.$$
Obviously, $1.8 \not\leq 0.05$, so the inequality is false.
