Consequences of Proposition 1.1 on Semi-inner-product $A$-modules

Proposition 1.1 If $E$ is a semi-inner-product $A$-module and $x,y \in E$, then $\langle y,x \rangle \langle x,y \rangle \le \|\langle x,x \rangle\| \langle y,y \rangle$

I am trying to verify that $N := \{x \in E \mid \langle x,x \rangle = 0 \}$ is an inner-product $A$-module (i.e., a sub-$A$-module of $E$). I need to verify that it is a linear subspace of $E$ and that it is closed under multiplication on the right by elements of $A$. Clearly it is closed under scalar multiplication. If $\alpha \in \Bbb{C}$ and $x \in N$, then

$$\langle \alpha x , \alpha x \rangle = \langle x,x\rangle |\alpha|^2 \langle x , x\rangle = |\alpha|^2 0 = 0,$$ thereby proving $\alpha x \in N$.

Similarly, it is closed under multiplication on the right by $A$. If $x \in N$ and $a \in A$, then

$$\langle xa,xa \rangle = a^* \langle x,x \rangle a = a^*0a=0,$$ thereby provint $xa \in N$.

However, I am having trouble verifying that it is closed under vector addition. The book I'm working through claims that it follows from proposition 1.1; but I don't see it.

EDIT: I am also having trouble verifying the following corollaries of proposition 1.1

(1) $\|\langle x,y \rangle \| \le \|x\| \cdot \|y\|$, where $\|x\| = \|\langle x,x\rangle\|^{\frac{1}{2}}$

(2) $|\langle x,y \rangle| \le \|x\| \cdot |y|$, where $|y|:= \langle y,y \rangle^{\frac{1}{2}}$

If $\langle x,x\rangle=\langle y,y\rangle=0$, then \begin{align} \langle x+y,x+y\rangle&=\langle y,x\rangle+\langle x,y\rangle = 2\operatorname{Re}\langle x,y\rangle. \end{align} Now, by Proposition 1.1 you have $$ (\langle x,y\rangle)^*\langle x,y\rangle=\langle y,x\rangle\langle x,y\rangle\leq \|\langle x,x\rangle\|\,\langle y,y\rangle=0. $$ So $\langle x,y\rangle=0$ and then $\langle x+y,x+y\rangle=0$.

You also have $$ |\langle x,y\rangle|^2=\langle y,x\rangle\langle x,y\rangle\leq \|\langle x,x\rangle\|\,\langle y,y\rangle. $$ As the square root is operator-monotone, $$ |\langle x,y\rangle|\leq \|x\| \,\langle y,y\rangle^{1/2}=\,\|x\|\,|y|, $$ giving $(2)$. For $(1)$, you just take norms in $(1)$.