On the approximate Birkhoff orthogonality In an inner product space $(X,\langle \cdot \vert \cdot \rangle)$, an approximate orthogonality with $\varepsilon \in [0,1)$ is defined as follows: $$x\perp^{\varepsilon}y  \iff |\langle\,x\,|\,y\,\rangle| \leq \varepsilon \| x\|\,\|y\|.$$
On the other hand, in a normed space $(X,\Vert \cdot \Vert )$, an approxmiate Birkhoff orthogonality with $\varepsilon \in [0,1)$ is defined as follows: $x\perp_B^{\varepsilon} y$ iff $$\forall \lambda \in \mathbb{K}\; : \; \Vert x+\lambda y\Vert^2  \geq \Vert x\Vert^2 -2\varepsilon \Vert x\Vert \Vert \lambda y\Vert .$$ How can I show that $\perp^{\varepsilon}_B =\perp^{\varepsilon}$ in an inner product space?
 A: Suppose that $x\perp^\varepsilon y$. Then
\begin{align*}
\|x+\lambda y\|^2
&=\|x\|^2+\|\lambda y\|^2+2\operatorname{Re}\langle x,\lambda y\rangle\\[0.3cm]
&\geq\|x\|^2+\|\lambda y\|^2-2|\langle x,\lambda y\rangle|\\[0.3cm]
&\geq\|x\|^2-2\varepsilon\,\| x\|\,\|\lambda y\|.
\end{align*}
Conversely, suppose that $x\perp_B^\varepsilon y$. We have
$$
\|x\|^2-2\varepsilon\|x\|\,\|\lambda y\|\leq\|x+\lambda y\|^2=\|x\|^2+\|\lambda y\|^2+2\operatorname{Re}\langle x,\lambda y\rangle.
$$
The left-hand-side does not change if we replace $x$ with $-x$. Thus
$$
\|x\|^2-2\varepsilon\|x\|\,\|\lambda y\|\leq\|-x+\lambda y\|^2=\|x\|^2+\|\lambda y\|^2-2\operatorname{Re}\langle x,\lambda y\rangle.
$$
This gives us
$$\tag1
2\operatorname{Re}\langle x,\lambda y\rangle\leq2\varepsilon\|x\|\,\|\lambda y\|+|\lambda|^2\,\|y\|.
$$
Write $\langle x,y\rangle=\beta\,|\langle x,y\rangle|$, with $|\beta|=1$. Putting $\lambda=\beta\,t$, with $t>0$, we have
$$\tag2
2\operatorname{Re}\langle x,\lambda y\rangle=2t\,|\langle x,y\rangle|,
$$
while
$$\tag3
2\varepsilon\|x\|\,\|\lambda y\|+|\lambda|^2\,\|y\|
=2t\,\varepsilon\|x\|\,\| y\|+t^2\,\|y\|
$$
Using $(2)$ and $(3)$, now $(1)$ reduces to
$$
|\langle x,y\rangle|\leq\varepsilon\|x\|\,\|y\|+\frac t2\,\|y\|.
$$
As this works for all $t>0$,
$$\tag4
|\langle x,y\rangle|\leq\varepsilon\|x\|\,\|y\|.
$$
