For two nonzero vectors $x$ and $y$ in $\mathbb{R}^n$ is there an inequality similar to \begin{align*} \langle x , y \rangle > \Vert y \Vert - \Vert x \Vert . \end{align*}By similar I mean that it would be also ok to have, for example, \begin{align*}\langle x , y \rangle > \Vert y \Vert^2 - \Vert x \Vert^2. \end{align*}
And, if not, are there any special cases where the inequality holds? For example if I assume $\Vert y_i \Vert > \Vert x \Vert$? I have locked for similar inequalities, but somehow none seems to fit my problem.
No. Fix $x$ and let $y$ be orthogonal to $x$ and of norm one. If the inequality would hold, then for $c\in \mathbb{R}$ we would have $$0 = \langle x,cy \rangle > \|cy\| - \|x\| = |c| - \|x\| $$
Choosing $c > \|x\|$, we obtain a contradiction. The same would hold if we were looking at an inequality of this type with the squares of the norms.
EDIT: As Yanko points out in the comments, you can also find counterexamples by making $x$ and $y$ "almost" orthogonal.
You might be interested in the polarization identity, which looks a bit similar to what you asked. It says
$$\langle x,y\rangle = \frac{\|x+y\|^2 - \|x-y\|^2}{4} = \frac{\|x+y\|^2 - \|x\|^2 - \|y\|^2}{2}$$