Strengthened Cauchy-Schwarz inequality

Question

I'm looking for some simple proof of the following consequence of the "strengthened" Cauchy-Schwarz inequality:

Let $\mathcal{H}$ be a real Hilbert space such that $\mathcal{H}=\mathcal{V}\oplus\mathcal{W}$ with an inner product $(\cdot,\cdot)$ and the induced norm $\|\cdot\|$. Let $\alpha\in[0,1)$ be the smallest constant such that $$\tag{1} (v,w)\leq\alpha\|v\|\|w\|\quad\forall v\in\mathcal{V},\;w\in\mathcal{W}. $$ Then for any $v\in\mathcal{V}$, $$\tag{2} \|v\|\leq\frac{1}{\sqrt{1-\alpha^2}}\inf_{w\in\mathcal{W}}\|v+w\|. $$

Obviously, (2) is equivalent to $$\tag{3} (1-\alpha^2)(v,v)\leq (v+w,v+w) \quad\Leftrightarrow\quad 0\leq\alpha^2(v,v)+2(v,w)+(w,w) $$ for any $v\in\mathcal{V}$ and $w\in\mathcal{W}$.

I'm however not sure how to verify that (3) holds using (1).

Thanks in advance.

Answer 1

$$\alpha^2(v,v) + 2(v,w) + (w,w) \geqslant \alpha^2(v,v) - 2\alpha\lVert v\rVert\,\lVert w\rVert + (w,w) = \left(\alpha \lVert v\rVert - \lVert w\rVert\right)^2 \geqslant 0,$$

since by $(1)$, we have $(v,w) \geqslant - \alpha\lVert v\rVert\,\lVert w\rVert$.