Let $p\in (0,1)$. How can I prove that there exists $c>0$ such that
$$\prod_{i=0}^{\infty}\frac{1-(1-p)^2\bigg(1-\frac{1}{\sqrt 2}\bigg)^{2i}}{\Bigg(1+(1-p)\bigg(1-\frac{1}{\sqrt 2}\bigg)^{i}\Bigg)^2}\geq c.$$ Is it possible to extimate this constant $c$?
I think that we can bound from below the previous product with
$$\prod_{i=0}^{\infty}\frac{1-(1-p)^2\bigg(1-\frac{1}{\sqrt 2}\bigg)^{2i}}{1+(1-p)^2\bigg(1-\frac{1}{\sqrt 2}\bigg)^{2i}}.$$ But after how can I conclude?
Thank you very much.