How does $\left(p+\frac{2-p^2}{r}\right)^2<2$ simplify to $2pr+2-p^2<r^2$?

Question

How the inequality given below $$\left(p+\frac{2-p^2}{r}\right)^2<2$$ will eventually be simplified to the inequality $$2pr+2-p^2<r^2$$ I got the reference from this quora Link .

Answer 1

As given $$\left(p+\frac{2-p^2}{r}\right)^2<2$$ we can simplify it as $$ (pr+2-p)^2-2r^2<0$$ $$\Rightarrow p^2r^2+4+p^4-4p^2+4pr-2p^3r-2r^2<0$$ $$\Rightarrow r^2(p^2-2)+2(2-p^2)-p^2(2-p^2)-2pr(p^2-2)<0$$ $$\Rightarrow -r^2(2-p^2)+2(2-p^2)-p^2(2-p^2)+2pr(2-p^2)<0$$ $$\Rightarrow (2-p^2)(-p^2+2+2pr-r^2)<0$$ From the link we knows that $$(2-p^2)>0$$ $$\Rightarrow -p^2+2+2pr-r^2<0$$ $$\Rightarrow 2pr+2-p^2<r^2$$