$(2x+\sqrt{4x^2+1})(\sqrt{y^2+4}-2)>=y>0$, minimum vale of $x+y$

Question

The answer is 2. How to consider such an inequality? I think it like parabola or some assumption with special skills. Please help and thanks.

Answer 1

$g(y)=\frac y{\sqrt{y^2+4}-2}$ is monotonic decreasing and $g(y)\ge 1$ for $y\ge 0$, and function $f(x)=2x+\sqrt{4x^2+1}$ is monotonic increasing and $f(x)\ge 1 $ requires $x\ge 0$.

First we could easily find out that when x+y is minimized, $f(x)=g(y)=z\ge 1$ so $x=f^{-1}(z)=\frac{z^2-1}{4z}$ and $y=g^{-1}(z)=\frac{4z}{z^2-1}$.

So $x+y=\frac{z^2-1}{4z}+\frac{4z}{z^2-1}\ge 2$,where $z\ge 1$

The minimal value is reached when $z^2-1=4z$.