Repeating Radicals

Let $y$ be similar to the given expression $x$ but with the addition and subtraction signs switched. Then we have that

$x = \sqrt{1 + \dfrac{1}{y}}$ and $y = \sqrt{1 - \dfrac{1}{x}}$

$\Longrightarrow x^{2} - y^{2} = \dfrac{1}{y} + \dfrac{1}{x} \Longrightarrow (x - y)(x + y) = \dfrac{x + y}{xy} \Longrightarrow x - y = \dfrac{1}{xy},$

since clearly $x + y \ne 0,$ (as both $x,y \gt 0$ ). We then have that

$(x - y)xy = 1 \Longrightarrow x^{2}y - xy^{2} = 1$

$\Longrightarrow \left(1 + \dfrac{1}{y}\right)y - x\left(1 - \dfrac{1}{x}\right) = 1 \Longrightarrow y + 1 - (x - 1) = 1 \Longrightarrow x - y = 1$

$\Longrightarrow xy = 1 \Longrightarrow x(x - 1) = 1 \Longrightarrow x^{2} - x - 1 = 0$

$\Longrightarrow x = \dfrac{1 + \sqrt{5}}{2} = 1.618033989.....,$ since $x \gt 0.$

Thus $\lfloor 1000x \rfloor = \boxed{1618}.$

Here is a dirty little solution:

$y=\sqrt{1+\frac1{\sqrt{{1-\frac1y}}}}$ Which when rearranging gets us $y=(y-1)^3(y+1)^2$ Let $u=y-1$ then $u+1=u^3(u+2)^2\\ \implies 0=u^5 + 4u^4 + 4u^3-u-1$ Now Setting up a Newton-Raphson approximation: $x_{n+1}=x_n-\frac{x_n^5+4x_n^4+4x_n^3-x_n-1}{5x_n^4+16x_n^3+12x_n^2-1}$ Iterate until about $n=4$ where it should stop changing digits after 3 d.p. Then add one and you should get an answer of $\boxed{1618}$ .

This is under the assumption that a calculator is allowed, which I take as implied due to the nature of the answer.

Mathematica does it in one line:

Floor[1000 x] /. Solve[Sqrt[1+1/Sqrt[1-1/x]]==x,{x}]

-a fool with a tool...