A golden question $\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$ How would you solve this problem for real $x$?
$$\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$

It can be easily shown that both equations
$$x=\sqrt{2+\sqrt{2-x}}\tag{1}$$
and $$x=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}\tag{2}$$
have the same real solution, i.e. the golden ratio, $$x = \frac {\sqrt 5 + 1}2=\phi.$$
The main question is derived by combining (1) and (2) into one equation. The final solution is still the same, but the approach might not be as straightforward.

(The following is about the other roots in ConMan's answer, but it was too long for a comment.)
Courtesy WA, the original equation can be brought to polynomial form with groebnerBasis[ {u - v - w, x - 2 + (u^2 - 2)^2, v^2 x - x^2 + 1, w^2 x - x + 1}, {x}, {u, v, w}  ] which returns and factors:
$$
x^6 - 10 x^5 + 39 x^4 - 22 x^3 - 63 x^2 + 32 x + 32 = (x^2 - x - 1) (x^4 - 9 x^3 + 31 x^2 - 32)
$$

*

*The positive root $\,\varphi\,$ of the quadratic and the negative real root $\,\simeq -0.8959\,$ of the quartic satisfy the original equation $\,\sqrt{2 + \sqrt{2-x}} = x = \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\,$.


*The negative root $\,-\dfrac{1}{\varphi}\,$ of the quadratic satisfies $\,\sqrt{2 - \sqrt{2-x}} = -x = - \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\,$.


*The positive real root $\,\simeq 1.2197\,$ of the quartic satisfies $\,\sqrt{2 - \sqrt{2-x}} = \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\,$.


*The two complex roots $\,\simeq 4.3381 \pm 3.2349 i\,$ of the quartic satisfy $\,\sqrt{2 - \sqrt{2-x}} = \sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\,$.
