Equation with radicals and reciprocals Find all $x\in\mathbb{R}$ satisfying $$x=\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}.$$

Multiply both sides by $x^{1/2}$ to get $$x^{3/2} = \sqrt{x^2-1} + \sqrt{x-1}.$$Making the substitution $a = \sqrt{x^2 - 1}$, $b=\sqrt{x-1}$, we have $a+b = x^{3/2}$ and $a^2 - b^2 = x^2 - x$, so $a-b = (a^2-b^2)/(a+b) = \frac{x-1}{x^{1/2}}$. Then solving for $b$ gives $b = \frac{x^{3/2} - x^{1/2} + x^{-1/2}}{2}$. Then, we get the equation $$\frac{x^{3/2} - x^{1/2} + x^{-1/2}}{2} = \sqrt{x-1},$$which we can expand/rearrange to get $$x^4 - 2x^3 - x^2 + 2x + 1 = 0,$$which factors as $$\left(x^2 - x - 1\right)^2 = 0 \implies x^2 - x - 1 = 0,$$which has solutions $x = \frac{1\pm\sqrt{5}}{2}$, but note that we must have $x \ge 1$ by the problem, so $\boxed{x = \frac{1+\sqrt{5}}{2}}$.
I would be interested in seeing a slicker solution, though. 

Square both sides first: $x^2 = x-\dfrac{2}{x} + 1+ 2\sqrt{x-1-\dfrac{1}{x}+\dfrac{1}{x^2}}\Rightarrow x\left(x-1-\dfrac{1}{x}+\dfrac{1}{x^2}\right)+\dfrac{1}{x} =2\sqrt{x-1-\dfrac{1}{x}+\dfrac{1}{x^2}}$. but by AM-GM inequality, the $LHS \geq RHS$. Thus equality occurs or the equation holds if $x\left(x-1-\dfrac{1}{x}+\dfrac{1}{x^2}\right) = \dfrac{1}{x}$. From this you have it back that $x^2 - x - 1 = 0 \Rightarrow x = ...$
