Calculate the number of solutions

Hints:

• Your observation "The LHS is concave, the RHS is convex" puts an upper bound on the number of solutions
• The square roots on the LHS require $x \le \frac14$
• Consider the difference between the two sides for some sensible values of $x$: $\frac14$ and $0$ look obvious possibilities and you might also think about a very negative value

Hint : If $X= \sqrt{a}+ \sqrt{b}+\sqrt{c}$ ... keep squaring and rearranging ... \begin{eqnarray*} X &=& \sqrt{a}+ \sqrt{b}+\sqrt{c} \\ \frac{X^2-a-b-c}{2} &=& \sqrt{ab}+ \sqrt{bc}+\sqrt{ca} \\ \left(\frac{X^2-a-b-c}{2} \right)^2 &=& 2\sqrt{abc}(\sqrt{a}+ \sqrt{b}+\sqrt{c})=2\sqrt{abc}X. \\ \end{eqnarray*} Now square one final time & what order is the polynomial ?

I think the argumentation becomes easier if you transform the given equation to:

$$x+\frac{1}{1+\sqrt{1-x}}+\frac{2}{1+\sqrt{1-2x}}+\frac{4}{1+\sqrt{1-4x}}=\frac{1}{x}$$

The left side is almost linear (beside a small rest), growing from the $3^{rd}$ till to the $1^{st}$ part of the coordinate plane. The right side has one falling curve in the $3^{rd}$ part and one falling curve in the $1^{st}$ part. So there is one negative (we can see this using $x\to -\infty$ and $x=-2$) and one positive solution (we can see this using $x\downarrow 0$ and $x=1/4$) .