<s>[INST] Solve this equation $\frac{1}{x+\sqrt{x^2-1}}=\frac{1}{4x}+\frac{3x}{2x^2+2}$ 
Solve this equation
  $$\frac{1}{x+\sqrt{x^2-1}}=\frac{1}{4x}+\frac{3x}{2x^2+2}$$

This equation has solutions $x=\pm 1$ but I cannot find a solution.
Do you think using an inequality will is the way forward?
 [/INST] LHS part is equal to $x - \sqrt{x^2-1}$ and RHS is $\frac{1+ 7x^2}{4x^3 +4x}$. Thus,
$$ x - \sqrt{x^2-1} =  \frac{1+ 7x^2}{4x^3 +4x} \Rightarrow x^2 -1 = \left(\frac{1+7x^2}{4x^3 +4x} -x \right)^2 \Rightarrow (x^2-1)(4x^3+4x)^2 = (-4x^4 +3x^2+1)^2 \Rightarrow 40x^6 -17x^4 -22x^2 -1 = 0.$$ I think you can solve now.
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