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

Question

Find all real values of $ x>1$ which satisfy$$ \frac{x^2}{x-1} + \sqrt{x-1} +\frac{\sqrt{x-1}}{x^2} = \frac{x-1}{x^2} + \frac{1}{\sqrt{x-1}} + \frac{x^2}{\sqrt{x-1}}$$

Well I first observe that each factor has its own inverse involve in the equation and rewriting

gives $$\left( \frac{x^2}{x-1}-\frac{x-1}{x^2} \right) + \left(\sqrt{x-1} -\frac{1}{\sqrt{x-1}} \right)+\left(\frac{\sqrt{x-1}}{x^2} -\frac{x^2}{\sqrt{x-1}}\right) =0 $$

However after expanding all the terms it does not look wise to do so. Does anyone has an idea

Answer 1

Let $$A=\frac{x^2}{x-1},\qquad B=\sqrt{x-1},\qquad C=\frac{\sqrt{x-1}}{x^2}$$

Then, we have $$A+B+C=\frac 1A+\frac 1B+\frac 1C\qquad\text{and}\qquad ABC=1$$ from which we have $$A+B+C=AB+BC+CA\qquad\text{and}\qquad ABC=1$$

Let $s:=A+B+C=AB+BC+CA$.

It follows from these that $A,B,C$ are the solutions of $$t^3-st^2+st-1=0,$$ i.e. $$(t-1)(t^2+t+1-st)=0$$

So, we see that either $A,B$ or $C$ has to be equal to $1$.

$A=\frac{x^2}{x-1}=1\implies x^2-x+1=0$ has no real solutions.

$B=\sqrt{x-1}=1\implies x=2$ which is sufficient.

$C=\frac{\sqrt{x-1}}{x^2}=1\implies x^4-x+1=0$.

Let $f(x):=x^4-x+1$. Then, $f'(x)=4x^3-1=0$. So, $f'(x)\gt 0$ for $x\gt 1$.

Since $f(1)=1\gt 0$, we see that $f(x)=0$ has no real solutions.

Therefore, $\color{red}{x=2}$ is the only solution.

Answer 2

Let $\sqrt {x-1}=t \implies x= 1+t^2 \quad x^2=(1+t^2)^2$

$$\frac{(1+t^2)^2}{t^2} + t +\frac{t}{(1+t^2)^2} = \frac{t^2}{(1+t^2)^2} + \frac{1}{t} + \frac{(1+t^2)^2}{t}$$

$$(1+t^2)^4 + t^3(1+t^2)^2 +t^3 = t^4 + t(1+t^2)^2 + t(1+t^2)^4$$

$$(t - 1)(t^2 - t + 1) (t^2 + t + 1) (t^4 + 2 t^2 - t + 1)=0 \implies t=1$$