fonction inverse of $f(x)=\sqrt{x+2}+\sqrt{x}$

Question

How can i show that the fonction inverse of $f(x)=\sqrt{x+2}+\sqrt{x}$ is $\dfrac{x^{4}-4x^{2}+4}{4x^{2}}$ My attempt: $$\begin{cases} f^{-1}(x)=y \\ x\in[\sqrt{2};+\infty) \end{cases}\iff \begin{cases} x=f(y) \\ x\in[0;+\infty) \end{cases} $$ \begin{align} x=f(y) &\iff x=\sqrt{y+2}+\sqrt{y}\\ &\iff x^2=y+2+y+2\sqrt{(y+2)y}\\ &\iff x^2=2y+2+2\sqrt{(y+2)y}\\ &\iff \frac{x^2}{2}=y+1+\sqrt{(y+2)y}\\ &\iff \frac{x^2}{2}-1=y+\sqrt{(y+2)y}\\ \end{align}

Answer 1

With $x\ge0$ and $y\ge\sqrt2$, $$(y-\sqrt x)^2=y^2-2\sqrt xy+x=(\sqrt{x+2})^2=x+2$$ gives

$$\sqrt x=\frac{y^2-2}{2y}.$$

Answer 2

You have reached a good point: $$\frac{x^2}{2}-1=y+\sqrt{(y+2)}y\implies \frac{x^2}{2}-1-y=\sqrt{(y+2)}y\implies \frac{x^4}{4}+1+y^2-x^2-x^2 y+2y=y^2+2y$$ Here, we have simply squared both sides, knowing that LHS and RHS are always greater or equal to $0$ with your contrainst.

Now, we have to solve for $y$. We obtain: $$\implies \frac{x^4}{4}+1-x^2-x^2 y=0 \implies \frac{x^4+4-4x^2}{4}=x^2 y\implies y=f^{-1}(x)=\frac{x^4-4x^2+4}{4x^2}$$

Answer 3

Per the request of the OP, here is solution based directly on his approach. His last line is: $$ \frac x2-1=y+\sqrt{(y+2)y}=\sqrt y(\sqrt y+\sqrt{y+2})=\sqrt y\cdot x $$ Squaring both sides: $$ \frac{(x^2-2)^2}4=x^2y $$ or: $$ \frac{(x^2-2)^2}{4x^2}=y $$