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Question is as follows:

Prove that, when $$y=\frac{\sqrt{(1+x)}+\sqrt{1+2x}+\sqrt{x}}{\sqrt{(1+x)}-\sqrt{1+2x}+\sqrt{x}}$$

then $$4y^2(1+2x)^2=(y-1)^4x(x+1)$$

Despite several attempts at this, I have been unable to derive the required result. My main line of attack has been to use:

If $\frac{a}{b}=\frac{c}{d}$ then $\frac{a+b}{a-b}=\frac{c+d}{c-d}$. So from above:

$$\frac{y}{1}=\frac{\sqrt{(1+x)}+\sqrt{1+2x}+\sqrt{x}}{\sqrt{(1+x)}-\sqrt{1+2x}+\sqrt{x}}$$ gives $$\frac{y+1}{y-1}=\frac{\sqrt{(1+x)}+\sqrt{x}}{\sqrt{1+2x}}$$

I've tried squaring etc after this but can't seem to get the result.

I also tried working backwards from the required result. After taking positive square roots and rearranging I was left with:

$$\frac{2y}{(y-1)^2}=\frac{\sqrt{(1+x)}*\sqrt{x}}{\sqrt{1+2x}}$$

I'm finding this more than a little tricky so if anyone can offer any help I would be extremely grateful.

There is a second line of a attack using the following substitutions:

$$1+x=u^2, 1+2x=v^2, x=w^2$$ This method apparently dispenses with the root but I can't get the result using this method either.

Many thanks in advance for any advice given and I hope I've given enough information in my attempted solution.

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  • $\begingroup$ You might want to give a more descriptive title. $\endgroup$ Aug 17, 2023 at 21:21
  • $\begingroup$ @NDB Thank you for your reply. $\endgroup$
    – GR L
    Aug 18, 2023 at 20:25

1 Answer 1

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Here's one approach: Noticing that the result we're trying to prove has a $(y-1)$ term, let's see what that can be written as:

$$\begin{eqnarray} y - 1 & = & \frac{\sqrt{1+x} + \sqrt{1+2x} + \sqrt{x}}{\sqrt{1+x} - \sqrt{1+2x} + \sqrt{x}} - 1 \\ & = & \frac{(\sqrt{1+x} + \sqrt{1+2x} + \sqrt{x}) - (\sqrt{1+x} - \sqrt{1+2x} + \sqrt{x})}{\sqrt{1+x} - \sqrt{1+2x} + \sqrt{x}} \\ & = & \frac{2\sqrt{1+2x}}{\sqrt{1+x} - \sqrt{1+2x} + \sqrt{x}} & (1) \\ & = & \frac{2\sqrt{1+2x}}{\sqrt{1+x} - \sqrt{1+2x} + \sqrt{x}} \frac{\sqrt{1+x} + \sqrt{1+2x} + \sqrt{x}}{\sqrt{1+x} + \sqrt{1+2x} + \sqrt{x}} \\ & = & \frac{(2\sqrt{1+2x})(\sqrt{1+x} + \sqrt{1+2x} + \sqrt{x})}{(\sqrt{1+x} + \sqrt{x})^2 - (\sqrt{1+2x})^2} \\ & = & \frac{(2\sqrt{1+2x})(\sqrt{1+x} + \sqrt{1+2x} + \sqrt{x})}{((1+x) + 2\sqrt{(1+x)x} + x) - (1 + 2x)} \\ & = & \frac{(2\sqrt{1+2x})(\sqrt{1+x} + \sqrt{1+2x} + \sqrt{x})}{(1 + 2x + 2\sqrt{(1+x)x}) - (1 + 2x)} \\ & = & \frac{(2\sqrt{1+2x})(\sqrt{1+x} + \sqrt{1+2x} + \sqrt{x})}{2\sqrt{(1+x)x}} \\ & = & \frac{(\sqrt{1+2x})(\sqrt{1+x} + \sqrt{1+2x} + \sqrt{x})}{\sqrt{(1+x)x}} & (2) \end{eqnarray}$$

Where going from $(1)$ to $(2)$ involves applying one step of the usual technique to rationalise the denominator using a difference of squares trick.

And from here we're basically done - we can multiply $(1)$ and $(2)$ together, giving an expression for $(y-1)^2$ that essentially looks like $\frac{4(1+2x)}{x(1+x)} \times y^2$.

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    $\begingroup$ Yep, that's it: (+1 :-), basic underlying is $N\cdot D= 2\sqrt{(1+x)x}$ with $N,D$ abbreviating the numerator and denominator, respectively. So luckily it is not as difficult as announced. $\endgroup$
    – Hanno
    Aug 18, 2023 at 6:32
  • $\begingroup$ Your last expression should contain a root in the denominator, and a $y$ instead of its square. $\endgroup$
    – Hanno
    Aug 18, 2023 at 6:33
  • $\begingroup$ @Hanno Thank you for your reply. $\endgroup$
    – GR L
    Aug 18, 2023 at 20:26
  • $\begingroup$ @ConMan Thank you for your worked solution. $\endgroup$
    – GR L
    Aug 18, 2023 at 20:27

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