$\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$

Question

Find the value of $\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$, if $x\in \left(0,\frac{1}{2}\right)$. I know it is equal to $\sqrt{2}+1$, but I don't know how to prove it?

Answer 1

$x$ can be set to $\sin^22\theta$ as $0<x<\frac 12\implies 0<\sin2\theta<\frac1{\sqrt 2}$

$\implies 0< 2\theta< \frac\pi4$

$1+\sqrt x=1+\sin2\theta=(\cos\theta+\sin\theta)^2\implies \sqrt{1+\sqrt x}=\cos\theta+\sin\theta$

$1-\sqrt x=1-\sin2\theta=(\cos\theta-\sin\theta)^2\implies \sqrt{1+\sqrt x}=\cos\theta-\sin\theta$ as $\cos\theta>\sin\theta>0$ as $0< \theta< \frac\pi8$

$1-x=\cos^2\theta, \sqrt{1-x}=\cos2\theta,$

As $\cos2\theta=2\cos^2\theta-1=1-2\sin^2\theta$

$1+\sqrt{1-x}=1+\cos2\theta=2\cos^2\theta, 1-\sqrt{1-x}=1-\cos2\theta=2\sin^2\theta$

$$\text{ Then, }\frac{\sqrt{1+\sqrt x}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt x}+\sqrt{1-\sqrt{1-x}}}$$

$$=\frac{\cos\theta+\sin\theta+\sqrt2\cos\theta }{\cos\theta-\sin\theta+\sqrt2\sin\theta}$$

$$=\frac{(\sqrt2+1)\cos\theta+\sin\theta }{\cos\theta+(\sqrt2-1)\sin\theta}$$

$$=(\sqrt2+1)\cdot\frac{(\sqrt2+1)\cos\theta+\sin\theta }{(\sqrt2+1)\cos\theta+(\sqrt2+1)(\sqrt2-1)\sin\theta}$$

$$=(\sqrt2+1)$$ as $(\sqrt2+1)\cos\theta+\sin\theta\ne 0$ as $\cos\theta>\sin\theta>0$ as $0< \theta< \frac\pi8$

Answer 2

This is the result of the Taylor series of the expression at $x=0$ given by Maple:

Answer 3

Hint: I'm not quite sure what exactly you need to get, but anyway - a good place to start with those kind of expresions is to use the following $$(a+b)(a-b)=a^2-b^2 \Rightarrow a+b=(a+b)\frac{a-b}{a-b}=\frac{a^2-b^2}{a-b}$$ Plug $a=\sqrt{1+\sqrt{x}}$ and $b=\sqrt{1+\sqrt{1-x}}$, that will simplify the numerator a little. Then you can repeat this.

Answer 4

a$\sqrt{1-\sqrt{x}}=a$

$\sqrt{1+\sqrt{x}}=b$

$ab=\sqrt{1-x}$

Now your expression will look something like :

$\frac{b+\sqrt{1+ab}}{a +\sqrt{1-ab}}$

Answer 5

Let $a=\sqrt{1-\sqrt{x}}, b=\sqrt{1+\sqrt{x}}$, and $b>a>0$ for all $x\in\left(0,\frac{1}{2}\right)$. One could get that $ab=\sqrt{1-x}$, $b^2-a^2=2\sqrt{x}$, $1-a^2b^2=\sqrt{x}$, and $b^2+a^2=2$. Then, one could obtain: $b+a=\sqrt{2}\cdot\sqrt{1+ab}$ and $b-a=\sqrt{2}\sqrt{1-ab}$. Now, when one would prove that the value of the above function is equal to $\sqrt{2}+1$ for all $x\in\left(0,\frac{1}{2}\right)$. Given $y=f(x)=\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}=\frac{b+\sqrt{1+ab}}{a+\sqrt{1-ab}}\equiv \sqrt{2}+1$. So, it only required to prove that $y^2-2y-1=0$. The proof is the same as to prove that $\left(\frac{b+\sqrt{1+ab}}{a+\sqrt{1-ab}}\right)^2-2\left(\frac{b+\sqrt{1+ab}}{a+\sqrt{1-ab}}\right)-\left(\frac{a+\sqrt{1-ab}}{a+\sqrt{1-ab}}\right)$ shall be equal to zero for all $x\in\left(0,\frac{1}{2}\right)$. In the case of a non-trivial solution being existed in, the portion of denominator shall not be equal to zero, and the numerator shall be zero. After expansion, one got the portion of numerator as follows: $(b+\sqrt{1+ab})^2-2(b+\sqrt{1+ab})(a+\sqrt{1-ab})-(a+\sqrt{1-ab})^2=b^2-a^2+2ab-2ab+2(b-a)\sqrt{1+ab}-2(b+a)\sqrt{1-ab}-2\sqrt{1-a^2b^2}\equiv 2\sqrt{x}+2\sqrt{2}\sqrt{1-ab}\sqrt{1+ab}-2\sqrt{2}\sqrt{1+ab}\sqrt{1-ab}-2\sqrt{1-a^2b^2}=2\sqrt{x}+2\sqrt{2}\sqrt{1-a^2b^2}-2\sqrt{2}\sqrt{1-a^2b^2}-2\sqrt{x}=0$. So, the proof was done, and we show that $y=f(x)=\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}\equiv\sqrt{2}+1$, $\forall x\in\left(0,\frac{1}{2}\right)$.