Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:

$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$$

(original image)
I think we need to simplify it writing it in summation sign as you can see here:
$$\frac{\sum\limits_{n=1}^{99} \sqrt{10 + \sqrt{n}}}{\sum\limits_{n=1}^{99} \sqrt{10 - \sqrt{n}}}$$
or in Wolfram Alpha input in comments.
I can compute it too! It's easy to write a script for this kind of question.
I need a way to solve it. How would you solve it on a piece of paper?
 A: I'd like to add one more answer "by color-coding".
First, observe that
$$
\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}} = \sqrt{2}\sqrt{n-\sqrt{n^2-k}}.
$$
Now the problem.
Designate
$$
x = \frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}},
$$
and calculate
$$
x-1 =
\frac{\overbrace{\sqrt{10+\sqrt{1}}-\sqrt{10-\sqrt{1}}}^{\color{red}{\sqrt{2}\sqrt{10-\sqrt{99}}}} 
+ \cdots + 
\overbrace{\sqrt{10+\sqrt{99}}-\sqrt{10-\sqrt{99}}}^{\color{blue}{\sqrt{2}\sqrt{10-\sqrt{1}}}} }{\color{blue}{\sqrt{10-\sqrt{1}}}+\cdots+\color{red}{\sqrt{10-\sqrt{99}}}}.
$$
Everything simplifies to $\sqrt{2}$, thus
$$
x = 1 + \sqrt{2}
$$
A: Hint: Show that for all relevant $n$
$$
\frac{\sqrt{10+\sqrt{100-n}}+\sqrt{10+\sqrt{n}}}{\sqrt{10-\sqrt{100-n}}+
\sqrt{10-\sqrt{n}}}=\sqrt2 +1.
$$
IOW pair up terms in the numerator and the denominator starting from both ends.

[Edit:]
Claim. Assume that $a,b,c$, all positive, are the lengths of the sides of a right angled triangle - a Pythagorean triple if you like - $c$ is the hypotenuse. Then
$$
\frac{\sqrt{c+a}+\sqrt{c+b}}{\sqrt{c-a}+\sqrt{c-b}}=1+\sqrt2.
$$
Proof. The left hand side of the claim is clearly immune to scaling. We can adjust the scale so that $c-b=2$. Then a calculation (familiar to enthusiasts of Pythagorean triples) shows that for some positive real number $m$ we have
$$c=m^2+1,\quad b=m^2-1,\quad a=2m.$$ (IOW instead of the usual integer parametrization in terms of $(m,n)$ we set $n=1$, and let $m$ be arbitrary.) We then see that the numerator is $m+1+\sqrt2 m=m(1+\sqrt2)+1$, and the denominator is $m-1+\sqrt2$. Because $(\sqrt2+1)^{-1}=\sqrt2-1$ the claim follows. Q.E.D.
Leaving it to the reader to derive the identity of my hint as a corollary of the claim. 
A: For the Calculation of  $$\displaystyle \frac{\sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}}{\sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}} = $$ 
Let $$\displaystyle A_{n} = \sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}$$ and $$\displaystyle B_{n} = \sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}$$ , where $n>1$
Now $$\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right)^2 = 2n-2\sqrt{n^2-k}$$
So $$\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right) = \sqrt{2}\cdot \sqrt{n-\sqrt{n^2-k}}$$
So $$\displaystyle \sum_{k=1}^{n^2-1}\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right) = \sum_{k=1}^{n^2-1}\sqrt{2}\cdot \sqrt{n-\sqrt{n^2-k}}$$
So So $$\displaystyle \sum_{k=1}^{n^2-1}\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right) = \sum_{k=1}^{n^2-1}\sqrt{2}\cdot \sqrt{n-\sqrt{k}}$$
So $$A_{n}-B_{n} = B_{n}\sqrt{2}$$
So $$A_{n} = B_{n}\left(1+\sqrt{2}\right)$$
So $$\displaystyle \frac{A_{n}}{B_{n}} = 1+\sqrt{2}$$
Now Put $\displaystyle n^2-1 = 99\Rightarrow n= 10\;,$ So we get $$\displaystyle \frac{\sum_{k=1}^{99}\sqrt{n+\sqrt{k}}}{\sum_{k=1}^{99}\sqrt{n-\sqrt{k}}} =\frac{A_{10}}{B_{10}} = 1+\sqrt{2}.$$
A: Let $f(n)=\frac{\sum_{k=1}^{n} \sqrt{\sqrt{n+1}+\sqrt{k}}}{\sum_{k=1}^{n} \sqrt{\sqrt{n+1}-\sqrt{k}}}$, and the original problem is to calculate $f(99)$.
Now we claim that $f(n)=\sqrt{2}+1$ for $\forall n>0$.
$f(n)=\frac{\sqrt{\sqrt{2}+1}}{\sqrt{\sqrt{2}-1}} \cdot \frac{\sum_{k=1}^{n} \sqrt{(\sqrt{n+1}+\sqrt{k})(\sqrt{2}-1)}}{\sum_{k=1}^{n} \sqrt{(\sqrt{n+1}-\sqrt{k})(\sqrt{2}+1)}}$
$=(\sqrt{2}+1) \cdot \frac{\sum_{k=1}^{n} \sqrt{A(k)+B(k)}}{\sum_{k=1}^{n} \sqrt{A(k)-B(k)}}$, where $A(k)=\sqrt{2(n+1)}-\sqrt{k}$, $B(k)=\sqrt{2k}-\sqrt{n+1}$.
Notice that:
$\left(\sqrt{A(k)+B(k)}-\sqrt{A(k)-B(k)}\right)^2=2A(k)-2\sqrt{A(k)^2-B(k)^2}\\=2\left(\sqrt{2(n+1)}-\sqrt{k}-\sqrt{n+1-k}\right)$
That means 
$\left(\sqrt{A(n+1-k)+B(n+1-k)}-\sqrt{A(n+1-k)-B(n+1-k)}\right)^2\\=2\left(\sqrt{2(n+1)}-\sqrt{n+1-k}-\sqrt{k}\right)$
So that 
$\sqrt{A(k)+B(k)}-\sqrt{A(k)-B(k)}\\=\sqrt{A(n+1-k)-B(n+1-k)}-\sqrt{A(n+1-k)+B(n+1-k)}$
So that 
$\sum_{k=1}^{n} \sqrt{A(k)+B(k)}=\sum_{k=1}^{n} \sqrt{A(k)-B(k)}$
So that $f(n)=\boxed{\sqrt{2}+1}$
A: The hint
$$
\frac{ \sqrt{1+\sin(x)}+\sqrt{1+\cos(x)} }{ \sqrt{1-\sin(x)}+\sqrt{1-\cos(x)} }=1+\sqrt2 \quad \text{for } x\in [0,\pi/2]
$$
is spot on.
We have
$$
\begin{align}
\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{99}}
&=
\sqrt{10}\left(\sqrt{1+\sqrt{0.01}}+\sqrt{1+\sqrt{0.99}}\right)
\\&=
\sqrt{10}\left(\sqrt{1+\sin(t)}+\sqrt{1+\cos(t)}\right)
\\&=
\sqrt{10}(1+\sqrt 2)\left(\sqrt{1-\sin(t)}+\sqrt{1-\cos(t)}\right)
\\&=
\sqrt{10}(1+\sqrt 2)\left(\sqrt{1-\sqrt{0.01}}+\sqrt{1-\sqrt{0.99}}\right)
\\&=
(1+\sqrt 2)\left(\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{99}}\right)
\end{align}
$$
for some $t$.
Therefore, we can rearrange the numerator to be $1+\sqrt 2$ times the denominator. (You need to handle the middle terms separately but it works out the same.)
A: Let the numerator and the denominator 
$$N= \sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\ldots+\sqrt{10+\sqrt{99}}$$
$$D =\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\ldots+\sqrt{10-\sqrt{99}}$$
Apply the denesting formulas 
$$\sqrt{a\pm\sqrt c} = \frac{1}{\sqrt2} \left( \sqrt{a+\sqrt{a^2-c}} \pm 
\sqrt{a-\sqrt{a^2-c}} \right)$$
to get
$$\sqrt{10\pm\sqrt n} = \frac{1}{\sqrt2} \left( \sqrt{10+\sqrt{100-n}} \pm 
\sqrt{10-\sqrt{100-n}} \right)$$
where $n=1,2,...99$. As a result,
$$N= \frac{1}{\sqrt2}(N+D), \>\>\>\>\>D= \frac{1}{\sqrt2}(N-D)$$
Take the ratio,
$$\frac ND=\frac{N+D}{N-D}$$
or,
$$\left(\frac ND\right)^2 - 2\frac ND -1 =0$$
Solve to obtain,
$$\frac ND = 1+\sqrt2$$
A: Rewrite the sum as
$$\frac{\sum_{n=1}^{99} \sqrt{10+\sqrt{n}}}{\sum_{n=1}^{99} \sqrt{10-\sqrt{n}}} = \frac{a}{b}$$
then let
$$\Delta_n = \sqrt{10+\sqrt{n}} - \sqrt{10-\sqrt{n}}$$
By squaring $\Delta_n$ and simplifying we get
$$\Delta_n = \sqrt{2} \sqrt{10 - \sqrt{100-n}}$$
Now summing all those $\Delta_n$s can be done by letting the index $n$ run from $1$ to $99$ and replacing $\sqrt{100-n}$ with $\sqrt{n}$ since both cases will yield the same summands. Hence
$$\sum_{n=1}^{99} \Delta_n = \sqrt{2} \sum_{n=1}^{99}\sqrt{10 - \sqrt{n}} = \sqrt{2} \cdot b$$
thus
$$\frac{a-b}{b} = \frac{a}{b} - 1 = \sqrt{2} \implies \frac{a}{b} = \sqrt{2} + 1$$
