<s>[INST] Find value of $\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}$ Find value of 

$$S=\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}$$

I started with $$S+\frac{1}{S}=\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}+\frac{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}$$
$\implies$
$$S+\frac{1}{S}=\frac{\sum_{i=1}^{399}(20+\sqrt{i})+2S_1+\sum_{i=1}^{399}(20-\sqrt{i})+2S_2}{\left(\sum_{k=1}^{399}\sqrt{20+\sqrt{k}}\right) \times \left(\sum_{k=1}^{399}\sqrt{20-\sqrt{k}}\right)}$$ $\implies$
$$S+\frac{1}{S}=\frac{15960+2S_1+2S_2}{\left(\sum_{k=1}^{399}\sqrt{20+\sqrt{k}}\right) \times \left(\sum_{k=1}^{399}\sqrt{20-\sqrt{k}}\right)}$$
where $$S_1=\sum_{i \ne j=1}^{399}\left(\sqrt{20+\sqrt{i}}\right)\left(\sqrt{20+\sqrt{j}}\right)$$ and like wise
$$S_2=\sum_{i \ne j=1}^{399}\left(\sqrt{20-\sqrt{i}}\right)\left(\sqrt{20-\sqrt{j}}\right)$$
Any way to proceed from here?
 [/INST] Amazingly, it appears that 
$$ \dfrac{\sum_{k=1}^{m^2-1} \sqrt{m + \sqrt{k}}}{\sum_{k=1}^{m^2-1} \sqrt{m - \sqrt{k}}} = 1 + \sqrt{2} $$
for all integers $m \ge 2$.
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