<s>[INST] Proof $\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt2} + \frac{\sqrt{2 + \sqrt2}}{\sqrt2 \sqrt{2 - \sqrt{2 - \sqrt2}}} =\sqrt{2-\sqrt{2+\sqrt2}}$ I want to prove the following equation
$$\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt{2}} + \frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2} \sqrt{2 - \sqrt{2 - \sqrt{2}}}}$$
I know this must be equal to $\sqrt{2-\sqrt{2+\sqrt{2}}}$, but I find it pretty hard to prove it manually, without requiring a calculator nor estimates (since the expression should be equal to one of the roots of the polinomyal $f(x)=x^8-8x^6+20x^4-16x^2+2$, and all of them are real and distinct).
Thanks in advance for your help.
 [/INST] Note that the second term of the LHS can be expressed as
$$\begin{align}
\frac{\sqrt{2 + \sqrt{2} }}{\sqrt{2} \sqrt{2 - \sqrt{2 - \sqrt{2}}}}
=\frac{\sqrt{2 + \sqrt{2 - \sqrt{2}}}}
{\sqrt{2}}  
\end{align}$$
which can be verified by cross multiply. Then,
$$LHS = \sqrt{ \frac{2 + \sqrt{2 - \sqrt{2}}} {2} }
- \sqrt{ \frac{2 - \sqrt{2 - \sqrt{2}}} {2} }=\sqrt{2-\sqrt{2+\sqrt{2}}}=RHS $$
where the denest formula below is used
$$\sqrt{a-\sqrt c} =\sqrt{\frac{a+\sqrt {a^2-c}}{2}}  -\sqrt{\frac{a-\sqrt {a^2-c}}{2}} $$
with $a=2$ and $c=2+\sqrt2$.
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