Proving $\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a.....}}}}-\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{a....}}}}=1$ I checked many values of this inequality, but I don't have the complete proving.
$$\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a.....}}}}-\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{a....}}}}=1$$ if $$a>1$$
 A: this is 
$$
\lim U_n - \lim u_n
$$
where
$$
U_{n+1 } = \sqrt{a + U_n}
\\
u_{n+1 } = \sqrt{a - u_n}
$$
Provided that both limits exists, there are solutions of
$$
U^2 = a + U\\
u^2 = a - u
$$
the part $b^2 - 4ac $ is the same, so it remains only the $-\frac b {2a}$ part of both quadratic equations, giving
$$
U - u = \frac 12 - \left(-\frac 12\right) = 1
$$

Another 'proof':
\begin{align} \delta &=
 \sqrt{a + \sqrt{a + \sqrt{a +\cdots}}}- \sqrt{a - \sqrt{a - \sqrt{a -\cdots}}}
\\&= 
\frac{a + \sqrt{a + \sqrt{a +\cdots}} - \left[a - \sqrt{a - \sqrt{a -\cdots}}\right]}
{\sqrt{a + \sqrt{a + \sqrt{a +\cdots}}}+ \sqrt{a - \sqrt{a - \sqrt{a -\cdots}}}}
\\&= 
\frac{ \sqrt{a + \sqrt{a +\cdots}} + \sqrt{a - \sqrt{a -\cdots}}}
{\sqrt{a + \sqrt{a + \sqrt{a +\cdots}}}+ \sqrt{a - \sqrt{a - \sqrt{a -\cdots}}}}
\\&=1
\end{align}
A: Define $x = \sqrt{a + \sqrt{a + \sqrt{a +\cdots}}}$ and $y = \sqrt{a - \sqrt{a - \sqrt{a -\cdots}}}.$ Then we have $x^2 = a+ x$ and $y^2 = a-y$.
Now compute $(x+y)(x-y) = x^2 - y^2 = x+y$. Cancelling $x+y$ from both sides gives $x-y = 1$, which is the result you're after.
(Of course, I've ignored convergence issues here and just worked formally).
