How to solve system of equations involving square roots How to solve the following system of equations? I've tried some basic techniques like adding/substracting and squaring but with no effect.
$$
\left\{ 
\begin{array}{c}
\sqrt{1 + x_1} + \sqrt{1 + x_2} + \sqrt{1 + x_3} + \sqrt{1 + x_4} = 2\sqrt{5} \\ 
\sqrt{1 - x_1} + \sqrt{1 - x_2} + \sqrt{1 - x_3} + \sqrt{1 - x_4} = 2\sqrt{3}
\end{array}
\right. 
$$

Denote 
$$\sqrt{1+x_i}=a_i,~~~\sqrt{1-x_i}=b_i, ~~~(i=1,2,3,4)$$
It's clear that $$a_i,b_i \geq 0,$$and$$a_i^2+b_i^2=2.\tag1$$
Moreover, the system of equations could be rewritten as $$\sum_{i=1}^4a_i=2\sqrt{5},~~~~\sum_{i=1}^4b_i=2\sqrt{3}.\tag2$$
From $(1)$, we have $$\sum_{i=1}^4a_i^2+\sum_{i=1}^4b_i^2=8.$$
From $(2)$, we have $$\left(\frac{1}{4}\sum_{i=1}^4a_i\right)^2+\left(\frac{1}{4}\sum_{i=1}^4b_i\right)^2=2.\tag3$$
But by the inequality $A_n \leq Q_n$, we have 
$$\left(\frac{1}{4}\sum_{i=1}^4a_i\right)^2+\left(\frac{1}{4}\sum_{i=1}^4b_i\right)^2 \leq \frac{1}{4}\sum_{i=1}^4a_i^2+\frac{1}{4}\sum_{i=1}^4b_i^2=\frac{1}{4}\left(\sum_{i=1}^4a_i^2+\sum_{i=1}^4b_i^2\right)=2.\tag4   $$
Form $(3)$, we see that the equality holds in $(4)$. Recall the condition for equality in $A_n \leq Q_n$, which is $$a_1=a_2=a_3=a_4,~~~b_1=b_2=b_3=b_4.$$
Thus, we may obtain $$a_i=\frac{\sqrt{5}}{2},~~~b_i=\frac{\sqrt{3}}{2}.$$
As a result, $$x_i \equiv\frac{1}{4},$$ where $i=1,2,3,4.$
