Sum of All the Roots

By AM-GM inequality , we have:

$\begin{cases} \sqrt{1+x_1} + \sqrt{1+x_2} + \sqrt{1+x_3} + \cdots + \sqrt{1+x_{2020}} & \ge \displaystyle 2020 \sqrt[2020]{\prod_{n=1}^{2020}\sqrt{1+x_n}} = \sqrt{2020 \times 2021} \\ \sqrt{1-x_1} + \sqrt{1-x_2} + \sqrt{1-x_3} + \cdots + \sqrt{1-x_{2020}} & \ge \displaystyle 2020 \sqrt[2020]{\prod_{n=1}^{2020}\sqrt{1-x_n}} = \sqrt{2020 \times 2019} \end{cases}$

Equality in both cases occur when $x_1 = x_2 = x_3 = \cdots = x_{2020} = \dfrac 1{2020}$ . Therefore they are the roots of the system of equations, and the sum of all roots is $\dfrac {2020}{2020} = \boxed 1$ .

This solution is inspired in Hana's one. First of all, note that, for any $a>0$ ,

(1) $\sqrt{1+x} = a^{-1/2}\sqrt{(1+x)a} \leq a^{-1/2}(\frac{1+x+a}{2})$ .

Therefore,

(2) $\sqrt{2020\times 2021} = \sum_{i=1}^{2020}\sqrt{1+x_i} \leq a^{-1/2}\sum_{i=1}^{2020}(\frac{1+x_i+a}{2}) = \frac{1}{2}a^{-1/2}[2020(1+a) + \sum_{i=1}^{2020}x_i]$ ,

so that

(3) $\sum_{i=1}^{2020}x_i \geq 2a^{1/2}\sqrt{2020\times 2021} - 2020(1+a)$ .

Likewise, for any $b>0$ ,

(4) $\sqrt{1-x} = b^{-1/2}\sqrt{(1-x)b} \leq b^{-1/2}(\frac{1-x+b}{2})$ ,

so that

(5) $\sqrt{2020\times 2019} = \sum_{i=1}^{2020}\sqrt{1-x_i} \leq b^{-1/2}\sum_{i=1}^{2020}(\frac{1-x_i+b}{2}) = \frac{1}{2}b^{-1/2}[2020(1+b) - \sum_{i=1}^{2020}x_i]$ ,

from which follows

(6) $\sum_{i=1}^{2020}x_i \leq 2020(1+b) - 2b^{1/2}\sqrt{2020\times 2019}$ .

Combining (3) and (6) we obtain

(7) $2a^{1/2}\sqrt{2020\times 2021} - 2020(1+a) \leq \sum_{i=1}^{2020}x_i \leq 2020(1+b) - 2b^{1/2}\sqrt{2020\times 2019}$ .

Now, the magic: if we choose $a = \frac{2021}{2020}$ and $b = \frac{2019}{2020}$ , the double inequality (7) becomes

(8) $1 \leq \sum_{i=1}^{2020}x_i \leq 1$ ,

which implies $\sum_{i=1}^{2020}x_i = 1$ . $\square$

\[\begin{cases}\sqrt{1+x_1} + \sqrt{1 +x_2} + \sqrt{1 +x_3} +\dots+\sqrt{ 1 + x_{2020}} = \sqrt{ 2020(2020+1)} \\

\sqrt{1 - x_1} + \sqrt{1 - x_2} + \sqrt{1 -x_3} +\dots+\sqrt{ 1 - x_{2020}} = \sqrt{2020(2020-1)}\end{cases}\]

Applying Cauchy's Inequality , we have:

\[\begin{cases}\sqrt {(1+x_1)\frac{2020+1}{2020}} \le \frac{1}{2} ( 1+x_1+\frac{2020+1}{2020})\\

\hspace{3cm}\dots \\

\hspace{3cm}\dots\\

\sqrt {(1+x_{2020})\frac{2020+1}{2020}} \le \frac{1}{2} ( 1+x_{2020}+\frac{2020+1}{2020})\end{cases}\]

It follows:

$\sqrt{\frac{2020+1}{2020}} \sqrt{2020(2020+1)} = 2020 +1 \le \frac{1}{2} \Big( 2( 2020 ) + 1+x_1+x_2+\dots+x_{2020}\Big) \dots (1)$

Similarly we have:

$\sqrt{\frac{2020-1}{2020}} \sqrt{2020(2020-1)} = 2020 - 1 \le \frac{1}{2} \Big( 2( 2020 ) - 1 - x_1 - x_2 - \dots - x_{2020}\Big)\dots (2)$

Form $(1) \text{ and } (2)$ , we get:

$2 (2020) \le \frac{1}{2} \Big( 4(2020) +1+x_1+x_2+x_3+\dots+x_{2020}-1-x_1-x_2-\dots-x_{2020}\Big) = 2(2020)$

This equality happens if $x_1=x_2=\dots=x_{2020}=\frac{1}{2020}\implies \sum_{i=1}^{2020} x_i = 2020\Big(\frac{1}{2020}\Big) = 1$