20 unknowns in one equation?

Note that $\sqrt{x-1}\leq \frac {x}{2}$ , with equality for $x=2$ . Now $\sum_{k=1}^{20} k\sqrt{x_k-k^2}=\sum_{k=1}^{20} k^2\sqrt{\frac{x_k}{k^2}-1}\leq \sum_{k=1}^{20} k^2\frac{x_k}{2k^2} =\sum_{k=1}^{20}\frac{x_k}{2}$ with equality for $x_k=2k^2$ and $\sum_{k=1}^{20}x_k=2\sum_{k=1}^{20}k^2=2*2870=\boxed{5740}$

Given,

$\large\sum _{ m=1 }^{ 20 } m\sqrt { x_{ m }-m^{ 2 } } =\frac { 1 }{ 2 } \large\sum _{ m=1 }^{ 20 } x_{ m }\\ \Rightarrow \sum _{ m=1 }^{ 20 } x_{ m }-\sum _{ m=1 }^{ 20 } 2m\sqrt { x_{ m }-m^{ 2 } } =0\\ \Rightarrow \sum _{ m=1 }^{ 20 } (x_{ m }-2m\sqrt { x_{ m }-m^{ 2 } } )=0\\ \Rightarrow \sum _{ m=1 }^{ 20 } \{ (x_{ m }-m^{ 2 })-2m\sqrt { x_{ m }-m^{ 2 } } +{ m }^{ 2 }\} =0\\ \Rightarrow \sum _{ m=1 }^{ 20 } (\sqrt { x_{ m }-m^{ 2 } } -m)^{ 2 }=0$

Therefore every term is equal zero. That means,

$\sqrt { x_{ m }-m^{ 2 } } -m=0$ , for $\\ m=1,2,3,.......,20$

$\Rightarrow x_{ m }=2m^{ 2 }$

$\Rightarrow \large\sum _{ m=1 }^{ 20 }x_{ m }=\large\sum _{ m=1 }^{ 20 }2m^{ 2 }=2\times \frac { 20(20+1)(2\times 20+1) }{ 6 }=\boxed{5740}$

Define $y_i = x_i - i^2$ .

Then the equation can be re-expressed as $\sum_{i=1}^{20}i\sqrt{y_i} = \sum_{i=1}^{20}\frac{y_i+i^2}{2}$

It is obvious, then, that each term of the sum corresponds to a term in the other sum in the AM-GM inequality: $\frac{y_i+i^2}{2} \geq \sqrt{y_i \cdot i^2} = i\sqrt{y_i}$

Therefore equality only occurs when $y_i = i^2$ , and hence the sought for sum is equivalent to: $\sum_{i=1}^{20}y_i+i^2 = 2 \sum_{i=1}^{20}i^2$ $= 2 \dfrac{(20)(20+1)(2\cdot 20 +1)}{6}$ $= 5740$