Answer
$$800^{2}+200^{2} \gt 600^{2}+400^{2} $$
Work Step by Step
Given $$f(x)=(1,000-x)^{2}+x^{2}$$
Since
$$f'(x)= -2(1,000-x)+2x =4x-2000$$
Then $f'(x)=0$ for $x= 500$ and
\begin{align*}
f'(100)&\lt0\\
f'(600)&\lt 0
\end{align*}
Hence, $f(x)$ is decreasing on $(-\infty ,500 )$ and increasing on $(500, \infty)$.
For $800^{2}+200^{2}$, by comparing with $f(x)$, $x_1= 200$ and for $ 600^{2}+400^{2}$, $x_2=400$.
Since $f(x)$ is decreasing for $x <500$, then for
$$x_1\lt x_2 \ \ \to\ \ \ f(x_1)\gt f(x_2) $$
Hence $$800^{2}+200^{2} \gt 600^{2}+400^{2} $$
