## Calculus (3rd Edition)

$$800^{2}+200^{2} \gt 600^{2}+400^{2}$$
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 $$800^{2}+200^{2} \gt 600^{2}+400^{2}$$