Team Selection Test

Let $s=x^2+y^2$ and $t=x+y$. We want to show that $$s^3\ge 8t(3s-t^2)(t^2-2t-s)$$ for $2s\ge t^2$ and $t\ge 0$. Now let $s=rt$. This transforms the inequality to $$r^3\ge 8(3r-t)(t-2-r)$$ for $2r\ge t\ge 0$. Since $r^3-8(3r-t)(t-2-r)=8(t-(2r+1){)}^2+r^3-8r^2+16r-8\ge r^3-8r^2+16r=r(r-4{)}^2\ge 0$ for $2r\ge t\ge 0$, we are done.