Selection Examination

Without loss of generality, let $a\ge b\ge c\ge d$ and we will prove that $a+b\ge 2$. We have that $ab\ge c^2$ and $ab\ge d^2$, so by adding them we have: $2ab\ge c^2+d^2$. Therefore, $$(a+b{)}^2=a^2+b^2+2ab\ge a^2+b^2+c^2+d^2=4,$$ so $a+b\ge 2$.