### 2005 IMO SL #A3

Four real numbers $p$, $q$, $r$, $s$ satisfy $p+q+r+s = 9$ and $p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $\left(a,b,c,d\right)$ of $\left(p,q,r,s\right)$ such that $ab-cd \geq 2$.

Without loss of generality, assume that $p \ge q \ge r \ge s$.

We have $$pq+rs+pr+qs+ps+qr=\frac{(p+q+r+s)^2-(p^2+q^2+r^2+s^2)}{2}=30.$$

Now by the Rearrangement Inequality, we know $pq+rs\ge pr+qs\ge ps+qr$, so $pq+rs\ge10$.

Then we have

$$(p+q)+(r+s)=9$$

$$(p+q)^2+(r+s)^2=(p^2+q^2+r^2+s^2)+2(pq+rs)\ge 21+2\cdot 10=41.$$

Now because $p+q\ge r+s$, we must have $p+q\ge5$.

Lastly, $$25\le(p+q)^2+(r-s)^2=21+2(pq-rs),$$ so we must have $pq-rs \ge 2$, as desired. $\square$.

1. Anonymous5/11/2022

cool solution!!