2005 IMO Shortlist Problems/A3

Problem

(Czech Republic) Four real numbers $\displaystyle p,q,r,s$ satisfy

$\displaystyle p+q+r+s = 9$ and $\displaystyle p^2 + q^2 + r^2 + s^2 = 21$ .

Prove that $ab-cd \ge 2$ holds for some permutation $\displaystyle (a,b,c,d)$ of $\displaystyle (p,q,r,s)$ .

Solutions

Solution 1

Lemma. $\max (p,q,r,s) \le 3$ .

Proof. Suppose on the contrary that $p > 3$ . Then $(p,q,r,s)$ majorizes $(3,2,2,2)$ , and since $f : x \mapsto x^2$ is a convex function, by Karamata's Inequality, $21 = p^2+q^2+r^2+s^2 \ge 3^2 + 2^2 + 2^2 + 2^2 = 21$ . But since $f$ is strictly convex, equality occurs only when $(p,q,r,s)$ is a permutation of $(3,2,2,2)$ , a contradiction, since we assumed $p>3$ . ∎

Without loss of generality, let $p \ge q,r,s$ . Now, since $f : x \mapsto x(9-x)$ is increasing on the interval $( -\infty, 9/2 ]$ ,

$pq + pr + ps = p(9-p) \ge 3(9-3) = 18$ .

Also, we note

$pq+pr+ps+rs+qs+rq = \frac{1}{2}[ (p+q+r+s)^2 - (p^2+q^2+r^2+s^2) ] = \frac{1}{2} ( 9^2 - 21) = 30$ .

Hence

$(pq-rs) + (pr-qs) + (ps-rq) = 2(pq+pr+ps) - (pq+pr+ps+rs+qs+rq) \ge 2\cdot 18 - 30 = 6$ .

It follows that at least one of $(pq-rs), (pr-qs), (ps-rq)$ must be at least $2$ .

Solution 2

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

Lemma. $p + q \ge 5$ .

Proof 1. By the Rearrangement Inequality,

$pq + sr \ge pr + sq = pr + qs \ge ps + qr$ .

As in the first solution, we see $\displaystyle (pq+rs) + (pr+sq) + (ps+qr) = 30$ , so $pq + rs \ge 10$ . It follows that

$(p+q)^2 + [9-(p+q)]^2 = (p+q)^2 + (r+s)^2 = p^2+q^2+r^2+s^2 + 2(pq+rs) \ge 41$ .

This is equivalent to

$(p+q-4)(p+q-5) \ge 0$ ,

so either $p+q \ge 5$ or $p+q \le 4$ . But since $2(p+q) \ge p+q+r+s = 9$ , $\displaystyle p+q \ge 9/2$ , so $p+q \ge 5$ . ∎

Proof 2. From the identity $\displaystyle (x+y)^2 + (x-y)^2 = 2(x^2+y^2)$ we have

$\displaystyle (p+q+r+s)^2+ (p+q-r-s)^2 + (p+r-q-s)^2 + (p+s-q-r)^2$ $\displaystyle = 2[(p+q)^2 + (r+s)^2 + (p-q)^2 + (r-s)^2]$
$\displaystyle = 2[ 2(p^2+q^2) + 2(r^2+s^2) ] = 84$ .