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Difference between revisions of "2005 IMO Shortlist Problems/A3"

 
(→‎Solution 1)
 
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'''Lemma.''' <math> \max (p,q,r,s) \le 3 </math>.
 
'''Lemma.''' <math> \max (p,q,r,s) \le 3 </math>.
  
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''Proof.''  Suppose on the contrary that <math> \displaystyle p > 3 </math>.  Then <math> \displaystyle (p,q,r,s) </math> [[majorize]]s <math> \displaystyle (3,2,2,2) </math>, and since <math> f : x \mapsto x^2 </math> is a [[convex function]], by [[Karamata's Inequality]], <math> 21 = p^2+q^2+r^2+s^2 \ge 3^2 + 2^2 + 2^2 + 2^2 = 21 </math>.  But since <math> \displaystyle f </math> is strictly convex, equality occurs only when <math> \displaystyle (p,q,r,s) </math> is a permutation of <math> \displaystyle (3,2,2,2) </math>, a contradiction, since we assumed <math> \displaystyle p>3 </math>.  {{Halmos}}
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''Proof.''  Suppose on the contrary that <math>p > 3 </math>.  Then <math>(p,q,r,s) </math> [[majorize]]s <math>(3,2,2,2) </math>, and since <math> f : x \mapsto x^2 </math> is a [[convex function]], by [[Karamata's Inequality]], <math> 21 = p^2+q^2+r^2+s^2 \ge 3^2 + 2^2 + 2^2 + 2^2 = 21 </math>.  But since <math>f </math> is strictly convex, equality occurs only when <math>(p,q,r,s) </math> is a permutation of <math>(3,2,2,2) </math>, a contradiction, since we assumed <math>p>3 </math>.  {{Halmos}}
  
 
Without loss of generality, let <math> p \ge q,r,s </math>.  Now, since <math> f : x \mapsto x(9-x) </math> is increasing on the interval <math> ( -\infty, 9/2 ] </math>,
 
Without loss of generality, let <math> p \ge q,r,s </math>.  Now, since <math> f : x \mapsto x(9-x) </math> is increasing on the interval <math> ( -\infty, 9/2 ] </math>,
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<math> (pq-rs) + (pr-qs) + (ps-rq) = 2(pq+pr+ps) - (pq+pr+ps+rs+qs+rq) \ge 2\cdot 18 - 30 = 6 </math>.
 
<math> (pq-rs) + (pr-qs) + (ps-rq) = 2(pq+pr+ps) - (pq+pr+ps+rs+qs+rq) \ge 2\cdot 18 - 30 = 6 </math>.
 
</center>
 
</center>
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It follows that at least one of <math> \displaystyle (pq-rs), (pr-qs), (ps-rq) </math> must be at least 2.
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It follows that at least one of <math>(pq-rs), (pr-qs), (ps-rq) </math> must be at least <math>2</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Latest revision as of 16:09, 19 December 2018

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

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

Since $\displaystyle (p+q+r+s)^2 = 81$, this implies

$(p+q-r-s)^2 + (p+r-q-s)^2 + (p+s-q-r)^2 \ge 3$,

and since $(p+q-r-s)^2 \ge (p+r-q-s)^2, (p+s-q-r)^2$, this gives us $(p+q-r-s)^2 \ge 1$, or $p+q -r-s \ge 1$. Thus

$2(p+q) -9 = p+q+ (p+q-9) = p+q-r-s \ge 1$,

so $p+q \ge 5$. ∎

Now,

$21 + 2(pq-rs) = (p+q)^2 + (r-s)^2 \ge 25$,

so $pq-rs \ge 2$, as desired.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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