Difference between revisions of "2005 IMO Shortlist Problems/A3"
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'''Lemma.''' <math> \max (p,q,r,s) \le 3 </math>. | '''Lemma.''' <math> \max (p,q,r,s) \le 3 </math>. | ||
− | ''Proof.'' Suppose on the contrary that <math> | + | ''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> | ||
− | It follows that at least one of <math> | + | 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 17:09, 19 December 2018
Problem
(Czech Republic)
Four real numbers satisfy
and
.
Prove that holds for some permutation
of
.
Solutions
Solution 1
Lemma. .
Proof. Suppose on the contrary that . Then
majorizes
, and since
is a convex function, by Karamata's Inequality,
. But since
is strictly convex, equality occurs only when
is a permutation of
, a contradiction, since we assumed
. ∎
Without loss of generality, let . Now, since
is increasing on the interval
,
.
Also, we note
.
Hence
.
It follows that at least one of must be at least
.
Solution 2
Without loss of generality, we assume .
Lemma. .
Proof 1. By the Rearrangement Inequality,
![$pq + sr \ge pr + sq = pr + qs \ge ps + qr$](http://latex.artofproblemsolving.com/3/8/2/38248a6142b99372688b17e111923a3c07c84be8.png)
As in the first solution, we see , so
. It follows that
.
This is equivalent to
![$(p+q-4)(p+q-5) \ge 0$](http://latex.artofproblemsolving.com/c/4/c/c4c2b0c9c0278c5f6d97544c8b52a4928550ca58.png)
so either or
. But since
,
, so
. ∎
Proof 2. From the identity we have
.
Since , this implies
![$(p+q-r-s)^2 + (p+r-q-s)^2 + (p+s-q-r)^2 \ge 3$](http://latex.artofproblemsolving.com/5/c/8/5c8f793450ad69adc4c686854eb608b0a425aa5f.png)
and since , this gives us
, or
. Thus
![$2(p+q) -9 = p+q+ (p+q-9) = p+q-r-s \ge 1$](http://latex.artofproblemsolving.com/f/0/d/f0db7d0de2194493f32a54ddc69c20ff6092a0ef.png)
so . ∎
Now,
,
so , as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.