# 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 16:09, 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,

As in the first solution, we see , so . It follows that

.

This is equivalent to

so either or . But since , , so . ∎

*Proof 2.* From the identity we have

.

Since , this implies

and since , this gives us , or . Thus

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.*