Točno
10. listopada 2013. 01:11 (10 godine, 9 mjeseci)
Upozorenje: Ovaj zadatak još niste riješili!
Kliknite ovdje kako biste prikazali rješenje.
Kliknite ovdje kako biste prikazali rješenje.
![\sum\limits_{cyc}{ (a-b)^2 } \geqslant 18](/media/m/6/4/d/64d276fb87e461a445c41d2fb4129384.png)
![\frac {\sum\limits_{cyc}{ (a-b)^2 }}{6} \geqslant 3](/media/m/1/3/0/1302087021cd941e454cd538129395f4.png)
![\sqrt{\frac {\sum\limits_{cyc}{ (a-b)^2 }}{6}} = K \geqslant \sqrt{3}](/media/m/e/6/c/e6c25e1c01993b647c1af832a1ad1324.png)
![K \geqslant A = \frac {\sum\limits_{cyc}{ |a-b| }}{6}](/media/m/3/9/c/39c581bde5df860ea0cf83bb34b76610.png)
Ako zamislimo brojeve na brojevnom pravcu, preko svakog segmenta cemo preci paran broj puta.
Dakle,
![A \geqslant \frac{2 \cdot (max - min)}{6} = \frac{max - min}{3}](/media/m/9/d/5/9d515aaef79b316bb122e5f893bc10f3.png)
Ako je
![max - min \geqslant 6](/media/m/7/c/b/7cb1bb8df351447e87d509853f3278cb.png)
![A > \sqrt{3}](/media/m/4/7/3/4734ffd1be6823d1162f04b3958d9076.png)
Preostaje slucaj
![max - min = 5](/media/m/9/d/8/9d848e94d906a25d6225f4534f27812c.png)
Ako brojevi nisu strogo rastuci pa strogo padajuci, onda cemo preko nekog segmenta preci barem 4 puta, sto opet znaci
![A > \sqrt{3}](/media/m/4/7/3/4734ffd1be6823d1162f04b3958d9076.png)
Mozemo pretpostaviti
![min = 0](/media/m/e/6/7/e674f30ba8c411282aa39ee4efcec4c3.png)
![max = 5](/media/m/3/d/5/3d50e7bda1ceda2f5812c383032d91b4.png)
Dovoljno je provjeriti slucajeve kad imamo
![0](/media/m/7/b/8/7b8b0b058cf5852d38ded7a42d6292f5.png)
![1](/media/m/a/9/1/a913f49384c0227c8ea296a725bfc987.png)
![2](/media/m/e/e/e/eeef773d19a3b3f7bdf4c64f501e0291.png)
![0](/media/m/7/b/8/7b8b0b058cf5852d38ded7a42d6292f5.png)
![5](/media/m/e/a/3/ea36c795dac330f34d395d8364d379b6.png)
Jednakost
![(0, 2, 4, 5, 3, 1)](/media/m/0/2/6/026a73a04a2d40e892bbe838b5605037.png)
Ocjene: (2)
Komentari:
ikicic, 14. listopada 2013. 21:49