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B. 3802. Given are seven real numbers in such a way that the sum of any three of them is less than the sum of the remaining four. Show that each number is positive.

(3 points)

Deadline expired on 15 April 2005.

Google Translation (Sorry, the solution is published in Hungarian only.)

Megoldás. Legyen a 7 szám . Ekkor

a2+a3+a4a5+a6+a7.

Ha tehát a10 lenne, akkor

a1+a2+a3+a4a5+a6+a7

is teljesülne, ami viszont ellentmond a feltételeknek. Ezért hát a1>0, következésképpen aia1>0 is igaz minden 1i7 esetén.

Statistics on problem B. 3802.
 171 students sent a solution. 3 points: 100 students. 2 points: 60 students. 1 point: 7 students. 0 point: 4 students.

• Problems in Mathematics of KöMaL, March 2005

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