Problem B. 3802. (March 2005)
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 pont)
Deadline expired on April 15, 2005.
Sorry, the solution is available only in Hungarian. Google translation
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:
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