Problem B. 4867. (April 2017)
B. 4867. The sum of the real numbers \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle c\) and \(\displaystyle d\) is \(\displaystyle 0\). Let \(\displaystyle M=ab+bc+cd\) and \(\displaystyle N=ac+ad+bd\). Prove that at least one of the sums \(\displaystyle 20M+17N\) and \(\displaystyle 20N+17M\) is non-positive.
(Bulgarian problem)
(4 pont)
Deadline expired on May 10, 2017.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. A \(\displaystyle 20M+17N\) és \(\displaystyle 20N+17M\) számok összege
\(\displaystyle 37(M+N)=37(ab+ac+ad+bc+bd+cd)=\)
\(\displaystyle =37\cdot \frac{(a+b+c+d)^2-(a^2+b^2+c^2+d^2)}{2}=-\frac{37}{2}(a^2+b^2+c^2+d^2)\leq 0,\)
így legalább az egyikük nem pozitív.
Statistics:
84 students sent a solution. 4 points: 78 students. 3 points: 2 students. 2 points: 1 student. 1 point: 1 student. Unfair, not evaluated: 2 solutionss.
Problems in Mathematics of KöMaL, April 2017