Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# 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 solutions.

Problems in Mathematics of KöMaL, April 2017