Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem B. 4931. (February 2018)

B. 4931. Prove that if $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ are the sides of a triangle then

$\displaystyle \frac{a^2(b+c)+b^2(a+c)}{abc}>3.$

(3 pont)

Deadline expired on March 12, 2018.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. A háromszög-egyenlőtlenség szerint $\displaystyle a+b>c$, ezt használva:

$\displaystyle \frac{a^2(b+c)+b^2(a+c)}{abc}=\frac{ab(a+b)+(a^2+b^2)c}{abc}>$

$\displaystyle >\frac{abc+(a^2+b^2)c}{abc}=\frac{abc+(a-b)^2c+2abc}{abc}\geq \frac{3abc}{abc}=3,$

hiszen $\displaystyle (a-b)^2\geq 0$. Ezzel az állítást bizonyítottuk.

### Statistics:

 141 students sent a solution. 3 points: 138 students. 1 point: 2 students. 0 point: 1 student.

Problems in Mathematics of KöMaL, February 2018