Mathematical and Physical Journal
for High Schools
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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.


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

Problems in Mathematics of KöMaL, February 2018