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.
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141 students sent a solution. 3 points: 138 students. 1 point: 2 students. 0 point: 1 student.
Problems in Mathematics of KöMaL, February 2018