Mathematical and Physical Journal
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Problem C. 1011. (December 2009)

C. 1011. Prove that the value of the expression a3-3ab2+2b3 is non-negative if a and b are non-negative real numbers.

(5 pont)

Deadline expired on January 11, 2010.

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

Megoldás. \(\displaystyle a^3 - 3ab^2 + 2b^3 = a(a^2 - b^2) - 2b^2(a - b) = (a - b)\left(a(a+b)-2b^2\right)\). Ha \(\displaystyle a\ge b\), akkor egyrészt \(\displaystyle a-b\ge 0\), másrészt mivel

\(\displaystyle b\ge 0 \quad : \quad a^2\ge b^2; \ \ ab\ge b^2\)

alapján \(\displaystyle a(a+b)\ge 2b^2\), azaz a második tényező is nemnegatív: szorzatuk nemnegatív. Ha \(\displaystyle a<b\), akkor \(\displaystyle a-b<0\) és \(\displaystyle a\ge 0\) miatt \(\displaystyle a^2<b^2\) ill. \(\displaystyle ab<b^2\), azaz \(\displaystyle a(a+b)< 2b^2\). Mivel a szorzat mindkét tényezője negatív, szorzatuk pozitív. Tehát az eredeti összeg minden nemnegatív \(\displaystyle a\), \(\displaystyle b\)-re nemnegatív.


271 students sent a solution.
5 points:131 students.
4 points:88 students.
3 points:16 students.
2 points:11 students.
1 point:11 students.
0 point:12 students.
Unfair, not evaluated:2 solutions.

Problems in Mathematics of KöMaL, December 2009