Problem C. 1401. (February 2017)
C. 1401. The positive numbers \(\displaystyle x\), \(\displaystyle y\) satisfy the equation \(\displaystyle x^3+y^3=x-y\). Prove that \(\displaystyle x^2+y^2<1\).
(5 pont)
Deadline expired on March 10, 2017.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. Mivel \(\displaystyle x\) és \(\displaystyle y\) pozitív számok, így
\(\displaystyle x^3+y^3=x-y=\frac{(x-y)(x^2+xy+y^2 )}{x^2+xy+y^2}=\frac{x^3-y^3}{x^2+xy+y^2}.\)
Átrendezve:
\(\displaystyle x^2+xy+y^2=\frac{x^3-y^3}{x^3+y^3}.\)
Ebből következik, hogy
\(\displaystyle x^2+y^2<x^2+xy+y^2=\frac{x^3-y^3}{x^3+y^3}<1,\)
mert a tört számlálója kisebb, mint a nevezője.
Tehát \(\displaystyle x^2+y^2<1\).
Statistics:
119 students sent a solution. 5 points: 96 students. 4 points: 6 students. 3 points: 4 students. 2 points: 5 students. 0 point: 6 students. Unfair, not evaluated: 2 solutionss.
Problems in Mathematics of KöMaL, February 2017