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