Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?
I want the old design back!!! :-)

Problem B. 4376. (September 2011)

B. 4376. Prove that if x, y are non-negative numbers then

x^4 + y^3 + x^2 + y + 1 > \frac{9}{2}xy .

Suggested by J. Szoldatics, Dunakeszi

(4 pont)

Deadline expired on October 10, 2011.

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

Megoldás. Tetszőleges \(\displaystyle a,b\) valós számokra \(\displaystyle (a-b)^2\ge 0\) miatt teljesül \(\displaystyle a^2+b^2\ge 2ab\). Ezért \(\displaystyle x^4+1\ge 2x^2\), továbbáha \(\displaystyle y\) nemnegatív, akkor \(\displaystyle y^3+y=y(y^2+1)\ge y(2y)=2y^2\). Mindezek alapján

\(\displaystyle x^4 + y^3 + x^2 + y + 1\ge 3x^2+2y^2\ge 2(\sqrt{3}x)(\sqrt{2}y)> \frac{9}{2}xy,\)

hiszen \(\displaystyle xy\) nemnegatív és \(\displaystyle 2\sqrt{6}>9/2\).


93 students sent a solution.
4 points:70 students.
3 points:7 students.
2 points:1 student.
1 point:5 students.
0 point:8 students.
Unfair, not evaluated:2 solutions.

Problems in Mathematics of KöMaL, September 2011