Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem B. 4376. (September 2011)

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

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 $\displaystyle 2\sqrt{6}>9/2$.

### Statistics:

 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