JBTST V 2007, Problema 2

[Laurian Filip]

Sa se arate ca

\( \frac{x^{3}+y^{3}+z^{3}}{3}\ge xyz+\frac{3}{4}|(x-y)(y-z)(z-x)| \) ,

oricare ar fi numerele reale \( x,y,z \geq 0 \).

Viorel Vajaitu

[Marius Dragoi]

Avem
\( |(x-y)(y-z)(z-x)|=|x-y||y-z||z-x| \),
asadar putem presupune fara a restange generalitatea \( x \geq y \geq z \).
Atunci vom avea
\( \frac {1}{3} \) \( \sum_{cyc}{} {x^3} \geq xyz + \frac {3}{4} (\sum_{cyc}{} {x^2y-x^2z}) \Leftrightarrow \)
\( \Leftrightarrow 4 \sum_{cyc}{} {x^3} + 3 \sum_{cyc}{} {x^2z} \geq 12xyz + 3 \sum_{cyc}{} {x^2y} \).

Dar
\( \sum_{cyc}{} {x^3} \geq \sum_{cyc}{} {x^2y} \Rightarrow 4 \sum_{cyc}{} {x^3} + 3 \sum_{cyc}{} {x^2z} \geq 12xyz + 3 \sum_{cyc}{} {x^2y} \Leftrightarrow \)
\( \Leftrightarrow \sum_{cyc}{} {x^3} + 3 \sum_{cyc}{} {x^2z} \geq 12xyz \)
ceea ce rezulta din medii.