Inegalitate conditionata

[Marius Mainea]

Fie x,y,z numere pozitive cu x+y+z=1. Demonstrati ca:

\( \frac{x^2-yz}{x^2+x} \)+\( \frac{y^2-zx}{y^2+y} \)+\( \frac{z^2-xy}{z^2+z} \)\( \leq \)\( 0. \)

Marius Mainea (rev. Arhimede 7-8 /2004)

[Marius Dragoi]

\( \frac{x^2-yz}{x^2+x} \)+\( \frac{y^2-zx}{y^2+y} \)+\( \frac{z^2-xy}{z^2+z} \)\( \leq \)\( 0 \Leftrightarrow \sum_{cyc}{} {\frac {x}{x+1}} \) \( \leq \) \( \sum_{cyc}{} {\frac {xy}{x(x+1)}} \)

Aducand la acelasi numitor, mai ramane de demonstrat ca: \( 3{(xyz)}^2+ xyz + 2xyz \sum_{cyc}{} {xy} \leq \sum_{cyc}{} {x^3y^3} + \sum_{cyc}{} {(x^3y^2+x^3z^2)} + \sum_{cyc}{} {x^2y^2} \)

dar: \( \sum_{cyc}{} {x^3y^3} \geq 3{(xyz)}^2 \) (1)

\( \sum_{cyc}{} {(x^3y^2+x^3z^2)} = \sum_{cyc}{} {x^3(y^2+z^2)} \geq \sum_{cyc}{} {2x^3yz} = 2xyz \sum_{cyc}{} {x^2} \geq 2xyz \sum_{cyc}{} {xy} \) (2)

\( \sum_{cyc}{} {x^2y^2} = \sum_{cyc}{} {\frac {x^2(y^2+z^2)}{2}} \geq \sum_{cyc}{} {x^2yz} = xyz(x+y+z)=xyz \) (3)

Din (1) , (2) si (3) obtinem concluzia.

[Marius Mainea]

\( \sum\frac{x+gz}{x^2+x}=\sum\frac{x(x+y+z)+yz}{x(x+1)}=\sum\frac{(x+y)(x+z)}{x(x+y+x+z)}=\sum\frac{1}{\frac{x}{x+y}+\frac{x}{x+z}}\geq\sum\frac{(1+1+1)^2}{\sum\frac{x}{x+y}+\frac{x}{x+z}}=\frac{9}{3}=3 \)

Deci \( LHS=\sum\frac{x^2+x}{x^2+x}-\frac{x+gz}{x^2+x}\leq3-3=0 \)