Fie \( a,b,c\ge 0 \) astfel incat \( ab+bc+ca=3 \) . Demonstrati ca :
\( \frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2}\le 1 \)
ab+bc+ca=3
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ab+bc+ca=3
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Marius Mainea
- Gauss
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Se aplica CBS 
\( LHS=\frac{3}{2}-\frac{1}{2}\sum\frac{a^2}{a^2+2}\le RHS \)
\( LHS=\frac{3}{2}-\frac{1}{2}\sum\frac{a^2}{a^2+2}\le RHS \)
Last edited by Marius Mainea on Thu Jun 11, 2009 10:32 pm, edited 1 time in total.
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Mateescu Constantin
- Newton
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Inegalitatea e echivalenta cu \( a^2b^2+b^2c^2+c^2a^2+a^2b^2c^2\ge 4 \)
Notam \( x=bc,\ y=ca,\ z=ab \) si ramane \( x^2+y^2+z^2+xyz\ge 4 \) cu \( x+y+z=3 \)
Fie \( x=\mbox min\{x,\ y,\ z\},\ x\le 1 \)
\( \Longrightarrow x^2+y^2+z^2+xyz-4=x^2+(y+z)^2+yz(x-2)-4\ge x^2+(y+z)^2+\frac{1}{4}(y+z)^2(x-2)-4=x^2+\frac{x+2}{4}(y+z)^2-4=x^2+\frac{x+2}{4}(3-x)^2-4=\frac{1}{4}(x-1)^2(x+2)\ge 0 \).
Notam \( x=bc,\ y=ca,\ z=ab \) si ramane \( x^2+y^2+z^2+xyz\ge 4 \) cu \( x+y+z=3 \)
Fie \( x=\mbox min\{x,\ y,\ z\},\ x\le 1 \)
\( \Longrightarrow x^2+y^2+z^2+xyz-4=x^2+(y+z)^2+yz(x-2)-4\ge x^2+(y+z)^2+\frac{1}{4}(y+z)^2(x-2)-4=x^2+\frac{x+2}{4}(y+z)^2-4=x^2+\frac{x+2}{4}(3-x)^2-4=\frac{1}{4}(x-1)^2(x+2)\ge 0 \).