a) Aratati ca oricare ar fi \( x,y,z \in \mathbb{R} \) au loc inegalitatile:
I)\( x^2+y^2+z^2 \geq xy+yz+zx \)
II)\( x^4+y^4+z^4\geq xyz(x+y+z) \).
b) Daca \( x,y,z>0 \) si \( x+y+z=1 \), atunci: \( \frac{x+y+1}{x^4+y^4+1}+\frac{y+z+1}{y^4+z^4+1}+\frac{z+x+1}{z^4+x^4+1} \leq \frac{1}{xyz}. \)
Marius Damian, Braila
Concursul "Congruente", problema 2
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Claudiu Mindrila
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Concursul "Congruente", problema 2
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Marius Mainea
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a) cunoscuta.
b) Aplicam a) de doua ori :
\( x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyyz+yzzx+zxxy=xyz(x+y+z) \)
c) Aplicam b) si ipoteza x+y+z=1 :
\( \frac{x+y+1}{x^4+y^4+1}+\frac{y+z+1}{y^4+z^4+1}+\frac{z+x+1}{z^4+x^4+1}\le \frac{z+y+1}{xy1(x+y+1)}+\frac{y+z+1}{yz1(y+z+1)}+\frac{z+x+1}{zx1(z+x+1)}=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}=\frac{1}{xyz} \)
b) Aplicam a) de doua ori :
\( x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyyz+yzzx+zxxy=xyz(x+y+z) \)
c) Aplicam b) si ipoteza x+y+z=1 :
\( \frac{x+y+1}{x^4+y^4+1}+\frac{y+z+1}{y^4+z^4+1}+\frac{z+x+1}{z^4+x^4+1}\le \frac{z+y+1}{xy1(x+y+1)}+\frac{y+z+1}{yz1(y+z+1)}+\frac{z+x+1}{zx1(z+x+1)}=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z+x+y}{xyz}=\frac{1}{xyz} \)