Demonstrati ca daca \( x,y,z > 0
\) si \(
x^2 + y^2 + z^2 = 3
\), atunci
\( \sqrt {\frac{{x^3 }}{{x + y}} + \frac{{y^3 }}{{y + z}} + \frac{{z^3 }}{{z + x}}} \ge \sqrt{\frac{3}{2}} \).
Claudiu Mindrila
Inegalitate...
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Claudiu Mindrila
- Fermat
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Inegalitate...
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste
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Marius Dragoi
- Thales
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- Location: Bucharest
Este cunoscuta urmatoarea inegalitate:
\( \frac {{x^2}}{{a}} + \frac{{y^2}}{{b}} + \frac{{z^2}}{{c}} \geq \frac {{{(x+y+z)}^2}}{{a+b+c}} \) unde \( x,y,z,a,b,c \) strict pozitive.
Atunci vom avea:
\( \ sqrt {\frac {{x^4}}{{x^2 +xy}} + \frac{{y^4}}{{y^2 +yz}} + \frac{{z^4}}{{z^2 + zx}}} \geq \sqrt{\frac {{(x^2 +y^2 +z^2)}^2}{{x^2 +y^2+z^2+xy+yz+zx}}} = \sqrt{\frac{9}{{3+xy+yz+zx}} \)
\( \sqrt{\frac{9}{{3+xy+yz+zx}}} \geq \sqrt{\frac{3}{2}} \Leftrightarrow \sqrt {\frac {6}{{3+xy+yz+zx}}} \geq 1 \Leftrightarrow \)
\( \Leftrightarrow 3 \geq xy+yz+zx \Leftrightarrow x^2+y^2+z^2 \geq xy+yz+zx \) ceea ce este adevarat.
\( \frac {{x^2}}{{a}} + \frac{{y^2}}{{b}} + \frac{{z^2}}{{c}} \geq \frac {{{(x+y+z)}^2}}{{a+b+c}} \) unde \( x,y,z,a,b,c \) strict pozitive.
Atunci vom avea:
\( \ sqrt {\frac {{x^4}}{{x^2 +xy}} + \frac{{y^4}}{{y^2 +yz}} + \frac{{z^4}}{{z^2 + zx}}} \geq \sqrt{\frac {{(x^2 +y^2 +z^2)}^2}{{x^2 +y^2+z^2+xy+yz+zx}}} = \sqrt{\frac{9}{{3+xy+yz+zx}} \)
\( \sqrt{\frac{9}{{3+xy+yz+zx}}} \geq \sqrt{\frac{3}{2}} \Leftrightarrow \sqrt {\frac {6}{{3+xy+yz+zx}}} \geq 1 \Leftrightarrow \)
\( \Leftrightarrow 3 \geq xy+yz+zx \Leftrightarrow x^2+y^2+z^2 \geq xy+yz+zx \) ceea ce este adevarat.
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
The Faculty of Automatic Control and Computers