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Claudiu Mindrila
Fermat
Joined: 01 Oct 2007
Posts: 517
Location: Targoviste
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 Posted: Thu Feb 04, 2010 6:41 pm Post subject: Inegalitate simpluta, 1 |
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Fie  . Sa se arate ca: }{z+yz}+\frac{y\left(z+x\right)}{y+zx}+\frac{z\left(x+y\right)}{z+xy}\le2\left(\frac{x^{2}}{x+yz}+\frac{y^{2}}{y+zx}+\frac{z^{2}}{z+xy}\right) "\frac{x\left(y+z\right)}{z+yz}+\frac{y\left(z+x\right)}{y+zx}+\frac{z\left(x+y\right)}{z+xy}\le2\left(\frac{x^{2}}{x+yz}+\frac{y^{2}}{y+zx}+\frac{z^{2}}{z+xy}\right)")
Stefan Gavril
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elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste |
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Marius Mainea
Gauss
Joined: 26 May 2008
Posts: 1075
Location: Gaesti (Dambovita)
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| Posted: Fri Feb 05, 2010 1:44 pm Post subject: |
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\le\sum\(\frac{x^2}{x+yz}+\frac{y^2}{y+zx}\) "\sum\(\frac{xy}{x+yz}+\frac{yx}{y+zx}\)\le\sum\(\frac{x^2}{x+yz}+\frac{y^2}{y+zx}\)") |
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