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Cho x,y,z thỏa mãn: x+y+z=1


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#1
jackboy225

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Cho x,y,z thỏa mãn: x+y+z=1. TÌm Min:
A= $\frac{1}{x^{2}+y^{2}+z^{2}} + \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}$

#2
Tru09

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Cho x,y,z thỏa mãn: x+y+z=1. TÌm Min:
A= $\frac{1}{x^{2}+y^{2}+z^{2}} + \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}$

CHém nhanh nào
A =$\frac{1}{x^{2}+y^{2}+z^{2}} + \frac{1}{3xy} + \frac{1}{3yz} + \frac{1}{3zx}$+$ \frac{2}{3xy} + \frac{2}{3yz} + \frac{2}{3zx} \geq \frac{16}{x^2+y^2+z^2+3xy+3xz+3yz}+\frac{2}{3}. \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} \geq \frac{16}{1+\frac{1}{3}} +\frac{6}{xy+xz+yz} \geq \frac{16}{1+\frac{1}{3}} +\frac{6}{\frac{1}{3}}=30$
Dấu= sảy ra khi $x=y=z=\frac{1}{3}$

Edited by Tru09, 28-07-2012 - 09:45.





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