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$\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\geq 3(\frac{3}{x+2y}+\frac{3}{y+2z


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#1
hoangmanhquan

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Cho x,y,z>0. Chứng minh rằng:

$\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\geq 3(\frac{3}{x+2y}+\frac{3}{y+2z}+\frac{3}{z+2x}-2)$


Edited by hoangmanhquan, 23-10-2013 - 06:08.

:icon1: Sống là cho, đâu chỉ nhận riêng mình :icon1: 

 

 


#2
kfcchicken98

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$\frac{1}{x^{3}}\geq \frac{3}{x}-2$
$\frac{1}{y^{3}}\geq \frac{3}{y}-2$

$\frac{1}{z^{3}}\geq \frac{3}{z}-2$

$3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 3(\frac{3}{x+2y}+\frac{3}{y+2z}+\frac{3}{z+2x})$
cộng vào suy ra đpcđpcm



#3
Tsubasa

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$\frac{1}{x^3}+1+1\geqslant \frac{3}{x}$

 

$\frac{1}{y^3}+1+1\geqslant \frac{3}{y}$

 

$\frac{1}{z^3}+1+1\geqslant \frac{3}{z}$ 

 

$\frac{1}{x}+\frac{1}{y}+\frac{1}{y}\geqslant \frac{9}{x+2y}$

 

$\frac{1}{y}+\frac{1}{z}+\frac{1}{z}\geqslant \frac{9}{y+2z}$

 

$\frac{1}{z}+\frac{1}{x}+\frac{1}{x}\geqslant \frac{9}{z+2x}$

 

=> đpcm


Edited by Tsubasa, 23-10-2013 - 08:39.





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