chứng minh bất đẳng thức.Khá dễ

L

l94

cho 3 số thực dương x,y,z thỏa mãn : 1/x+1/y+1/z=4
CMR: 1/(2x+y+z) +1/(x+2y+z) +1/(x+y+2z) =< 1
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Bày này kinh điển.
$$ \sum_{cyc} \frac{1}{2x+y+z} \le \sum_{cyc}\frac{1}{4}(\frac{1}{x+y}+\frac{1}{x+z}) \le \sum_{cyc}\frac{1}{16}(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}) \le 1$$
 
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V

vy000

$\dfrac1{2x+y+z} \le \dfrac14(\dfrac 1{x+y}+\dfrac1{x+z}) \le \dfrac 1{16}(2\dfrac1x+\dfrac1y+\dfrac1z)$

$\dfrac1{x+2y+z} \le \dfrac14(\dfrac 1{x+y}+\dfrac1{y+z})\le \dfrac 1{16}(\dfrac1x+2\dfrac1y+\dfrac1z)$

$\dfrac1{x+y+2z} \le \dfrac14(\dfrac 1{x+z}+\dfrac1{y+z})\le \dfrac 1{16}(\dfrac1x+\dfrac1y+2\dfrac1z)$

$\Rightarrow \sum \dfrac1{2x+y+z} \le 1$
 
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