[Toán 10] $\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+b}\ge \frac{3}{a+b+c}$

N

nttthn_97

Áp dụng BĐT
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$[TEX]\geq[/TEX]$\frac{9}{a+b+c}$

VT[TEX]\geq[/TEX]$\frac{9}{2a+b+2b+c+2c+a}=\frac{3}{a+b+c}$
 
H

hoangtrongminhduc

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V

vansang02121998

Áp dụng Cauchy

$\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b} \ge \dfrac{9}{2a+b}$

$\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c} \ge \dfrac{9}{2b+c}$

$\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a} \ge \dfrac{9}{2c+a}$

$\Rightarrow \dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c} \ge \dfrac{9}{2a+b}+\dfrac{9}{2b+c}+\dfrac{9}{2c+a}$

$\Leftrightarrow \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \ge \dfrac{3}{2a+b}+\dfrac{3}{2b+c}+\dfrac{3}{2c+a}$
 
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