BDT khó mọi người giải giúp mình nha!!

V

vansang02121998

Áp dụng bất đẳng thức Cauchy Schwarz có 3 số, ta có

$\dfrac{1^2}{a}+\dfrac{1^2}{b}+\dfrac{1^2}{b} \geq \dfrac{(1+1+1)^2}{a+b+b} = \dfrac{9}{a+2b}$

$\dfrac{1^2}{b}+\dfrac{1^2}{c}+\dfrac{1^2}{c} \geq \dfrac{(1+1+1)^2}{b+c+c} = \dfrac{9}{b+2c}$

$\dfrac{1^2}{c}+\dfrac{1^2}{a}+\dfrac{1^2}{a} \geq \dfrac{(1+1+1)^2}{c+a+a} = \dfrac{9}{c+2a}$

Cộng vế với vế, ta có

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

$\Leftrightarrow \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 3(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a})$
 
N

nhok111o0o

ta có
12a+12b+12b≥(1+1+1)2a+b+b=9a+2b

12b+12c+12c≥(1+1+1)2b+c+c=9b+2c

12c+12a+12a≥(1+1+1)2c+a+a=9c+2a

Cộng vế với vế, ta có

3a+3b+3c≥9a+2b+9b+2c+9c+2a

⇔1a+1b+1c≥3(1a+2b+1b+2c+1c+2a)
 
B

baothi125

12a+12b+12b≥(1+1+1)2a+b+b=9a+2b

12b+12c+12c≥(1+1+1)2b+c+c=9b+2c

12c+12a+12a≥(1+1+1)2c+a+a=9c+2a

Cộng vế với vế, ta có

3a+3b+3c≥9a+2b+9b+2c+9c+2a

⇔1a+1b+1c≥3(1a+2b+1b+2c+1c+2a)[/QUOTE]
 
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