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B

braga

$\fbox{1}$Theo $Cauchy-Schwarz$ ta có:
$\dfrac{2}{a+3b}+\dfrac{1}{b+3c}+\dfrac{4}{c+3a} \\ \ge \dfrac{(2+1+4)^2}{2(a+3b)+(b+3c)+4(c+3a)} \\ = \dfrac{7}{2a+b+c}$
 
B

braga

$\fbox{3}$. Theo $Cauchy-Schwarz$ thì:
$$\sum \dfrac{4}{2a+b+c}\ge \dfrac{(2+2+2)^2}{4(a+b+c)}=\dfrac{9}{a+b+c}$$
 
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