$A=\frac{1}{\sqrt[3]{a+3b}}+\frac{1}{\sqrt[3]{b+3c}}+\frac{1}{\sqrt[3]{c+3a}}=\frac{12}{3.\sqrt[3]{8.8.(a+3b)}}+\frac{12}{3.\sqrt[3]{8.8.(b+3c)}}+\frac{12}{3.\sqrt[3]{8.8(c+3a)}}$Theo Cosi 3 số:$A\geq \frac{12}{16+a+3b}+\frac{12}{16+b+3c}+\frac{12}{16+c+3a}$$\Rightarrow A\geq \frac{12^2}{12(16+a+3b)}+\frac{12^2}{12(16+b+3c)}+\frac{12^2}{12(16+c+3a)}$Theo Bđt CBS dạng Engel thì $A\geq \frac{(12+12+12)^2}{12[4(a+b+c)+48]}=\frac{3}{2}$Dấu =xảy ra khi $a=b=c=2$
$A=\frac{1}{\sqrt[3]{a+3b}}+\frac{1}{\sqrt[3]{b+3c}}+\frac{1}{\sqrt[3]{c+3a}}=\frac{12}{3.\sqrt[3]{8.8.(a+3b)}}+\frac{12}{3.\sqrt[3]{8.8.(b+3c)}}+\frac{12}{3.\sqrt[3]{8.8(c+3a)}}$Theo Cosi 3 số:$A\geq \frac{12}{16+a+3b}+\frac{12}{16+b+3c}+\frac{12}{16+c+3a}$$\Rightarrow A\geq \frac{12^2}{12(16+a+3b)}+\frac{12^2}{12(16+b+3c)}+\frac{12^2}{12(16+b+3c)}$Theo Bđt CBS dạng Engel thì $A\geq \frac{(12+12+12)^2}{12[4(a+b+c)+48]}=\frac{3}{2}$Dấu =xảy ra khi $a=b=c=2$
$A=\frac{1}{\sqrt[3]{a+3b}}+\frac{1}{\sqrt[3]{b+3c}}+\frac{1}{\sqrt[3]{c+3a}}=\frac{12}{3.\sqrt[3]{8.8.(a+3b)}}+\frac{12}{3.\sqrt[3]{8.8.(b+3c)}}+\frac{12}{3.\sqrt[3]{8.8(c+3a)}}$Theo Cosi 3 số:$A\geq \frac{12}{16+a+3b}+\frac{12}{16+b+3c}+\frac{12}{16+c+3a}$$\Rightarrow A\geq \frac{12^2}{12(16+a+3b)}+\frac{12^2}{12(16+b+3c)}+\frac{12^2}{12(16+
c+3
a)}$Theo Bđt CBS dạng Engel thì $A\geq \frac{(12+12+12)^2}{12[4(a+b+c)+48]}=\frac{3}{2}$Dấu =xảy ra khi $a=b=c=2$