C2)bđt \Leftrightarrow \frac{\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} \leq1
\Leftrightarrow \sum \frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+ \frac{1}{b} +\frac{1}{c})}}\leq1
ta có \frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} =\sqrt[3]{\frac{1}{3}.\frac{a}{a+b+c}.\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}
\leq \frac{1}{3}(\frac{1}{3}+\frac{a}{a+b+c}+\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})
TT \Rightarrow VT\leq \frac{1}{3}(1+1+1)=1
dấu '=" \Leftrightarrow a=b=c