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 '=" $\Leftrightarrow a=b=c$
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 '=" $\Leftrightarrow a=b=c$