[tex][tex]\frac{1}{\sqrt[3]{a+3b}}+\frac{1}{\sqrt[3]{b+3c}}+\frac{1}{\sqrt[3]{c+3a}}\geq \frac{9}{\sqrt[3]{a+3b}+\sqrt[3]{b+3c}+\sqrt[3]{c+3a}}\geq \frac{9}{\frac{a+3b+1+1}{3}+\frac{b+3c+1+1}{3}+\frac{c+3a+1+1}{3}}=\frac{9}{\frac{4(a+b+c)+6}{3}}=\frac{9}{\frac{4.\frac{3}{4}+6}{3}}=\frac{9}{\frac{9}{3}}=3[/tex][/tex]