[tex]\frac{a}{c}+\frac{a}{c}+\frac{c}{b}\geq 3\sqrt[3]{\frac{a^2}{bc}}=3\sqrt[3]{\frac{a^3}{abc}}=3.\frac{a}{\sqrt[3]{abc}}\\TT:\frac{b}{a}+\frac{c}{b}+\frac{c}{b}\geq 3\frac{c}{\sqrt[3]{abc}}\\\frac{a}{c}+\frac{b}{a}+\frac{b}{a}\geq 3.\frac{b}{\sqrt[3]{abc}}\\\rightarrow \frac{a}{c}+\frac{b}{a}+\frac{c}{b}\geq \frac{1}{\sqrt[3]{abc}}\\\rightarrow \frac{a}{c}+\frac{b}{a}+\frac{c}{b}+\sqrt[3]{abc}\geq \frac{1}{\sqrt[3]{abc}}+\sqrt[3]{abc}= \frac{8}{9\sqrt[3]{abc}}+\frac{1}{9\sqrt[3]{abc}}+\sqrt[3]{abc}\geq \frac{8}{3(a+b+c)}+\frac{2}{3}=\frac{10}{3}=\frac{10}{3(a+b+c)^2}\geq \frac{10}{9(a^2+b^2+c^2)}[/tex]