3*(1/√3)^2-3x меньше 1/9 решите плиз срочно

[latex]3\cdot(\frac{1}{\sqrt{3}})^2-3x\ \textless \ \frac{1}{9}, \\ 3\cdot\frac{1}{3}-3x\ \textless \ \frac{1}{9}, \\ 1-3x\ \textless \ \frac{1}{9}, \\ -3x\ \textless \ \frac{1}{9}-1, \\ -3x\ \textless \ -\frac{8}{9}, \\ x\ \textgreater \ -\frac{8}{9}\cdot(-\frac{1}{3}), \\ x\ \textgreater \ \frac{8}{27}.[/latex]