We have, $x={\log }_{5}3+{\log }_{7}5+{\log }_{9}7$

Using $\mathrm{AM}\ge \mathrm{GM}$ for three positive real numbers $\frac{a+b+c}{3}\ge {\left(abc\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$. Equality holds when $a=b=c.$

$\therefore \frac{{\log }_{5}3+{\log }_{7}5+{\log }_{9}7}{3}>{\left({\log }_{5}3·{\log }_{7}5·{\log }_{9}7\right)}^{1/3}$$$$\Rightarrow x>3{\left(\frac{\log 3}{\log 5}\times \frac{\log 5}{\log 7}\times \frac{\log 7}{\log 9}\right)}^{1/3}$

$\Rightarrow x>3{\left(\frac{1}{2}\right)}^{1/3}\Rightarrow x>\frac{3}{\sqrt[3]{2}}$