${5^{\frac{1}{4}\left( {\log _2^5x} \right)}} \ge 5{x^{\frac{1}{5}\left( {{{\log }_5}x} \right)}}$

Taking logarithm on base $5,$ then

$\frac{1}{4}\left(\log _{5}^{2} x\right) \geq 1+\frac{1}{5}\left(\log _{5} x\right)\left(\log _{5} x\right)$

$\Rightarrow \frac{1}{20} \log _{5}^{2} x \geq 1$

or $\left(\log _{5}^{2} x\right) \geq 20$

or $\log _{5} x \geq 2 \sqrt{5}$ and $ \log _{5} x \leq-2 \sqrt{5}$

or $x \geq 5^{2 \sqrt{5}} $ and $ x \leq 5^{-2 \sqrt{5}}$

But $x>0$

$\therefore $ $x \in\left(0,5^{-2 \sqrt{5}}\right] \cup\left[5^{2 \sqrt{5}}, \infty\right)$