Đặt $\log_3 x = t \Rightarrow x=3^t$
BPT $\Leftrightarrow (\sqrt{10}+1)^t -(\sqrt{10}-1)^t \ge \dfrac{2}{3}.3^t$
$\Leftrightarrow \bigg (\dfrac{\sqrt{10}+1}{3} \bigg)^t-\bigg (\dfrac{\sqrt{10}-1}{3} \bigg)^t \ge \dfrac{2}{3}$
Đặt $\bigg (\dfrac{\sqrt{10}+1}{3} \bigg)^t = u \Rightarrow \bigg (\dfrac{\sqrt{10}-1}{3} \bigg)^t=\dfrac{1}{u}$
Vì $ \bigg (\dfrac{\sqrt{10}+1}{3} \bigg)^t .\bigg (\dfrac{\sqrt{10}-1}{3} \bigg)^t =1$
BPT $\Leftrightarrow u-\dfrac{1}{u} \ge \dfrac{2}{3}$
$\Leftrightarrow 3u^2 -2u -3 \ge 0$
$\Leftrightarrow \left [ \begin{matrix} u \ge \dfrac{\sqrt{10}+1}{3} \\ u\le - \dfrac{\sqrt{10}-1}{3} \end{matrix} \right.$
Tự giải nốt nhé