Điều kiện: $\begin{cases}x>0 \\ x^2-1>0 \end{cases} \Leftrightarrow x>1$. Đặt $x=\frac{1}{\cos \alpha}, \alpha \in (0;\frac{\pi}{2})$
BPT $\Leftrightarrow \frac{1}{\cos \alpha}+\frac{1}{\sin \alpha}>\frac{3\sqrt{5}}{2}$
$\Leftrightarrow 2(\sin \alpha+\cos \alpha)>3\sqrt{5}\sin \alpha\cos \alpha$ (*)
Đặt $t=\sin \alpha+\cos \alpha=\sqrt{2}\cos (t-\frac{\pi}{4}) \in(1;\sqrt{2}]$
$\Rightarrow \sin \alpha\cos \alpha=\frac{t^2-1}{2}$
Khi đó:
BPT (*) $\Leftrightarrow 2t>3\sqrt{5}\frac{t^2-1}{2} \Leftrightarrow 3\sqrt{5}t^2-4t-3\sqrt{5}<0 $
$\Leftrightarrow -\frac{\sqrt{5}}{3}<t<\frac{\sqrt{5}}{3} \Leftrightarrow \frac{\sqrt{2}}{2} <\cos (t-\frac{\pi}{4})<\frac{3}{\sqrt{10}}$
$\Leftrightarrow \left[ \begin{array}{l}\frac{\pi}{4}>t-\frac{\pi}{4}>\varphi\\-\frac{\pi}{4}<t-\frac{\pi}{4}<-\varphi \end{array} \right.$ ( với $\cos \varphi=\frac{3}{\sqrt{10}} , \varphi \in (0;\frac{\pi}{2})$
$\Leftrightarrow \left[ \begin{array}{l} \frac{\pi}{4}+\varphi<t<\frac{\pi}{2}\\0<t<\frac{\pi}{4}-\varphi\end{array} \right.$
Do hàm số $y=\cos x$ nghịch biến trên $[0;\pi]$ nên:
$\left[ \begin{array}{l}0<\cos t<\cos (\frac{\pi}{4}+\varphi)\\\cos (\frac{\pi}{4}-\varphi)<\cos t <1\end{array} \right.$
Hơn nữa : $\begin{cases}\cos (\frac{\pi}{4}+\varphi)=\frac{1}{\sqrt{2}}(\cos \varphi-\sin \varphi)=\frac{1}{\sqrt{2}}\left ( \frac{3}{\sqrt{10}}-\frac{1}{\sqrt{10}} \right )=\frac{1}{\sqrt{5}} \\ \cos (\frac{\pi}{4}-\varphi)=\frac{1}{\sqrt{2}}(\cos \varphi+\sin \varphi) =\frac{2}{\sqrt{5}}\end{cases}$
Do đó nghiệm BPT đã cho là $\left[ \begin{array}{l}x>\sqrt{5}\\1<x<\frac{\sqrt{5}}{2}\end{array} \right.$