\(\begin{array}{l}
\text{Ta có:}\\
\quad \sin C = \sin(A + B)\\
\Leftrightarrow \sin C = \sin A\cos B + \sin B\cos A\\
\Leftrightarrow \sin C = \cos A.\cos B\left(\dfrac{\sin A}{\cos A} + \dfrac{\sin B}{\cos B}\right)\\
\Leftrightarrow \sin C = \cos A.\cos B(\tan A + \tan B)\\
\Leftrightarrow \tan C = \dfrac{\cos A.\cos B}{\cos C}\cdot (\tan A + \tan B)\\
\Leftrightarrow \dfrac{\cos A.\cos B}{\cos C} = \dfrac{\tan C}{\tan A + \tan B}\\
\Leftrightarrow \sqrt{\dfrac{\cos A.\cos B}{\cos C}} = \sqrt{\dfrac{\tan C}{\tan A + \tan B}}\\
\text{Chứng minh tương tự ta được:}\\
\sqrt{\dfrac{\cos B.\cos C}{\cos A}} = \sqrt{\dfrac{\tan A}{\tan B + \tan C}}\\
\sqrt{\dfrac{\cos C.\cos A}{\cos B}} = \sqrt{\dfrac{\tan B}{\tan C + \tan A}}\\
Đặt\ \begin{cases}a = \tan A\\b = \tan B\\c = \tan C\end{cases}\\
\text{Ta cần chứng minh:}\\
\sqrt{\dfrac{a}{b+c}} + \sqrt{\dfrac{b}{c+a}} + \sqrt{\dfrac{c}{a+b}} >2\\
\text{Áp dụng bất đẳng thức $AM-GM$ ta được:}\\
\quad \sqrt{\dfrac{b+c}{a}} \leqslant \dfrac{\dfrac{b+c}{a}+1}{2}\\
\Leftrightarrow \sqrt{\dfrac{b+c}{a}} \leqslant \dfrac{a+b+c}{2a}\\
\Leftrightarrow \sqrt{\dfrac{a}{b+c}} \geqslant \dfrac{2a}{a+b+c}\\
\text{Hoàn toàn tương tự, ta được:}\\
\sqrt{\dfrac{b}{c+a}} \geqslant \dfrac{2b}{a+b+c}\\
\sqrt{\dfrac{c}{a+b}} \geqslant \dfrac{2c}{a+b+c}\\
\text{Cộng vế theo vế ta được:}\\
\sqrt{\dfrac{a}{b+c}} + \sqrt{\dfrac{b}{c+a}} + \sqrt{\dfrac{c}{a+b}} \geqslant \dfrac{2a}{a+b+c} +\dfrac{2b}{a+b+c} + \dfrac{2c}{a+b+c}\\
\Leftrightarrow \sqrt{\dfrac{a}{b+c}} + \sqrt{\dfrac{b}{c+a}} + \sqrt{\dfrac{c}{a+b}} \geqslant 2\\
\text{Dấu = xảy ra}\ \Leftrightarrow \begin{cases}a = b+ c\\b= c+a\\c = a + b\end{cases} \\\Leftrightarrow a +b + c =0\\
\Leftrightarrow \tan A + \tan B + \tan C = 0\quad \text{(vô lí)}\\
\Rightarrow \text{Dấu = không xảy ra}\\
\Rightarrow \sqrt{\dfrac{a}{b+c}} + \sqrt{\dfrac{b}{c+a}} + \sqrt{\dfrac{c}{a+b}} >2\\
Hay\ \ \sqrt{\dfrac{\cos A.\cos B}{\cos C}} + \sqrt{\dfrac{\cos B.\cos C}{\cos A}} + \sqrt{\dfrac{\cos C.\cos A}{\cos B}} > 2
\end{array}\)