Giúp em vs ạ Toán số

$\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}$