$bdt\Leftrightarrow \sqrt[n]{\frac{a}{(b+c)(n-1)}}+\sqrt[n]{\frac{b}{(c+a)(n-1)}}+\sqrt[n]{\frac{c}{(a+b)(n-1)}} > \frac{n}{n-1}$Ta có :
$\sqrt[n]{\frac{a}{(b+c)(n-1)}}=\frac{1}{\sqrt[n]{\left[\dfrac {b+c}a(n-1) \right].\overset{n-1 \hspace{1mm} \text{số} \hspace{1mm}1}{\overbrace{1.1...1}}}} \ge \frac{n}{\dfrac{b+c}a(n-1)+\overset{n-1 \hspace{1mm} \text{số} \hspace{1mm} 1}{\overbrace{1+1+1+...+1}}}$
$=\frac{an}{(a+b+c)(n-1)}$
Tương tự cộng lại $\Rightarrow$ dpcm (dấu = ko xảy ra)