Áp dụng bất đẳng thức $AM-GM$ cho hai số dương ta được:
$\begin{array}{l} \frac{{{a^2}}}{{b + c}} + \frac{{b + c}}{4} \ge 2\sqrt {\frac{{{a^2}}}{{b + c}}.\frac{{b + c}}{4}} = 2.\frac{a}{2} = a\\ \frac{{{b^2}}}{{a + c}} + \frac{{a + c}}{4} \ge 2\sqrt {\frac{{{b^2}}}{{a + c}}.\frac{{a + c}}{4}} = 2.\frac{b}{2} = b\\ \frac{{{c^2}}}{{a + b}} + \frac{{a + b}}{4} \ge 2\sqrt {\frac{{{c^2}}}{{a + b}}.\frac{{a + b}}{4}} = 2.\frac{c}{2} = c\\ \Rightarrow \frac{{{a^2}}}{{b + c}} + \frac{{{b^2}}}{{a + c}} + \frac{{{c^2}}}{{a + b}} \ge \left( {a + b + c} \right) - \frac{{a + b + c}}{2}\\ \Rightarrow \frac{{{a^2}}}{{b + c}} + \frac{{{b^2}}}{{a + c}} + \frac{{{c^2}}}{{a + b}} \ge \frac{{a + b + c}}{2} \end{array}$
Dấu bằng xảy ra khi $a=b=c$
