Cho a, b, c > 0. Chứng minh rằng:

\[\frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}} \ge \frac{3}{2}\].

Lời giải:

Đặt \[A = \frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}}\]

\[A + 3 = \frac{a}{{b + c}} + 1 + \frac{b}{{c + a}} + 1 + \frac{c}{{a + b}} + 1\]

\[A + 3 = \frac{{a + b + c}}{{b + c}} + \frac{{a + b + c}}{{c + a}} + \frac{{a + b + c}}{{a + b}}\]

\[A + 3 = \left( {a + b + c} \right)\left( {\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}}} \right)\]

Ta có: \[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \frac{9}{{a + b + c}}\]

\[A + 3 \ge \left( {a + b + c} \right)\left( {\frac{9}{{a + b + b + c + c + a}}} \right) = \frac{9}{2}\]

Suy ra \[A \ge \frac{3}{2}.\]