Ta chứng minh BĐT phụ:

\(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \frac{9}{{a + b + c}}\) (với a, b, c > 0)

Ta có: \(\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)\left( {a + b + c} \right)\)

\( = 1 + \frac{a}{b} + \frac{a}{c} + 1 + \frac{b}{a} + \frac{b}{c} + 1 + \frac{c}{a} + \frac{c}{b}\)

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

Áp dụng bất đẳng thức Cô-si ta có:

\(\frac{a}{b} + \frac{b}{a} \ge 2\); \(\frac{b}{c} + \frac{c}{b} \ge 2\); \(\frac{a}{c} + \frac{c}{a} \ge 2\).

Suy ra ta có: \(\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)\left( {a + b + c} \right) \ge 3 + 6 = 9\)

\( \Leftrightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \frac{9}{{a + b + c}}\)(*)

Áp dụng (*) ta có:

\(\frac{1}{{{a^2} + 2bc}} + \frac{1}{{{b^2} + 2ac}} + \frac{1}{{{c^2} + 2ab}} \ge \frac{9}{{{a^2} + {b^2}{ + ^2} + 2bc + 2ac + 2ab}}\)

\(\frac{1}{{{a^2} + 2bc}} + \frac{1}{{{b^2} + 2ac}} + \frac{1}{{{c^2} + 2ab}} \ge \frac{9}{{{{(a + b + c)}^2}}}\)

Mà a + b + c = 1 nên:

\(\frac{1}{{{a^2} + 2bc}} + \frac{1}{{{b^2} + 2ac}} + \frac{1}{{{c^2} + 2ab}} \ge 9\) (đpcm).

Dấu bằng xảy ra khi \(a = b = c = \frac{1}{3}\).