Cho a, b, c là độ dài 3 cạnh của tam giác. Chứng minh:

\(\frac{1}{{{a^2} + bc}} + \frac{1}{{{b^2} + ac}} + \frac{1}{{{c^2} + ab}} \le \frac{{a + b + c}}{{2abc}}\).

Áp dụng BĐT AM - GM:

• \({a^2} + bc \ge 2a\sqrt {bc} \Rightarrow \frac{1}{{{a^2} + bc}} \le \frac{1}{{2a\sqrt {bc} }}\)

• \({b^2} + ca \ge 2b\sqrt {ca} \Rightarrow \frac{1}{{{b^2} + ca}} \le \frac{1}{{2b\sqrt {ca} }}\)

\({c^2} + ab \ge 2c\sqrt {ab} \Rightarrow \frac{1}{{{c^2} + ab}} \le \frac{1}{{2c\sqrt {ab} }}\)

Khi đó:

\(\frac{1}{{{a^2} + bc}} + \frac{1}{{{b^2} + ac}} + \frac{1}{{{c^2} + ab}} \le \frac{1}{{2a\sqrt {bc} }} + \frac{1}{{2b\sqrt {ca} }} + \frac{1}{{2c\sqrt {ab} }}\)

\( = \frac{{\sqrt {bc} + \sqrt {ca} + \sqrt {ab} }}{{2abc}} \le \frac{{\frac{{b + c}}{2} + \frac{{c + a}}{2} + \frac{{a + b}}{2}}}{{2abc}}\)

\( = \frac{{a + b + c}}{{2abc}}\).

Vậy \(\frac{1}{{{a^2} + bc}} + \frac{1}{{{b^2} + ac}} + \frac{1}{{{c^2} + ab}} \le \frac{{a + b + c}}{{2abc}}\)