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

\(\frac{a}{{b + c - a}} + \frac{b}{{a + c - b}} + \frac{c}{{a + b - c}} \ge 3\).

Đặt: \(\left\{ \begin{array}{l}b + c - a = x\\a + c - b = y\\a + b - c = z\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x + y = 2c\\y + z = 2a\\z + x = 2b\end{array} \right.\)

Do a, b, c là độ dài 3 cạnh tam giác nên x, y, z > 0

Ta có:

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

\[ \Rightarrow 2A = \frac{{2a}}{{b + c - a}} + \frac{{2b}}{{a + c - b}} + \frac{{2c}}{{a + b - c}}\]

\[ = \frac{{y + z}}{x} + \frac{{z + x}}{y} + \frac{{x + y}}{z}\]

\[ = \frac{y}{x} + \frac{z}{x} + \frac{z}{y} + \frac{x}{y} + \frac{x}{z} + \frac{y}{z}\]

\[ = \left( {\frac{y}{x} + \frac{x}{y}} \right) + \left( {\frac{z}{x} + \frac{x}{z}} \right) + \left( {\frac{z}{y} + \frac{y}{z}} \right)\]

Dễ chứng minh \(\frac{a}{b} + \frac{b}{a} \ge 2\sqrt {\frac{a}{b}\,.\,\frac{b}{a}} = 2\;\left( {a;\;b > 0} \right)\) (BĐT AG – GM)

\( \Rightarrow 2A = \left( {\frac{y}{x} + \frac{x}{y}} \right) + \left( {\frac{z}{x} + \frac{x}{z}} \right) + \left( {\frac{z}{y} + \frac{y}{z}} \right) \ge 2 + 2 + 2 = 6\)

\( \Rightarrow A = \frac{a}{{b + c - a}} + \frac{b}{{a + c - b}} + \frac{c}{{a + b - c}} \ge 3\)(đpcm).