Cho a, b, c là 3 cạnh trong tam giác. Chứng minh rằng:

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

Đặt x = b + c – a

y = a + c – b

z = a + b – c

Suy ra \(\left\{ \begin{array}{l}x + z = b + c - a + a + b - c = 2b\\x + y = b + c - a + a + c - b = 2c\\y + z = a + c - b + a + b - c = 2{\rm{a}}\end{array} \right.\)

Ta có: \(2A = \frac{{y + z}}{x} + \frac{{x + z}}{y} + \frac{{x + y}}{z}\)

\( = \frac{y}{x} + \frac{z}{x} + \frac{x}{y} + \frac{z}{y} + \frac{x}{z} + \frac{y}{z}\)

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

Áp dụng bất đẳng thức Cô – si ta có:

\(\begin{array}{l}\left( {\frac{y}{x} + \frac{x}{y}} \right) \ge 2\sqrt {\frac{y}{x}.\frac{x}{y}} = 2\\\left( {\frac{z}{y} + \frac{y}{z}} \right) \ge 2\sqrt {\frac{z}{y}.\frac{y}{z}} = 2\\\left( {\frac{x}{z} + \frac{z}{x}} \right) \ge 2\sqrt {\frac{z}{x}.\frac{x}{z}} = 2\end{array}\)

Suy ra:

\(\left( {\frac{y}{x} + \frac{x}{y}} \right) + \left( {\frac{z}{y} + \frac{y}{z}} \right) + \left( {\frac{x}{z} + \frac{z}{x}} \right) \ge 6\)

\( \Leftrightarrow 2A = \frac{{y + z}}{x} + \frac{{x + z}}{y} + \frac{{x + y}}{z} \ge 6\)

\( \Leftrightarrow 2A = \frac{{2{\rm{a}}}}{{b + c - a}} + \frac{{2b}}{{a + c - b}} + \frac{{2c}}{{a + b - c}} \ge 6\)

\( \Leftrightarrow A = \frac{{\rm{a}}}{{b + c - a}} + \frac{b}{{a + c - b}} + \frac{c}{{a + b - c}} \ge 3\)

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