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

Đặt \(\left\{ {\begin{array}{*{20}{c}}{x = a + b - c > 0}\\{y = b + c - a > 0}\\{z = c + a - b > 0}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{a = \frac{{x + z}}{2}}\\{b = \frac{{x + y}}{2}}\\{c = \frac{{y + z}}{2}}\end{array}} \right.\)

Ta cần chứng minh: \(\frac{{{{\left( {x + y} \right)}^2}}}{{4z}} + \frac{{{{\left( {y + z} \right)}^2}}}{{4x}} + \frac{{{{\left( {z + x} \right)}^2}}}{{4y}} \ge x + y + z\)

Ta có: \(\frac{{{{\left( {x + y} \right)}^2}}}{{4z}} + \frac{{{{\left( {y + z} \right)}^2}}}{{4x}} + \frac{{{{\left( {z + x} \right)}^2}}}{{4y}} \ge \frac{{xy}}{z} + \frac{{yz}}{x} + \frac{{zx}}{y}{\rm{  }}\left( 1 \right)\)

Mặt khác: \(\frac{{xy}}{z} + \frac{{yz}}{x} \ge 2y;{\rm{ }}\frac{{yz}}{x} + \frac{{zx}}{y} \ge 2z;{\rm{ }}\frac{{xy}}{z} + \frac{{zx}}{y} \ge 2x\).

Khi đó \(\frac{{xy}}{z} + \frac{{yz}}{x} + \frac{{zx}}{y} \ge x + y + z{\rm{  }}\left( 2 \right)\)

Từ \(\left( 1 \right){\rm{, }}\left( 2 \right)\) ta có \(\frac{{{{\left( {x + y} \right)}^2}}}{{4z}} + \frac{{{{\left( {y + z} \right)}^2}}}{{4x}} + \frac{{{{\left( {z + x} \right)}^2}}}{{4y}} \ge x + y + z\)

Vậy \(\frac{{{a^2}}}{{b + c - a}} + \frac{{{b^2}}}{{c + a - b}} + \frac{{{c^2}}}{{a + b - c}} \ge a + b + c\)

Dấu bằng xãy ra khi \[a = b = c\]