Theo bất đẳng thức tam giác ta có:
$\left\{ \begin{array}{l} x + y - z > 0\\ y + z - x > 0\\ z + x - y > 0 \end{array} \right.$
$\begin{array}{l} \dfrac{1}{{x + y - z}} + \dfrac{1}{{y + z - x}} \ge \dfrac{{{{\left( {1 + 1} \right)}^2}}}{{x + y - z + y + z - x}} = \dfrac{4}{{2y}} = \dfrac{2}{y}\left( 1 \right)\\ \dfrac{1}{{y + z - x}} + \dfrac{1}{{x + z - y}} \ge \dfrac{{{{\left( {1 + 1} \right)}^2}}}{{y + z - x + x + z - y}} = \dfrac{4}{{2z}} = \dfrac{2}{z}\left( 2 \right)\\ \dfrac{1}{{x + z - y}} + \dfrac{1}{{x + y - z}} \ge \dfrac{{{{\left( {1 + 1} \right)}^2}}}{{x + z - y + x + y - z}} = \dfrac{4}{{2x}} = \dfrac{2}{x}\left( 3 \right)\\ \left( 1 \right) + \left( 2 \right) + \left( 3 \right)\\ \Rightarrow 2\left( {\dfrac{1}{{x + y - z}} + \dfrac{1}{{y + z - x}} + \dfrac{1}{{x + z - y}}} \right) \ge 2\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)\\ \Rightarrow \dfrac{1}{{x + y - z}} + \dfrac{1}{{y + z - x}} + \dfrac{1}{{x + z - y}} \ge \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \end{array}$
