cho x,y,z là ba cạnh của 1 tam giác. CMR:( 1/x+y-z)+(1/y+z-x)+(1/z+x-y)>= (1/x)+(1/y)+(1/z)

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1. Theo bất đẳng thức tam giác ta có:

   $\left\{ \begin{array}{l} x + y – z > 0\\ y + z – x > 0\\ z + x – y > 0 \end{array} \right.$

   $\begin{array}{l} \dfrac{1}{{x + y – z}} + \dfrac{1}{{y + z – x}} \ge \dfrac{{{{\left( {1 + 1} \right)}^2}}}{{x + y – z + y + z – x}} = \dfrac{4}{{2y}} = \dfrac{2}{y}\left( 1 \right)\\ \dfrac{1}{{y + z – x}} + \dfrac{1}{{x + z – y}} \ge \dfrac{{{{\left( {1 + 1} \right)}^2}}}{{y + z – x + x + z – y}} = \dfrac{4}{{2z}} = \dfrac{2}{z}\left( 2 \right)\\ \dfrac{1}{{x + z – y}} + \dfrac{1}{{x + y – z}} \ge \dfrac{{{{\left( {1 + 1} \right)}^2}}}{{x + z – y + x + y – z}} = \dfrac{4}{{2x}} = \dfrac{2}{x}\left( 3 \right)\\ \left( 1 \right) + \left( 2 \right) + \left( 3 \right)\\  \Rightarrow 2\left( {\dfrac{1}{{x + y – z}} + \dfrac{1}{{y + z – x}} + \dfrac{1}{{x + z – y}}} \right) \ge 2\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)\\  \Rightarrow \dfrac{1}{{x + y – z}} + \dfrac{1}{{y + z – x}} + \dfrac{1}{{x + z – y}} \ge \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \end{array}$

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