## Cho a>b>c>0. CMR: b/ (√a+b – √a-b) < c/ (√a+c - √a-c)

Question

Cho a>b>c>0. CMR: b/ (√a+b – √a-b) < c/ (√a+c - √a-c)

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1 năm 2021-08-09T01:46:34+00:00 1 Answers 4 views 0

## Answers ( )

1. Giải thích các bước giải:

Ta có:

$$\begin{array}{l} \dfrac{b}{{\sqrt {a + b} – \sqrt {a – b} }} – \dfrac{c}{{\sqrt {a + c} – \sqrt {a – c} }}\\ = \dfrac{{b.\left( {\sqrt {a + b} + \sqrt {a – b} } \right)}}{{\left( {\sqrt {a + b} – \sqrt {a – b} } \right)\left( {\sqrt {a + b} + \sqrt {a – b} } \right)}} – \dfrac{{c\left( {\sqrt {a + c} + \sqrt {a – c} } \right)}}{{\left( {\sqrt {a + c} – \sqrt {a – c} } \right)\left( {\sqrt {a + c} + \sqrt {a – c} } \right)}}\\ = \dfrac{{b.\left( {\sqrt {a + b} + \sqrt {a – b} } \right)}}{{\left( {a + b} \right) – \left( {a – b} \right)}} – \dfrac{{c\left( {\sqrt {a + c} + \sqrt {a – c} } \right)}}{{\left( {a + c} \right) – \left( {a – c} \right)}}\\ = \dfrac{{b.\left( {\sqrt {a + b} + \sqrt {a – b} } \right)}}{{2b}} – \dfrac{{c\left( {\sqrt {a + c} + \sqrt {a – c} } \right)}}{{2c}}\\ = \dfrac{{\sqrt {a + b} + \sqrt {a – b} }}{2} – \dfrac{{\sqrt {a + c} + \sqrt {a – c} }}{2}\\ = \dfrac{{\left( {\sqrt {a + b} – \sqrt {a + c} } \right) + \left( {\sqrt {a – b} – \sqrt {a – c} } \right)}}{2}\\ = \dfrac{{\dfrac{{\left( {\sqrt {a + b} – \sqrt {a + c} } \right)\left( {\sqrt {a + b} + \sqrt {a + c} } \right)}}{{\sqrt {a + b} + \sqrt {a + c} }} + \dfrac{{\left( {\sqrt {a – b} – \sqrt {a – c} } \right)\left( {\sqrt {a – b} + \sqrt {a – c} } \right)}}{{\sqrt {a – b} + \sqrt {a – c} }}}}{2}\\ = \dfrac{{\dfrac{{\left( {a + b} \right) – \left( {a + c} \right)}}{{\sqrt {a + b} + \sqrt {a + c} }} + \dfrac{{\left( {a – b} \right) – \left( {a – c} \right)}}{{\sqrt {a – b} + \sqrt {a – c} }}}}{2}\\ = \dfrac{1}{2}.\left[ {\dfrac{{b – c}}{{\sqrt {a + b} + \sqrt {a + c} }} + \dfrac{{ – b + c}}{{\sqrt {a – b} + \sqrt {a – c} }}} \right]\\ = \dfrac{1}{2}.\left( {b – c} \right).\left[ {\dfrac{1}{{\sqrt {a + b} + \sqrt {a + c} }} – \dfrac{1}{{\sqrt {a – b} + \sqrt {a – c} }}} \right]\\ \left. \begin{array}{l} \sqrt {a + b} > \sqrt {a – b} \\ \sqrt {a + c} > \sqrt {a – c} \end{array} \right\} \Rightarrow \sqrt {a + b} + \sqrt {a + c} > \sqrt {a – b} + \sqrt {a – c} \\ \Rightarrow \dfrac{1}{{\sqrt {a + b} + \sqrt {a + c} }} < \dfrac{1}{{\sqrt {a – b} + \sqrt {a – c} }}\\ a > b > c > 0 \Rightarrow \dfrac{1}{2}.\left( {b – c} \right).\left[ {\dfrac{1}{{\sqrt {a + b} + \sqrt {a + c} }} – \dfrac{1}{{\sqrt {a – b} + \sqrt {a – c} }}} \right] < 0\\ \Rightarrow \dfrac{b}{{\sqrt {a + b} – \sqrt {a – b} }} – \dfrac{c}{{\sqrt {a + c} – \sqrt {a – c} }} < 0\\ \Leftrightarrow \dfrac{b}{{\sqrt {a + b} – \sqrt {a – b} }} < \dfrac{c}{{\sqrt {a + c} – \sqrt {a – c} }} \end{array}$$