A) 0
B) 1
C) 2
D) 3
Correct Answer: D
Solution :
We know, that \[AM\ge GM\] \[\therefore \] \[\frac{{{x}^{\log y-\log z}}+{{y}^{\log z-\log x}}+{{z}^{\log x-\log y}}}{3}\ge \] \[\sqrt[3]{{{x}^{\log y-\log z}}.{{y}^{\log z-\log x}}.{{z}^{\log x-\log y}}}\] ?.(i) Now, \[\log \{{{x}^{\log y-\log z}}.{{y}^{\log z-\log x}}.{{z}^{\log x-\log y}}\}\] \[=\log \{{{x}^{\log y}}.{{x}^{-\log z}}.{{y}^{\log z}}.{{z}^{-\log x}}.{{z}^{\log x}}.{{z}^{-\log y}}\}\] \[=\log ({{x}^{\log y}})+\log ({{x}^{-\log z}})+\log ({{y}^{\log z}})\] \[+\log ({{y}^{-\log x}})+\log ({{z}^{\log x}})+\log ({{z}^{-\log y}})\] \[=\log y\log x-\log z\log x+\log z\log y\] \[-\log x\log y+\log x\log z-\log y\log z\] \[=0\] \[\Rightarrow \] \[{{x}^{\log y-\log z}}+{{y}^{\log z-\log x}}+{{z}^{\log x-\log y}}\ge 3\] \[\therefore \]FromEq. (i), \[{{x}^{\log y-\log z}}+{{y}^{\log z-\log x}}+{{z}^{\log x-\log y}}\ge 3\] Hence, minimum value is 3.
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