A) 1
B) 2
C) 3
D) 2009
Correct Answer: C
Solution :
We have, \[{{x}^{2006}}\,{{y}^{2007}}\,{{z}^{2008}},\,{{x}^{2007}}\,{{y}^{2008}}\,{{z}^{2009}},\,{{x}^{2008}}\,{{y}^{2009}}{{z}^{2010}}\]are in AP. So, \[1,\,xyz,\,{{x}^{2}}{{y}^{2}}{{z}^{2}}\,\] \[\Rightarrow \] \[2xyz=1+{{x}^{2}}{{y}^{2}}{{z}^{2}}\] \[\Rightarrow \] \[1+{{x}^{2}}{{y}^{2}}{{z}^{2}}-2xyz=0\] \[\Rightarrow \] \[2xyz=1+{{x}^{2}}{{y}^{2}}{{z}^{2}}\] \[\Rightarrow \] \[xyz=1\] \[\because \] \[AM\ge GM\] \[\therefore \] \[\frac{x+y+z}{3}\ge \,{{(xyz)}^{1/3}}=1\] \[\Rightarrow \] \[x+y+z\ge 3\] \[\therefore \] \[x+y+z\ge 3\]
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