A) \[{{a}^{2}}+{{c}^{2}}>{{b}^{2}}+{{d}^{2}}\]
B) \[{{a}^{2}}+{{d}^{2}}>{{b}^{2}}+{{c}^{2}}\]
C) \[ac+bd>{{b}^{2}}+{{c}^{2}}\]
D) \[ac+bd>{{b}^{2}}+{{d}^{2}}\]
Correct Answer: C
Solution :
[c] If a, b, c and d are in H.P., then \[\frac{1}{a};\frac{1}{b},\frac{1}{c}\]and \[\frac{1}{d}\]will be in A.P. Therefore, \[\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}=\frac{1}{d}-\frac{1}{c}\] \[\Rightarrow b=\frac{2ac}{a+c}\] G.M. between a and c=\[\sqrt{ac}\] Now, since G.M.>H.M., we have \[\sqrt{ac}>b\] Or \[ac>{{b}^{2}}\] ...(1) Similarly, \[\sqrt{bd}>c\] Or \[bd>{{c}^{2}}\] ...(2) Adding (1) and (2), we get \[ac+bd>{{b}^{2}}+{{c}^{2}}\]
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