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 :
\[a,\ b,\ c,\ d\] be in H.P., then \[\frac{1}{a},\ \frac{1}{b},\ \frac{1}{c},\ \frac{1}{d}\] 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 as \[G.M>H.M\]., so here \[\sqrt{ac}>b\] or \[ac>{{b}^{2}}\]. Similarly \[\sqrt{bd}>c\] or \[bd>{{c}^{2}}\] Adding, \[ac+bd>{{b}^{2}}+{{c}^{2}}\]
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