A) positive
B) negative
C) non-positive
D) non-negative
Correct Answer: B
Solution :
Since,\[AM>GM\] \[\therefore \]\[\frac{(b+c-a)+(c+a-b)}{2}\] \[>{{[b+c-a(c+a-b)]}^{1/2}}\] \[\Rightarrow \]\[c>{{[b+c-a(c+a-b)]}^{1/2}}\] ?(i) Similarly. \[b>[{{(a+b-c(b+c-a)]}^{1/2}}\] ?(ii) and\[a>{{[(a+b-c)(c+a-b)]}^{1/2}}\] ?(iii) On multiplying Eqs. (i), (ii) and (iii), we get \[abc>(a+b-c)(b+c-a)(c+a-b)\]Hence, \[(a+b-c)(b+c-a)(c+a-b)-abc<0\]You need to login to perform this action.
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