A) positive
B) negative
C) zero
D) None of these
Correct Answer: B
Solution :
Let\[x=b+c-a,y=c+a-b\]and\[z=a+b-c\] Since, a, b, c are different numbers. \[a=\frac{y+z}{2}>\sqrt{yz}\] \[b=\frac{x+z}{2}>\sqrt{xz}\] \[c=\frac{x+y}{2}>\sqrt{xy}\] \[\therefore \] \[abc>\sqrt{yz}.\sqrt{xz}.\sqrt{xy}\] \[\Rightarrow \] \[abc>xyz\] \[\Rightarrow \]\[(b+c-a)(c+a-b)(a+b-c)-abc<0\] \[\therefore \]The value of\[(b+c-a)(c+a-b)(a+b-c)\] \[-abc\] will be negative.
You need to login to perform this action.
You will be redirected in
3 sec
