A) Positive
B) negative
C) Non-positive
D) non- negative
Correct Answer: B
Solution :
| Given that \[a,b,c\]are distinct \[+ve~\]numbers. The expression whose sign is to be checked is\[(b+c-a)(c+a-b)(a+b-c)-abc.\] |
| As this expression is symmetric in \[a,b,c\]without loss of generality, we can assume that \[a<b<c.\] |
| Then \[c-a=+ve\]and\[c-b=+ve\] |
| \[\therefore \]\[b+c-a=+ve\]and \[c+a-b=+ve\] |
| But \[a+b-c\]may be + ve or - ve. |
| Case I: If \[a+b-c=+\]ve then we can say that \[a,b,c\] are such that sum of any two is greater than the 3rd. consider | ||
| \[x=a+b-c,\] | \[y=b+c-a,\] | \[z=c+a-b\] |
| Then \[x,y,z\] all are \[+ve\] | ||
| And then \[a=\frac{x+z}{2},b=\frac{y+x}{2},c=\frac{z+y}{2}\] | ||
| Now we know that A.M.>G.M. for distinct real numbers | ||
| \[\therefore \frac{x+y}{2}>\sqrt{xy,}\frac{y+z}{2}>\sqrt{yz,}\frac{z+x}{2}>\sqrt{zx}\] | ||
| \[\Rightarrow \]\[\left( \frac{x+y}{2} \right)\left( \frac{y+z}{2} \right)\left( \frac{z+x}{2} \right)>xyz\] | ||
| \[\Rightarrow \]\[abc>(a+b-c)(b+c-a)(c+a-b)\]\[\Rightarrow \]\[(b+c-a)(c+a-b)(a+b-c)-abc<0\] | ||
| Case II: If \[a+b-c=-\operatorname{ve}\,\]then |
| \[(b+c-a)(c+a-b)(a+b-c)-abc\] |
| \[=(+ve)(+ve)(-ve)-(+ve)\] |
| \[=(-ve)-(+ve)=(-ve)\] |
| \[\Rightarrow (b+c-a)(c+a-b)(a+b-c)<abc\] |
| Hence in either case given expression is -ve. |
You need to login to perform this action.
You will be redirected in
3 sec
