A) Positive
B) Negative
C) Non-positive
D) Non-negative
Correct Answer: B
Solution :
[b] \[(b+c-a)(c+a-b)(a+b-c)-abc\] |
Without loss of generality we can assume, |
\[a>b>c\] |
Applying A.M, G.M. pair wise |
\[(b+c)>2\sqrt{bc}\] (i) |
\[(a+c)>2\sqrt{ac}\] (ii) |
\[(a+b)>2\sqrt{(ac)}\] (iii) |
Multiplying equation (i), (ii), (iii), we get \[(a+b)(b+c)(c+a)>8abc\] |
Let us put \[b+c=2p;c+a=2q;a+b=2r\] |
\[\Rightarrow a=-p+r+q;b=p-q+r;c=p+q-r\] |
\[\Rightarrow 2p.2q.2r\ge r\ge 8(q+r+p);(p+r-q);(p+q-r)\] |
\[\Rightarrow pqr\ge (q+r-p)(p+r-q)(p+r-r)\] |
Replacing p, q, r by a b c |
Without loss of generality, we get |
\[abc\ge (a+b-c)(b+c-a)(c+a-b)\] |
\[\Rightarrow \] The required expression is always negative |
You need to login to perform this action.
You will be redirected in
3 sec