Equations And Inequalities Question 11
If a, b, c are distinct positive real numbers, then the expression $ ( b+c-a )( c+a-b )( a+b-c )-abc $ is
Options:
A) Positive (No change required)
B) Negative
C) Non-positive
D) Non-negative
Show Answer
Answer:
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. pairwise $ (b+c)>2\sqrt{bc} $ (i) $ (a+c)>2\sqrt{ac} $ (ii) $ (a+b)^2>4ac $ (iii) Multiplying equation (i), (ii), (iii), we get $ (a+b)(b+c)(c+a) \geq 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-q) $ Replacing p, q, r by a b c Without loss of generality, we obtain $ abc\ge (a+b-c)(b+c-a)(c+a-b) $
$ \Rightarrow $ The required expression is always positive
BETA
