Sequence And Series Question 430
Question: If a, b, c are the sides of a triangle, then the minimum value of $ \frac{a}{b+c-a}+\frac{b}{c+a-b}+ $ $ \frac{c}{a+b-c} $ is equal to
Options:
A) 3
B) 6
C) 9
D) 12
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Given expression is $ \frac{1}{2}\sum{\frac{2a}{b+c-a}} $ $ =\frac{1}{2}\sum{( \frac{2a}{b+c-a}+1 )}-\frac{3}{2}=\frac{1}{2}(a+b+c) $ $ \sum{\frac{1}{b+c-a}}-\frac{3}{2} $ Now, as $ (a+b+c)=\Sigma (b+c-a) $ Applying $ A.M.\ge H.M. $ Minimum value of the expression $ =\frac{1}{2}\times 9-\frac{3}{2}=3. $
BETA
