A) 3
B) 6
C) 9
D) 12
Correct Answer: A
Solution :
[a] Given expression is \[\frac{1}{2}\sum{\frac{2a}{b+c-a}}\] \[=\frac{1}{2}\sum{\left( \frac{2a}{b+c-a}+1 \right)}-\frac{3}{2}=\frac{1}{2}(a+b+c)\] \[\sum{\frac{1}{b+c-a}}-\frac{3}{2}\] Now, as \[(a+b+c)=\Sigma (b+c-a)\] Applying \[A.M.\ge H.M.\] Minimum value of the expression \[=\frac{1}{2}\times 9-\frac{3}{2}=3.\]
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