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If `a ,b ,c`are the sides of a triangle, then the minimum value of `a/(b+c-a)+b/(c+a-b)+c/(a+b-c)`is equal to`3``6``9``12`A. 3B. 6C. 9D. 12 |
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Answer» Correct Answer - A `2E = (2a)/(b + c - a) + (2b)/(c + a - b) + (2c)/(a + b - c)` `= (2a)/(b + c + a) + 1 + (2b)/(c + a + b) + 1 + (2c)/(a + b -c) + 1 - 3` `= (a + b + c) ((1)/(b + c -a) + (1)/(c + a - b) + (1)/(a + b - c)) - 3` Using A.M `ge` H.M we have `((1)/(b + c - a) + (1)/(c + a - b) + (1)/(a + b -c))/(3) ge (3)/(a + b + c)` Or `(a + b+ c) ((1)/(b + c -a)+(1)/(c+a+b) + (1)/(a + b+ c)) ge 9` or `(a + b+ c) ((1)/(b + c -a) + (1)/(c + a - b) + (1)/(a + b - c)) - 3 ge 6` `implies E ge 3` |
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