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If `(log)_2(a+b)+(log)_2(c+d)geq4.`Then find the minimum value of the expression `a+b+c+ddot` |
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Answer» `log_(2) (a + b) + log_(2) (c + d) ge 4` `:. Log_(2) {(a + b) (c + d)} ge 4` `:. (1 + b) (c + d) ge 2^(4)` Now `((a + b)+(c + d))/(2) ge sqrt((a + b)(c + d)) ge 2^(2)` or `a + b + c + d ge 8` Hence, the minimum value of a + b + c + d is 8 |
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