1.

If `(log)_2(a+b)+(log)_2(c+d)geq4.`Then find the minimum value of the expression `a+b+c+ddot`

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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