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If `xyz = 1` and `x, y, z gt 0` then the minimum value of the expression `(x+2y)(y+2z)(z+2x)` isA. `81`B. `9`C. `18`D. `27` |
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Answer» Correct Answer - D If `xyz = 1` and ………. apply `AM ge GM` `(x+y+y)/(3)ge(xy^(2))^((1)/(3))` ….(i) `(y+z+z)/(3)ge (yz^(2))^((1)/(3))` ….(ii) `(z+x+x)/(3)ge (zx^(2))^((1)/(3))` ….(iii) From equation (i), (ii), and (iii) `((x+2y)/(3))((y+2z)/(3))((z+2x)/(3))ge xyz` `implies (x+2y) (y+2z) (z+2x) ge 27` |
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