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Question

Statement-I : If 27 abc ≥ (a + b + c)3 and 3a + 4b + 5c = 12 then 1 over a squared plus 1 over b cubed plus 1 over c to the power of 5 equals 10 where a, b, c are positive real numbers.
Statement-II : For positive real numbers A.M. ≥ G.M.

  1. Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.    
  2. Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I.    
  3. Statement-I is true, Statement-II is false.    
  4. Statement-I is false, Statement-II is true.    

hintHint:

apply the AM >= GM rule and solve for a b c

The correct answer is: Statement-I is false, Statement-II is true.



    Statement-I is false, Statement-II is true.

    Given, a,b,c are positive real numbers
    =>  AM>= GM can be applied to these numbers.
    => (a+b+c)/3>=(abc)^(1/3)
    On solving we get
    (a+b+c)^3>=27abc
    From the problem, we know that (a+b+c)^3<=27abc

    Therefore, 27 abc = (a+b+c)^3
    Or, AM= GM. This happens only when a=b=c.
    Therefore, 3a+4b+5c=12
    =>  3a+4a+5a=12 => a=1=b=c
    Therefore, the required answer is 1/1 + 1/1+1/1 = 3, which is not equal to 10. Hence, statement 1 is false

    a,b,c are positive real numbers
    =>  AM>= GM can be applied to these numbers
    for real and positive numbers, we can use the property AM>=GM

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