Question
Statement-I : If 27 abc ≥ (a + b + c)3 and 3a + 4b + 5c = 12 then where a, b, c are positive real numbers.
Statement-II : For positive real numbers A.M. ≥ G.M.
- Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.
- Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I.
- Statement-I is true, Statement-II is false.
- Statement-I is false, Statement-II is true.
Hint:
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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