Question
Let p, q, r 
R+ and 27 pqr ≥ (p + q + r)3 and 3p + 4q + 5r = 12 then p3 + q4 + r5 is equal to -
- 2
- 6
- 3
- none of these
Hint:
apply the property of AM GM and solve
The correct answer is: 3
3
Given, p,q.r > 0
Therefore, we can use the property of AM>=GM
=> (p+q+r)/3 >= (pqr)^(1/3)
Cubing both sides we get
(p+q+r)3 /27>=pqr
(p+q+r)3 >=27pqr
Given, (p+q+r)3 <=27pqr
Therefore, (p+q+r)3 = 27pqr
This is only possible when p=q=r=1.
Therefore, p3+q4+r5=1+1+1
= 3
for real and positive numbers, we can use the property AM>=GM
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