Let p,q,r∈R+ such that 27pqr≥(p+q+r)3 and 3p+4q+5r=12. Then the value of p+q+r is
A
3
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B
6
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C
2
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D
12
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Solution
The correct option is A3 Since p,q,r∈R+
Applying A.M. ≥ G.M. ⇒p+q+r3≥(pqr)1/3 ⇒(p+q+r)3≥27pqr
But given, (p+q+r)3≤27pqr
which holds good if (p+q+r)3=27pqr
i.e., A.M. = G.M. ⇒p=q=r