If p, q, r are three positive integers such that, 27pqr ≥  (p + q + r)³ and 2p + q + r = 4 then 

Concept:

AM, GM, HM Formulas:

If A is the arithmetic mean 

⇔ \({\rm{A}} = \frac{{{\rm{a\;}} + {\rm{\;b}}}}{2}\)

If G is the geometric mean 

⇔ \({\rm{G}} = \sqrt {{\rm{ab}}} \)

Relation between AM, and GM

AM  ≥  GM

Calculation:

Given that,

2p + q + r = 4    ----(1)

We know that,

AM  ≥  GM

\(\frac{a+b+c}{3}\ge ({abc})^{\frac{1}{3}}\)

\(\frac{(a+b+c)^3}{3^3}\ge ({abc})\)

⇒ 27pqr ≥  (p + q + r)3 

This will be possible only if 

p = q = r

Hence, from equation (1)

2p + p + p = 4

⇒ 4p = 4 or p = 1

⇒ p = q = r = 1