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What is the minimum value of the expression (p^2 p 1)(q^2 q 1)(r^2 r 1)÷pqr, where p,q&r are all positive?
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What is the minimum value of the expression (p^2 p 1)(q^2 q 1)(r^2 r 1...
The given expression can be simplified as follows:

(p^2 p 1)(q^2 q 1)(r^2 r 1) ÷ pqr

= (p^2 + p + 1)(q^2 + q + 1)(r^2 + r + 1) ÷ pqr

= [(p^2 + p + 1)q^2 + (p^2 + p + 1)q + (p^2 + p + 1)](r^2 + r + 1) ÷ pqr

= [(p^2q^2 + pq^2 + q^2) + (p^2q + pq + q) + (p^2 + p + 1)](r^2 + r + 1) ÷ pqr

= [p^2q^2 + pq^2 + q^2 + p^2q + pq + q + p^2 + p + 1](r^2 + r + 1) ÷ pqr

= [p^2q^2 + p^2q + pq^2 + pq + p^2 + q^2 + q + p + 1](r^2 + r + 1) ÷ pqr

= [(p^2q^2 + p^2q + pq^2) + (pq + p^2 + q^2) + (p + q + 1)](r^2 + r + 1) ÷ pqr

= [(pq(p + q) + pq + (p + q)^2) + (p + q + 1)](r^2 + r + 1) ÷ pqr

= [(pq(p + q) + (p + q)^2) + (pq + p + q + 1)](r^2 + r + 1) ÷ pqr

= [(pq(p + q) + (p + q)^2) + (p + q + 1)](r^2 + r + 1) ÷ pqr

= [(p + q)(pq + (p + q)) + (p + q + 1)](r^2 + r + 1) ÷ pqr

= [(p + q)(p + q + 1)](r^2 + r + 1) ÷ pqr

Now, to find the minimum value of this expression, we need to minimize each factor.

The minimum value of (p + q) is 2√(pq) by the AM-GM inequality.

The minimum value of (p + q + 1) is 2√(pq) + 1 by adding 1 to the minimum value of (p + q).

The minimum value of (r^2 + r + 1) is 1 by completing the square or observing that it represents a quadratic with no real roots.

The minimum value of pqr is 2√(pqr) by the AM-GM inequality.

Therefore, the minimum value of the given expression is:

[(2√(pq))(2√(pq) + 1)](1) ÷ (2√(pqr))

= (4pq + 2√(pq))(
Community Answer
What is the minimum value of the expression (p^2 p 1)(q^2 q 1)(r^2 r 1...
If it is p square + P +1 then The answer is (2+1)(2+1)(2+1) which is 27 from AM>=GM
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What is the minimum value of the expression (p^2 p 1)(q^2 q 1)(r^2 r 1)÷pqr, where p,q&r are all positive?
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