Question

Who wants my 20 points..I need full explanation..I will report if you spam...if a b c are all positive then the minimum value of the expression (a2+a+1)(b2+b+1)(c2+c+1)/9abc​

Answer 1

Answer:

Assuming that a, b, c are positive.

It is a well know fact that A.M≥H.M.A.M≥H.M.

Applying this to a,b,ca,b,c

a+b+c3≥3(1a+1b+1c)a+b+c3≥3(1a+1b+1c)

Cross multiplying gives that

(a+b+c)(1a+1b+1c)≥9(a+b+c)(1a+1b+1c)≥9

Thus, the minimum value is 99, when a=b=c=1a=b=c=1

Or you could also do:

By AM-GM

(a+b+c)(1a+1b+1c)(a+b+c)(1a+1b+1c)

≥3(abc)13≥3(abc)13⋅3(1abc)13⋅3(1abc)13

≥9≥9

Equality occurs again when a=b=ca=b=c

Or by Cauchy-Schwarz

(a+b+c)(1a+1

Step-by-step explanation:

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