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so i had a question that i have managed to simplify this far, but i cant quite get it down far enough.

given that a*b*c=27, and that a,b and c are negative real numbers, find the minimum value for (1-a)(1-b)(1-c) i believe the minimum is when all of a,b,and are at -3 to give a minuimum value of 64, but i cant prove it. we learned the AM GM inequality in class so maybe that can help?

Jun 5, 2021

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nevermind i got it! for those of you that had the same or similar question, use AM GM on each of the (1-a)(1-b)(1-c) terms and set that to be less than or equal to the original, the answer will work itself out from there.

Jun 5, 2021
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nevermind, while i think im on the right track this shows it is >= 24sqrt3, but i cannot find any values that even get close to 24sqrt3 so i still havent proved it

Guest Jun 5, 2021
edited by Guest  Jun 5, 2021
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For this response, I will assume that abc = -27 instead of 27, as it is impossible for 3 negative numbers to multiply to a positive number

The minimum value is 64

Expanding (1-a)(1-b)(1-c), we have −abc+ab+ac+bc−a−b−c+1

Thus, to find what the minimum value for the desired expression is, we have to calculate the minimum of -a-b-c, and the minimum of ab+ac+bc.

Since a, b, and c are negative, we cannot use the AM-GM inequality for 3 variables on it, but after multiplying each of the variables by -1, making them positive, we have $$\frac{-a-b-c}{3}≥ \sqrt[3]{-abc}$$, simplifying to $$-(a+b+c)≥9$$, so the minimum value of -(a+b+c) is 9.

Since a, b, c are negative, ab, ca, and bc must all be positive, hence allowing for an easy AM-GM application.  $$\frac{ab+ac+bc}{3}≥\sqrt[3]{abc^2}$$, simplifying to $$ab+ac+bc≥ 27$$.  Both -(a+b+c) and ab+ac+bc reach the minimum value when the equality condition of the AM-GM inequality is met, namely, when -a=-b=-c, or when a=b=c, and when ab=ac=bc.  Both of these simplify to a=b=c.  Since abc=-27, a=b=c=-3.  Thus, the minimum value of −abc+ab+ac+bc−a−b−c+1 occurs when a=b=c=-3, rendering the minimum value of (1-a)(1-b)(1-c) as 64, as you assumed.