(a-6)(b+3)(c+2)≥48√abc

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вопрос

To solve this inequality, we can expand the left side of the equation:

(a-6)(b+3)(c+2) = abc + 2ab - 6ac - 12b + 3c + 18

Now the inequality is:

abc + 2ab - 6ac - 12b + 3c + 18 ≥ 48√abc

Let's square both sides of the inequality to get rid of the square root:

(abc + 2ab - 6ac - 12b + 3c + 18)^2 ≥ (48√abc)^2
(abc + 2ab - 6ac - 12b + 3c + 18)^2 ≥ 2304abc

Expanding the left side of the inequality:

(a^2b^2c^2) + (4a^2b^2c) - (12a^2bc^2) - (24ab^2c) + (6a^2bc) + (36a^2c) + (4ab^2c) + (12abc^2) - (36ab) - (48ac^2) - (72bc) + (9ac) + 108bc + 324 ≥ 2304abc

Now we combine like terms on the left side:

a^2b^2c^2 + 4a^2b^2c + 6a^2bc + 4ab^2c - 12a^2bc^2 - 24ab^2c - 36ab - 72bc + 3c + 9ac + 36a^2c + 12abc^2 + 12abc^2 - 48ac^2 ≥ 2304abc - 324

Simplify further:

a^2b^2c^2 + 10a^2b^2c + 6a^2bc + 4ab^2c - 12a^2bc^2 - 24ab^2c - 36ab - 72bc + 3c + 9ac + 36a^2c + 24abc^2 - 48ac^2 ≥ 2304abc - 324

Now we have our inequality in expanded form.