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Senpai908

Доказать неравенство: (a+6)(b+3)(c+2) ≥ 48√(abc), если a, b, c ≥ 0

Доказательство:

Для неотрицательных a,b,c применим неравенство Коши

a+6geq2sqrt{6a}\ b+3geq 2sqrt{3b}\ c+2geq 2sqrt{2c}

Перемножив все эти три неравенства, мы получаем

(a+6)(b+3)(c+2)geq 2sqrt{6a}cdot 2sqrt{3b}cdot 2sqrt{2c}=8cdot 6sqrt{abc}=48sqrt{abc}

Доказано.

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