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laskantaik
@laskantaik
August 2022
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Доказать, что если а≥0, b≥0, c≥0, то (a+b)(b+c)(a+c)≥8abc
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marusykaklimova
Используем неравенство Коши (a+b)/2≥√(ab) ⇒(a+b)≥2√(ab)
(b+c)/2≥√(bc) ⇒(b+c)≥2√(bc)
(a+c)/2≥√(ac) ⇒(a+c)≥2√(ac)
(a+b)(b+c)(a+c)≥8√(abbcac)
(a+b)(b+c)(a+c)≥8abc
3 votes
Thanks 10
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Answers & Comments
(b+c)/2≥√(bc) ⇒(b+c)≥2√(bc)
(a+c)/2≥√(ac) ⇒(a+c)≥2√(ac)
(a+b)(b+c)(a+c)≥8√(abbcac)
(a+b)(b+c)(a+c)≥8abc