Formula q_868 (Q675): Difference between revisions

Latest revision as of 13:25, 20 April 2020

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English	Formula q_868	No description defined

q_868

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3+{\big (}{\frac {a}{b}}+{\frac {b}{a}}{\big )}+{\big (}{\frac {a}{c}}+{\frac {c}{a}}{\big )}+{\big (}{\frac {b}{c}}+{\frac {c}{b}}{\big )}\geq 3+2+2+2=9=3^{2}\ \blacksquare

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Revision as of 13:25, 20 April 2020 SchuBot (talk \| contribs) SchuBot, Bots, Bureaucrats, flow-bot, Suppressors, Administrators 4,665 edits ‎Created claim: arqmath formula id (P8): q_868 ← Older edit	Latest revision as of 13:25, 20 April 2020 SchuBot (talk \| contribs) SchuBot, Bots, Bureaucrats, flow-bot, Suppressors, Administrators 4,665 edits ‎Created claim: Defining formula (P1): 3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) \ge 3 + 2 + 2 +2 = 9 = 3^2 \ \blacksquare
	Property / Defining formula
		$3+{\big (}{\frac {a}{b}}+{\frac {b}{a}}{\big )}+{\big (}{\frac {a}{c}}+{\frac {c}{a}}{\big )}+{\big (}{\frac {b}{c}}+{\frac {c}{b}}{\big )}\geq 3+2+2+2=9=3^{2}\ \blacksquare$ `3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) \ge 3 + 2 + 2 +2 = 9 = 3^2 \ \blacksquare`
	Property / Defining formula: $3+{\big (}{\frac {a}{b}}+{\frac {b}{a}}{\big )}+{\big (}{\frac {a}{c}}+{\frac {c}{a}}{\big )}+{\big (}{\frac {b}{c}}+{\frac {c}{b}}{\big )}\geq 3+2+2+2=9=3^{2}\ \blacksquare$ / rank
		Normal rank