Formula q_870 (Q677): Difference between revisions

Latest revision as of 13:25, 20 April 2020

No description defined

Language	Label	Description	Also known as
English	Formula q_870	No description defined

q_870

0 references

P(n)::{\big (}\sum _{i=1}^{n}a_{i}{\big )}{\big (}\sum _{i=1}^{n}{\frac {1}{a_{i}}}{\big )}\geq n^{2}

0 references

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_870 ← 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): P(n)::\big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big) \ge n^2
	Property / Defining formula
		$P(n)::{\big (}\sum _{i=1}^{n}a_{i}{\big )}{\big (}\sum _{i=1}^{n}{\frac {1}{a_{i}}}{\big )}\geq n^{2}$ `P(n)::\big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big) \ge n^2`
	Property / Defining formula: $P(n)::{\big (}\sum _{i=1}^{n}a_{i}{\big )}{\big (}\sum _{i=1}^{n}{\frac {1}{a_{i}}}{\big )}\geq n^{2}$ / rank
		Normal rank