sailorlyts14eh

sailorlyts14eh

Answered

2022-09-30

Assume n 3 , c 1 , d 1 are natural numbers such that c ² + d ² ( n 1 ) c d < 0. Show that ( n ³ n + 1 ) c ² + ( n + 1 ) d ² ( n ² + n 1 ) c d > 1

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2022-10-01Added 6 answers

Let c , d be positive real numbers and n > 1 (esp., n 3 n + 1 > 0). By the arithmetic-geometric inequality
( n 3 n + 1 ) c 2 + ( n + 1 ) d 2 2 ( n 3 n + 1 ) ( n + 1 ) c d .
One checks by multiplying out that
4 ( n 3 n + 1 ) ( n + 1 ) = ( n 2 + n 1 ) 2 + 3 + 3 n 2 ( n 2 1 ) + 2 n ( n 2 + 1 ) ,
hence 2 ( n 3 n + 1 ) ( n + 1 ) > n 2 + n 1 and finally
( n 3 n + 1 ) c 2 + ( n + 1 ) d 2 > ( n 2 + n 1 ) c d .

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