It's easy to prove that following upperbound is true: sum_(i=1)^N a_i ln a_i <= A \ln A, where sum_(i=1)^N a_i=A and a_i>=1 I'm wondering, is there stronger upperbound?

HypeMyday3m 2022-09-24 Answered
Upperbound for i = 1 N a i ln a i
It's easy to prove that following upperbound is true:
i = 1 N a i ln a i A ln A, where i = 1 N a i = A and a i 1
I'm wondering, is there stronger upperbound?
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Answers (1)

Katelyn Ryan
Answered 2022-09-25 Author has 11 answers
You can also use the following estimate
j = 1 N a j ln a j A ln ( A N + 1 ) .
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