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August 23rd, 2019, 02:12 AM   #1
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Prove inequality

Prove: $\displaystyle e^{n^2} \left(1^1 \cdot 2^2 \cdot 3^3 \cdot ...\cdot n^n \right)\leq n^{n^2 + n + 6}$.

Last edited by skipjack; August 23rd, 2019 at 06:16 AM.
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August 23rd, 2019, 09:35 AM   #2
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Prove: $\displaystyle e^{n^2} \left(1^1 \cdot 2^2 \cdot 3^3 \cdot ...\cdot n^n \right)\leq n^{n^2 + n + 6}$.
For $\displaystyle n\neq 1$.
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August 23rd, 2019, 09:49 AM   #3
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$\displaystyle e^{n^2 }(\underbrace{n^n •...•n^n }_{n} )<n^{n^2 +n+ 6}$.
$\displaystyle (en)^{n^2 } < n^{n^2 } n^{n} n^{6} $.

$\displaystyle e^{n^2 } <n^{n+6}\; $ , how to continue from here ?
Seems like the inequality is not true !

Last edited by idontknow; August 23rd, 2019 at 09:58 AM.
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August 23rd, 2019, 09:49 AM   #4
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Originally Posted by idontknow View Post
For $\displaystyle n\neq 1$.
Sounds like a mathematical induction problem. Can you set that up?

-Dan
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August 23rd, 2019, 01:18 PM   #5
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I've closed this thread as the original poster realizes he didn't get it right.
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