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October 7th, 2019, 11:27 PM   #1
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Prove that $\displaystyle a^b < b^a $, for $\displaystyle a>b\geq 3$.

Last edited by skipjack; October 9th, 2019 at 01:33 AM.
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October 8th, 2019, 01:19 AM   #2
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Let $a=b \times k$, where $k>1$.

$$(bk)^b < b^{bk} \Rightarrow bk<b^k \Rightarrow k< b^{k-1}$$
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October 8th, 2019, 01:53 AM   #3
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I done it this way : $\displaystyle b^{a^b } < b^{b^a }$ .
$\displaystyle b^{1/b^a }<b^{1/a^b }$ or $\displaystyle (b^{1/b})^{a} <(b^{1/a })^{b} $.
$\displaystyle b^{1/b} <b^b \: \Rightarrow \: b\geq 3>1/b$.
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October 8th, 2019, 02:46 AM   #4
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Originally Posted by idontknow View Post
I done it this way : $\displaystyle b^{a^b } < b^{b^a }$ .
$\displaystyle b^{1/b^a }<b^{1/a^b }$ or $\displaystyle (b^{1/b})^{a} <(b^{1/a })^{b} $.
$\displaystyle b^{1/b} <b^b \: \Rightarrow \: b\geq 3>1/b$.
Sth doesn't seem right.(?)
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October 8th, 2019, 03:15 AM   #5
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What do you mean ?
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October 8th, 2019, 04:43 PM   #6
SDK
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Originally Posted by tahirimanov19 View Post
Let $a=b \times k$, where $k>1$.

$$(bk)^b < b^{bk} \Rightarrow bk<b^k \Rightarrow k< b^{k-1}$$
This proof doesn't work. You proved the converse of the result you wanted to prove.
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October 9th, 2019, 06:49 AM   #7
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This proof doesn't work. You proved the converse of the result you wanted to prove.
It wasn't proof, just simplification...
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