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October 7th, 2019, 11:27 PM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 
Prove that $\displaystyle a^b < b^a $, for $\displaystyle a>b\geq 3$.
Last edited by skipjack; October 9th, 2019 at 01:33 AM. 
October 8th, 2019, 01:19 AM  #2 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle 
Let $a=b \times k$, where $k>1$. $$(bk)^b < b^{bk} \Rightarrow bk<b^k \Rightarrow k< b^{k1}$$ 
October 8th, 2019, 01:53 AM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 
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} $. $\displaystyle b^{1/b} <b^b \: \Rightarrow \: b\geq 3>1/b$. 
October 8th, 2019, 02:46 AM  #4 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle  
October 8th, 2019, 03:15 AM  #5 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 
What do you mean ? 
October 8th, 2019, 04:43 PM  #6 
Senior Member Joined: Sep 2016 From: USA Posts: 670 Thanks: 440 Math Focus: Dynamical systems, analytic function theory, numerics  
October 9th, 2019, 06:49 AM  #7 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle  

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calculus, inequality, prove 
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