My Math Forum Prove inequality without calculus

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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 $k>1$. $$(bk)^b < b^{bk} \Rightarrow bk  October 8th, 2019, 01:53 AM #3 Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 I done it this way : \displaystyle b^{a^b } < b^{b^a } . \displaystyle b^{1/b^a }1/b. October 8th, 2019, 02:46 AM #4 Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle Quote:  Originally Posted by idontknow I done it this way : \displaystyle b^{a^b } < b^{b^a } . \displaystyle b^{1/b^a }1/b. Sth doesn't seem right.(?)  October 8th, 2019, 03:15 AM #5 Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 What do you mean ? October 8th, 2019, 04:43 PM #6 Senior Member Joined: Sep 2016 From: USA Posts: 670 Thanks: 440 Math Focus: Dynamical systems, analytic function theory, numerics Quote:  Originally Posted by tahirimanov19 Let a=b \times k, where k>1.$$(bk)^b < b^{bk} \Rightarrow bk
This proof doesn't work. You proved the converse of the result you wanted to prove.

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
Quote:
 Originally Posted by SDK 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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