
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
April 12th, 2019, 06:26 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91  Inequality proof with two variables
Prove inequality without calculus. Given two natural numbers $\displaystyle x,y$ and $\displaystyle x^y > y^x \; \;$, show that $\displaystyle x<y$. Last edited by skipjack; April 14th, 2019 at 07:54 PM. 
April 12th, 2019, 11:00 AM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 
My approach is to find the equivalent function that satisfies the relation $\displaystyle x^y > y^x$ . That function can be as $\displaystyle F(t)=t^{\frac{1}{t}} \; $ , $\displaystyle t\in \mathbb{N}$ , where $\displaystyle F(x)>F(y)$ . The easiest possible way to prove it is to show that whether exists t such that $\displaystyle F(t)>F(t+1) $ , which gives $\displaystyle t>(1+\frac{1}{t} )^{t}$. (a) If $\displaystyle t>(1+1/t )^{t}$ and t has a lower bound then $\displaystyle x<y$ is proven . How to continue from (a) ? 
April 14th, 2019, 01:46 AM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 
If it is allowed : If $\displaystyle t$ must have a lower bound then $\displaystyle (1+1/t)^{t} $ must have an upper bound which is $\displaystyle \lim_{t\rightarrow \infty } (1+1/t)^{t}=e$. The nearest integer greater than e is 3 , so $\displaystyle 3\leq x <y$. Since the limit above converges it proves $\displaystyle x<y$ but also shows that $\displaystyle 3\leq x$. Last edited by idontknow; April 14th, 2019 at 02:00 AM. 
April 14th, 2019, 01:01 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,821 Thanks: 723 
$x^y\gt y^x$ is equivalent to $y\ln x \gt x\ln y$ or $\frac{y}{\ln y}\gt \frac{x}{\ln x}$ or $\frac{x}{\ln x}$ is an increasing function of $x$. This is straightforward.
Last edited by skipjack; April 14th, 2019 at 07:46 PM. 
April 16th, 2019, 06:30 PM  #5  
Senior Member Joined: Jun 2014 From: USA Posts: 564 Thanks: 43  Quote:
Then $5^1 > 1^5$ where $5 > 1$ is a counterexample. So is every other $(x,y)$ where $x > 1$ and $y = 1$. Finally, don't forget about $x = 3$ and $y = 2$. Then $3^2 > 2^3$, where $3 > 2$, is yet another counterexample. These are all of the counterexamples. Do these show up in the formula mathman gave?  
April 17th, 2019, 12:58 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,821 Thanks: 723 
My mistake x/ln(x) is not an increasing function of x. It has a minimum at x=e, decreasing for 0 < x < e and increasing thereafter.
Last edited by mathman; April 17th, 2019 at 01:00 PM. 

Tags 
inequality, proof, variables 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Inequality on three variables  Pell's fish  Real Analysis  1  July 7th, 2014 04:34 AM 
two variables polynomial inequality  ninhead  Algebra  7  October 31st, 2013 08:50 AM 
Inequality with radamacher variables  Anton29  Advanced Statistics  0  February 2nd, 2012 06:46 PM 
Inequality with two variables  bo1989  Algebra  12  March 12th, 2011 06:09 AM 
two variables polynomial inequality  ninhead  Abstract Algebra  2  December 31st, 1969 04:00 PM 