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 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 $\displaystyle x  April 12th, 2019, 11:00 AM #2 Senior Member Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 My approach is to find the equivalent function that satisfies the relation$\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
 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 $\displaystyle 3\leq x  April 14th, 2019, 01:01 PM #4 Global Moderator Joined: May 2007 Posts: 6,821 Thanks: 723$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$. This is straightforward. Thanks from topsquark and idontknow Last edited by skipjack; April 14th, 2019 at 07:46 PM. April 16th, 2019, 06:30 PM #5 Senior Member Joined: Jun 2014 From: USA Posts: 564 Thanks: 43 Quote:  Originally Posted by idontknow Prove inequality without calculus. Given two natural numbers$\displaystyle x,y$and$\displaystyle x^y > y^x \; \;$, show that$\displaystyle x
Let $x = 5$ and $y = 1$.

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. Thanks from topsquark and idontknow Last edited by mathman; April 17th, 2019 at 01:00 PM. Tags inequality, proof, variables Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Pell's fish Real Analysis 1 July 7th, 2014 04:34 AM ninhead Algebra 7 October 31st, 2013 08:50 AM Anton29 Advanced Statistics 0 February 2nd, 2012 06:46 PM bo1989 Algebra 12 March 12th, 2011 06:09 AM ninhead Abstract Algebra 2 December 31st, 1969 04:00 PM

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