My Math Forum Inequality proof with two variables

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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.

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