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October 8th, 2019, 04:39 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98  Inequality #2 without calculus
Given $\displaystyle x^y < y^x $, where $\displaystyle x>y \geq \lambda >0$, find the range of $\displaystyle \lambda $. Without calculus! Last edited by skipjack; October 9th, 2019 at 01:37 AM. 
October 8th, 2019, 05:33 AM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 
Here is the solution. $\displaystyle x^y < y^x \: $ ; $\displaystyle x^{1/x} <y^{1/y}$. Set $\displaystyle f(t)=t^{1/t}$ and $\displaystyle f(t)>f(1+t)\geq f(\lambda ) \: \Rightarrow \: t>(1+\frac{1}{t})^{t}\geq \lambda $. $\displaystyle t$diverges , $\displaystyle (1+1/t)^{t} $converges and is increasing. $\displaystyle \lambda \leq \sup\{ (1+1/t)^{t} \}=\lim_{t\rightarrow \infty }(1+1/t )^{t} =e <\lim_{t\rightarrow \infty} t =\infty$. $\displaystyle \lambda \in [e,\infty )$. Last edited by skipjack; October 9th, 2019 at 01:40 AM. 
October 8th, 2019, 04:24 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 670 Thanks: 440 Math Focus: Dynamical systems, analytic function theory, numerics 
How is your solution "without calculus"? The function $x^y$ is undefined without calculus. Then you claim that $f(t) > f(1 + t)$ which is probably impossible to prove without calculus. Then you talk about convergence, and take a limit. What does "without calculus" mean to you?


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