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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. $\displaystyle x^y < y^x \:$ ; $\displaystyle x^{1/x} f(1+t)\geq f(\lambda ) \: \Rightarrow \: t>(1+\frac{1}{t})^{t}\geq \lambda$. $\displaystyle t$-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? Thanks from idontknow

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