Let $m, n$ be positive integers and $a, b$ positive real numbers different from $1$ such thath $m > n$ and $$\frac{a^{m+1}-1}{a^m-1} = \frac{b^{n+1}-1}{b^n-1} = c$$. Prove that $a^m c^n > b^n c^{m}$ (Turkey)
Problem
Source: Balkan BMO Shortlist 2015 A5
Tags: inequalities, algebra, Exponential inequality, exponential, Exponential equation
20.02.2022 08:40
any solution
22.02.2022 06:21
ILOVEMYFAMILY wrote: any solution Very soon
22.02.2022 06:40
let p be a prime dividing c, then lte/fermats it?
22.02.2022 10:06
........
22.02.2022 16:21
Arrowhead575 wrote: let p be a prime dividing c, then lte/fermats it? I don't think it's related to number theory
04.05.2023 14:27
I use around 10 days to solve this problem. So now I will write it down as simple as possible. First, we know that $a, b$ are positive real numbers and $$c=\frac{a^{m+1}-1}{a^{m}-1}=\frac{a^{m}+a^{m-1}+\dots+1}{a^{m-1}+a^{m-2}+\dots+1}=\frac{a^m}{a^{m-1}+a^{m-2}+\dots+1}+1$$And finally we get $a^m=(c-1)(a^{m-1}+a^{m-2}+\dots+1)$ So we can see that $c>1$ Claim : $a>b$ Proof : We do the similar thing to $b$ so we get $b^n=(c-1)(b^{n-1}+b^{n-2}+\dots+1)$ Let $f(b)=b^{n}-(c-1)(b^{n-1}+b^{n-2}+\dots+1)$ We know that $f(0)=-(c-1) < 0$ And $f(a)=a^{n}-(c-1)(a^{n-1}+a^{n-2}+\dots+1)=\frac{(c-1)(a^{m-n-1}+\cdots+1)}{a^{m-n}}>0$ So by Intermediate value theorem we get $a>b>0$ Right now we are going to prove that $a^{m}>b^{n}c^{m-n}$ or $$a^{m-1}+a^{m-2}+\dots+1>c^{m-n}(b^{n-1}+b^{n-2}+\dots+1)$$Claim : For $n\leq k \leq m-1$ we get that $$a^{k} > (c-1)(a^{k-1}+a^{k-2}+\dots+1)$$Proof Because $$a^{k}=(c-1)(a^{k-1}+a^{k-2}+\dots+1+\dots+\frac{1}{a^{m-k}})>(c-1)(a^{k-1}+a^{k-2}+\dots+1)$$Let $m-n=l \geq 1$ So we get that $$a^{n+l-1}>(c-1)(a^{n+l-2}+\dots+1)$$$$a^{n+l-2}>(c-1)(a^{n+l-3}+\dots+1)$$$$\vdots$$$$a^{n}>(c-1)(a^{n-1}+a^{n-2}+\dots+1)$$If we let $S=a^{n-1}+a^{n-2}+\dots+1$ from easy induction we get $a^{n+l}>(c-1)c^{l}S$ Hence, $$a^{n+l-1}+a^{n+l-2}+\dots+a^{n}+a^{n-1}+\dots+1 > (c-1)(c^{l-1}+c^{l-2}+\dots+1)S+S=c^{l}S$$But from $a>b>0$ and $u$ is positive integer so we get $a^{u}>b^{u}$ So we get $a^{m-1}+a^{m-2}+\dots+1 > c^{m-n}(a^{n-1}+\dots+1)>c^{m-n}(b^{n-1}+\dots+1)$ as desired.