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So I got stuck on a little problem. Given $x,s,t>1$, show that

$1+(x^t-1)^s > (1+(x-1)^s)^t$

It seems true when I plot it, and I'm very convinced that it's true, but I cannot find a way to prove it. I have been stumped by this all day. Anyone know how to tackle these kind of inequalities?

I am trying to prove that $x\mapsto\frac{\log(1+(x-1)^s)}{\log x}$ is monotonely increasing for $x>1$, when $s>1$, if anyone cares how I arrived at the inequality.

EDIT: A commentor requested that I explain how I got from my main task to the inequality. Here goes: Let $f(x) = x\mapsto\frac{\log(1+(x-1)^s)}{\log x}$. Let $y>x>1$, then there exists $t>1$ such that $y=x^t$. $f(y)=\frac{\log(1+(x^t-1)^s)}{t\log x}=\frac{\log((1+(x^t-1)^s)^{1/t})}{\log x}$, which is larger than $f(x)=\frac{\log(1+(x-1)^s)}{\log x}$ if and only if $(1+(x^t-1)^s)^{1/t}>1+(x-1)^s$.

Asger Kjærulff Jensen
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1 Answers1

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Let $f(x) = 1 + (x-1)^s$, noting that $f(1) = 1$. The desired inequality is $$ f(x^t) > f(x)^t \,. $$ Now let $g(y) = \log f(e^y) = \log(1 + (e^y-1)^s)$ and note that $g(0) = 0$. The inequality becomes $$ g(ty) > tg(y) \quad \forall y > 0 $$ or $$ \frac{g(ty)}{ty} > \frac{g(y)}{y} \quad \forall y > 0 \,, $$ which is equivalent to the statement that $g(y)/y$ is strictly increasing. Note that since $g(y)/y$ is the average value of $g'$ from $0$ to $y$, it suffices to show $g'(y) > g(y)/y$ for all $y > 0$.

We have $$ g'(y) = \frac{s(e^y-1)^{s-1}e^y}{1 + (e^y-1)^s} = s\frac{u^{s-1}(1+u)}{1 + u^s} = s\frac{(1+u)/u}{(1+u^s)/u^s} $$ where $u = e^y-1$. Since $y = \log(1+u)$, we are trying to show $$ s\frac{(1+u)/u}{(1+u^s)/u^s} > \frac{\log(1+u^s)}{\log(1+u)} $$ or equivalently $$ \frac{(1+u)\log(1+u)}{u \log u} > \frac{(1+u^s)\log(1+u^s)}{u^s \log(u^s)} \,. $$ Since the functions $$ v \mapsto \frac{1+v}{v} \quad\text{and}\quad v \mapsto \frac{\log(1+v)}{\log v} $$ are decreasing on $(0,1)$ and on $(1,\infty)$, so is their product, and we are done since $u$ and $u^s$ lie on the same side of $1$.

eccheng
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