TITLE: Is $\lambda_1/(1+\lambda_1)^i \le \lambda_0/(1+\lambda_0)^i$ for $\lambda_1\ge \lambda_0>0$ and $i\ge 1/\log(1+\lambda_0)$?
QUESTION [0 upvotes]: Suppose $\lambda_1 \ge \lambda_0> 0$ and $i\in \mathbb{N}$ is such that $i\ge \frac{1}{\log(1+\lambda_0)}$.  Is it true that
$$
\frac{\lambda_1}{(1+\lambda_1)^i}\le \frac{\lambda_0}{(1+\lambda_0)^i}?
$$
I think that it is, but have been struggling to prove it to myself.  Thanks.
edit:  This is what I have tried.  Let $\lambda_1 = \sigma \lambda_0$ for $\sigma\in (1,\infty)$  Then we have
\begin{align*}
\frac{\lambda_1}{(1+\lambda_1)^i} 
&= 
\sigma \frac{\lambda_0}{(1+\sigma \lambda_0)^i} = \sigma \left( \frac{1+\lambda_0}{1+\sigma \lambda_0}  \right)^i \frac{\lambda_0}{(1+\lambda_0)^i}.
\end{align*}
So the result would follow if we can show that $$\sigma \left(\frac{1+ \lambda_0}{1+\sigma \lambda_0}\right)^i \le 1.$$  Numerical experiments are consistent with this being true, but this is still a dead end for me.  Any ideas are appreciated.
Edit #2

REPLY [1 votes]: It seems the following.
Put $f(\lambda)=\frac{\lambda}{(1+\lambda)^i}$. The simplest idea is to show that $f’(\lambda)\le 0$ for $\lambda\ge \lambda_0$. We have $f’(\lambda)=\frac{(1+\lambda)^i-i\lambda(1+\lambda)^{i-1} }{(1+\lambda)^{2i}}$. So there should be $i\ge\frac 1\lambda+1$, that is $i\ge\frac 1\lambda_0+1$. But we have only $i\ge \frac{1}{\log(1+\lambda_0)}$, so there should be a counterexample. Put $\lambda_0=e-1$, $i=1$, and $\lambda_1=e$. Then $f(\lambda_0)=1-\frac 1e<1-\frac 1{e+1}=f(\lambda_1)$.
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