Let P be a real number such that $1<P<2$, and $m,n\in\mathbb{N}$ where $m<n$. Let $y\in\mathbb{N}$, and suppose that $y>1+\frac{1}{\sqrt[n-m]{2/P}-1}$.
I am trying to show that $\left(\frac{y}{y-1}\right)^{n-m}<\frac{2}{P}$, but I always end up with $\left(\frac{y}{y-1}\right)^{n-m}>\frac{2}{P}$.
Here's what I did:
Since $y>1+\frac{1}{\sqrt[n-m]{2/P}-1}$, then $y-1>\frac{1}{\sqrt[n-m]{2/P}-1}$. Knowing that $y, y-1>0$, what I did is $\frac{y}{y-1}>\frac{1+\frac{1}{\sqrt[n-m]{2/P}-1}}{\frac{1}{\sqrt[n-m]{2/P}-1}}=\left(\frac{2}{P}\right)^{\frac{1}{n-m}}$. And since $\frac{y}{y-1}$,$\frac{2}{P}>1$, it follows that $\left(\frac{y}{y-1}\right)^{n-m}>\frac{2}{P}$.
I think I made a huge mistake in thinking that $\frac{y}{y-1}>\frac{1+\frac{1}{\sqrt[n-m]{2/P}-1}}{\frac{1}{\sqrt[n-m]{2/P}-1}}$. How do I prove that $\left(\frac{y}{y-1}\right)^{n-m}<\frac{2}{P}$ based on the given assumptions? Thanks in advance!
Your error: If $a>b$ and $c > d > 0$, then $\frac{1}{c} < \frac{1}{d}$ so you can't say that $\frac{a}{c} > \frac{b}{d}$ necessarily.
\begin{align} y &> 1 + \frac{1}{\sqrt[n-m]{2/P}-1} \\ y - 1 &> \frac{1}{\sqrt[n-m]{2/P}-1} \\ \frac{1}{y-1} &< \sqrt[n-m]{2/P}-1 \\ 1 + \frac{1}{y-1} &< \sqrt[n-m]{2/P}. \end{align} Note that the left-hand side equals $\frac{y}{y-1}$.