If $ m > 2$ and $t ∈ R$, find the integral part of $\left ( \frac{4|t|}{16+t^{2}} \right )^m$.
I have approached the question by dividing the problem into cases when $-1 <t <1$, $t>1$ and $t<-1$. So when $t>1$ and $t<-1$ then $16+t^2 > 4|t|$ them integralpart is $0$. Similarly, when $-1<t<1$ then let $t=1/n$ where $n$ be any integer. Therefore, again integral part is $0$. Please let me know is my answer correct.
Notice that for $t> 0$, $$t+\frac{16}{t}\ge 8\tag{By AM-GM}$$ However, as they've given $|t|$, we don't need to care about $t>0$ or $t<0$. So, the term inside bracket is always be between $(0,1)$ or more precisely $(0,1/2]$.