<s>[INST] Proving $x^{-1/2}\leq x^{n}y\leq x^{1/2}$ Let $x$ and $y$ be positive real numbers. Prove that there exists an integer $n$ such that $x^{-1/2}\leq x^{n}y\leq x^{1/2}$.
I have no idea how to do this.
 [/INST] For $x,y>0$ the required inequality is equivalent to write
$$1\le x^{n+\frac{1}{2}}y\le x.$$
The right hand side:
Let $f(x,y)=x-x^{n+\frac{1}{2}}y$ we want to show that $f(x,y)\ge 0$. So that 
\begin{align}
x-x^{n+\frac{1}{2}}y = x \left(1-x^{n-\frac{1}{2}}y\right) \ge 0 \qquad(?)
\end{align}
But since $x>0$ so we need $1-x^{n-\frac{1}{2}}y \ge 0$ which means
\begin{align}
x^{n-\frac{1}{2}}y \le 1
\end{align}
Taking $\ln(\cdot)$ for both sides (an increasing function), we get
\begin{align}
\ln x^{n-\frac{1}{2}}+ \ln y \le 0
\end{align}
i.e.,
\begin{align} 
\left\{ \begin{array}{l}
n\le \frac{1}{2}- \frac{\ln y}{\ln x}, \qquad x>1,y>0  \\ 
  \\ 
 n\ge\frac{1}{2}+ \frac{\ln y}{\ln x}, \qquad 1>x>0,y>0  \\ 
 \end{array} \right.
\qquad(*)
\end{align}
So that the value of such $n$ must chosen taking into account the above condition. The left hand side goes similarly so we can obtain 
\begin{align} 
\left\{ \begin{array}{l}
n\ge -\frac{1}{2}- \frac{\ln y}{\ln x}, \qquad x>1,y>0  \\ 
  \\ 
 n\le-\frac{1}{2}+ \frac{\ln y}{\ln x}, \qquad 1>x>0,y>0  \\ 
 \end{array} \right.
\qquad(**)
\end{align}
Combining the inequalities (*) and (**) we can say that such integer $n$ satisfies 
\begin{align}
\left\{ \begin{array}{l}
 \left| {n + \frac{{\ln y}}{{\ln x}}} \right| \le \frac{1}{2},\qquad x>1,y>0  \\ 
  \\ 
 \left| {n- \frac{{\ln y}}{{\ln x}}} \right| \le \frac{1}{2}, \qquad 1>x>0,y>0  
 \end{array} \right.
\end{align}
</s>