Show that A is compact Let $A = \{(x,y) \in \mathbb{R^2} : x&gt;0, y&gt;0, xy + \frac{1}{x} + \frac{1}{y} \leq \alpha \}$.

I need to prove that  A is compact, $\forall\alpha \in \mathbb{R}.$

My attempt

A is closed since the function $f: (0, \infty)\times(0, \infty) \to \mathbb{R}$, 
 $\quad f(x,y) = xy + \frac{1}{x} + \frac{1}{y}  $ is continuous.

I couldn't prove that A is bounded.

How can we do this? Any hints?

Thanks in advance.
If $\alpha\le0$ then $A=\varnothing$. For $\alpha &gt;0$, you can show that $A$ is bounded by finding an upper bound, let $R=\max\{\alpha,1\}$, then  $x\ge R$ or $y\ge R$  implies $(x,y)\notin A$. So, for all $(x,y)\in A$ we have $$x&lt;R\quad\text{and}\quad y&lt;R\qquad\qquad\implies\qquad\qquad \sqrt{x^2+y^2}&lt;\sqrt2R$$
$\displaystyle\alpha \ge xy+\frac{1}{x}+\frac{1}{y} \ge xy $ because $x,y&gt;0$

Now, for fixed $y&gt;0$, $\displaystyle xy\le \alpha \implies x \le \frac{\alpha}{y}$

similarly conclude for fixed $x$