Show that A is compact Let $A = \{(x,y) \in \mathbb{R^2} : x>0, y>0, xy + \frac{1}{x} + \frac{1}{y} \leq \alpha \}$.
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.
 A: If $\alpha\le0$ then $A=\varnothing$. For $\alpha >0$, 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 $$x<R\quad\text{and}\quad y<R\qquad\qquad\implies\qquad\qquad \sqrt{x^2+y^2}<\sqrt2R$$
A: As $ xy + x^{-1} + y^{-1} \leq \alpha $ and $ x,y> 0 $ you can observe that the individual terms should be less than $ \alpha $, that is $ xy,x^{-1},y^{-1} \leq \alpha $, thus $ x,y \geq \alpha^{-1} $. Now we can see that $ x\alpha^{-1} \leq xy \leq \alpha $ and thus $ x \leq \alpha^2 $. $ y \leq \alpha^2 $ follows by symmetry. Thus $ \|(x,y)\| \leq \sqrt{2}\alpha^2 $ for $ (x,y) \in A $.
A: If $\alpha\leq 0$ then $A=\emptyset$. We assume $\alpha>0$ and show that it must hold that, for any $(x,y) \in A$, $\|(x,y)\|_\infty \leq \alpha^2$. 
By symmetry it is enough to show that for all $(x,y) \in A$, $x \leq \alpha^2$. 
Given $(x,y)\in A$, suppose $x>\alpha^2$ i.e. $\frac{x}{\alpha}>\alpha$. We consider two cases. 
Case 1: $y \leq \frac{1}{\alpha}$
Then $xy+\frac{1}{x}+\frac{1}{y} > \frac{1}{y} \geq \alpha$. 
Case 2: $y > \frac{1}{\alpha}$
Then $xy+\frac{1}{x}+\frac{1}{y} > xy > \frac{x}{\alpha} > \alpha$.
So we are done.
A: $\displaystyle\alpha \ge xy+\frac{1}{x}+\frac{1}{y} \ge xy $ because $x,y>0$
Now, for fixed $y>0$, $\displaystyle xy\le \alpha \implies x \le \frac{\alpha}{y}$
similarly conclude for fixed $x$
