Minimum of a function $f(x,y)=x^2+y^2+\alpha/(xy)$

Question

Let $\alpha \in \mathbb{R}^{*+}$ and $\left(x,y\right) \in \left(\mathbb{R}^{*+}\right)^2$, I have the function $f$ given by $$ f\left(x,y\right)=x^2+y^2+\frac{\alpha}{xy} $$ I've found that there exists one critical point which is $\displaystyle \left(\sqrt[4]{\frac{\alpha}{2}},\sqrt[4]{\frac{\alpha}{2}}\right)$ and it values $2\sqrt{2\alpha}$. However I would like to prove it is a global minimum minimorum for $f$. I've tried to prove that $f(x,y)-2\sqrt{2\alpha} \geq 0$ but I think I dont minore my expression well ( I used $-2ab \geq -a^2-b^2$ ). How can I proceed ?

Answer 1

By $AM-GM$ we have $$x^2+y^2\geq 2xy$$ so $$x^2+y^2+\frac{\alpha}{xy}\geq 2xy+\frac{\alpha}{xy}\geq 2\sqrt{2xy\cdot\frac{\alpha}{xy}}$$

Answer 2

From $f(x,y)=(x-y)^2+2xy+\alpha/xy $ we conclude that $x=y$ since $x$ and $y$ are positive. Furthermore $2z+\alpha/z$ takes is global minimum at $z=\sqrt{\alpha/2}$.