Let be $f:]0,\infty[\times]0,\infty[\to\mathbb{R}$ where $f(x,y):=2xy+\frac{1}{x}+\frac{1}{y}$. We already know that $f$ has only one local minimum at $\left(\sqrt[3]{\frac{1}{2}},\sqrt[3]{\frac{1}{2}}\right)$ with $f\left(\sqrt[3]{\frac{1}{2}},\sqrt[3]{\frac{1}{2}}\right)=3\sqrt[3]{2}$.
Show that this is a global minimum by using Langrange multipliers.
(Our tutor gave us the hint to define sets and apply the Langrange mehtod on those sets)
My approach:
First, we rearrange the statement equivalently to$$2x^2y^2+x+y\overset{??}{\geq} 3\sqrt[3]{2}xy.$$Next, we choose an arbitrary but fixed $c>0$ and define the set$$M_c:=\left\{(x,y)\in]0,\infty[\times]0,\infty[~\mid 3\sqrt[3]{2}xy=c\right\}$$and look for extrema on $M_c$. Applying the Lagrange method we get $x=y=\frac{\sqrt{c}}{3\sqrt[3]{2}}$ as the only possible point. If I plug in this point into $2x^2y^2+x+y$ I get$$2\frac{c^2}{9(\sqrt[3]{2})^2}+2\frac{\sqrt{c}}{3\sqrt[3]{2}}\overset{??}{\geq}3\sqrt[3]{2}xy.$$At this point I am stuck. Maybe this is the wrong way to go?