Find local maximum, minimum and saddle points of $f(x,y) = x^4 + y^4 - 4xy + 2$

Question

Find maximum, minimum and saddle points of $f(x,y) = x^4 + y^4 - 4xy + 2$.

For critical points, $f_{x} = f_{y} = 0$, $f_{x} = 3x³ - 4y$ and $f_{y} = 3y³ - 4x$. Therefore $f_{xx} = 9x²$, $f_{yy} = 9y²$, and $f_{xy} = f_{yx} = -4$.

Hessian matrix determinant is positive for P($\frac{2}{\sqrt3},\frac{2}{\sqrt3},-14)$ and Q($\frac{-2}{\sqrt3}$,$\frac{-2}{\sqrt3}$,-14), so they are minimum points.

And R(0,0,2) is a saddle point.

Is this correct?

Answer 1

By AM-GM $$x^4+y^4+2\geq4\sqrt[4]{x^4y^4}=4|xy|\geq4xy.$$ The equality occurs for $x=y=1$, which gives that $0$ is a minimal value.