New approach to inequalities!

We can rewrite the given function as ${ ({ { x }^{ 2 }-25 }) }^{ 2 }+{ \left( { y }^{ 2 }-484 \right) }^{ 2 }+1$ . Thus, the global minimum is $1$ , when $x=5, -5$ and $y=22, -22$ .

Relevant wiki: Second Derivative Test

Let $f_x$ and $f_y$ denote the partial derivatives of $f(x,y)$ , respectively, with respect to $x$ and $y$ .

Then at extrema, $f_x = 4x^3 - 100x = 0$ , $f_y = 4y^3 - 1936 y = 0$ . Solving them gives $x = 0, \pm 5$ , and $y = 0, \pm 22$ .

Similarly, let $f_{xx}$ and $f_{yy}$ denote the second partial derivatives of $f(x,y)$ , respectively, with respect to $x$ and $y$ .

Then, $f_{xx} = 12x^2 - 100$ and $f_{yy} = 12y^2 - 1936$ .

We apply the theorem for the second derivative test discriminant as $D := f_{xx} f_{yy} - (f_{xy})^2$ .

Let $(x_0, y_0)$ denore the point of the local extremum.

If $D> 0$ , then $f_{xx} (x_0, y_0)$ is a local minimum.
If $D< 0$ , then $f_{xx} (x_0, y_0)$ is a local maximum.
If $D=0$ , the point is a saddle point.

Then $D = 12^2 \left( x^2 - \dfrac{100}{12} \right) \left(y^2 - \dfrac{1936}{12} \right) > 0$ is a local minimum when $(x^2,y^2 ) = ( 25, 484 )$ .

Thus $f(x,y)$ has a minimum value at $(x^2,y^2) = (25, 484 )$ , and the value of this number is $25^2 + 484^2 - 50(25)-968(484)+234882 =\boxed1$ .