Proving $\min\{x^4+x y+y^2-6 x-5 y\}=-9$

Question

I first found that

$$\min\{x^4+x y+y^2-6 x-5 y\} = -9$$

by knowing that if a point is a critical point, then the first partial derivatives with respect to $x$ and $y$ must be zero. Then I found the points that satisfies this constraint, which are just $(1,2)$. Then simply $f(1,2) = -9$. But critical point isn't necessarly minimum or maximum, so I have to prove that:

$$f(x,y)\ge f(x_0,y_0) = -9$$

for all $(x,y)$.

So basically I'm trying to prove that:

$$f(x,y)-f(x_0,y_0)\ge 0$$

We have that:

$$f(x,y)-f(x_0,y_0) = x^4+x y+y^2-6 x-5 y+9$$

Since I want to prove that this difference is greater than or equal to $0$ I tried to arrange everything in squares but it didn't end up very well. Any ideas to prove

$$f(x,y)-f(x_0,y_0) = x^4+x y+y^2-6 x-5 y+9>0 \text{ ?}$$

ps: just don't throw another method of proving that this is indeed a minimum point, I know other ways, but I wanted to do it this way for knowledge.

Answer 1

hint:

$P(y)=y^2+(x-5)y+x^4-6x=(y+\dfrac{x-5}{2})^2+x^4-(\dfrac{x-5}{2})^2-6x$ $Q(x)=x^4-(\dfrac{x-5}{2})^2-6x$

find $Q_{min}$, then you can find $P_{min}=Q_{min}$ when $y+\dfrac{x-5}{2}=0$