Finding minimum value of $x^2+y^2+xy+x-4y+9$ 
What is the minimum value of $f(x,y)=x^2+y^2+xy+x-4y+9$ ?

I tried completing squares,
$$x^2+y^2+xy+x-4y+9=\frac12(x^2+2xy+y^2+x^2+2x+1+y^2-8y+16+1)=\frac12[(x+y)^2+(x+1)^2+(y-4)^2+1]$$But not sure how to continue.

Hint:\begin{align}f(x+y,x-y)&=3x^2-3x+y^2+5y+9\\&=3\left(x-\frac12\right)^2+\left(y+\frac52\right)^2+2\end{align}
