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.

Your work is already very useful. Since the function is continuous and bounded from below (your calculations show that the expression is greater or equal than $\frac 12$), it has a global minimum. Since this is a differentiable function in an open set, this global minimum must be a stationary point, i.e.
$$
2x +y +1=0, \quad 2y + x-4 = 0.
$$
So, the global minimum is attained at $(-2,3)$. The minimum is 2.
