Constrained Optimisation Problem: Confusion with Algebra/Contradiction I completed the following constrained optimisation problem:

Maximum and minimum values of $f(x,y) = x^2 + 2xy + y^2$ on the ellipse $g(x,y) = 2x^2 + y^2 - xy - 4x = 0$.

$\nabla f = \lambda \nabla g$
$\therefore (2x + 2y, 2x + 2y) = \lambda(4x - y - 4, 2y - x)$
We have a system of 3 equations in 3 unknowns:
$2x + 2y = \lambda(4x - y - 4)$,
$2x + 2y = \lambda(2y - x)$,
$2x^2 + y^2 - xy - 4x = 0$.
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$2x + 2y = \lambda(4x - y - 4)$
$\implies \lambda = \dfrac{2x + 2y}{4x - y - 4} \forall \ 4x - y - 4 \not= 0$
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$2x + 2y = \lambda(2y - x)$
$\implies \dfrac{2x + 2y}{2y - x} \forall \ 2y - x \not= 0$
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$\therefore \dfrac{2x + 2y}{4x - y - 4} = \dfrac{2x + 2y}{2y - x} \forall \ y \not= \dfrac{x}{2}, y \not= 4x - 4$ Remember this part: This is where my confusion lies.
But if we continue with our reasoning, we get the following (correct) points of potential maxima/minima:
$(x,y) = (0,0)$, $(x,y) = (1, -1)$, $(x,y) = (2,2)$, and $(x,y) = \left(\dfrac{2}{7}, \dfrac{-6}{-7}\right)$.
And, finally, using these points we find that $f_{max} = 16$ at $(x,y) = (2,2)$, and $f_{min} = 0$ at $(x,y) = (0,0)$ and $(x,y) = (1,-1)$.
But here's the problem: We had above that $y \not= \dfrac{x}{2}$; therefore, $(x,y) = (0,0)$ would get us $0 \not= \dfrac{0}{2} \implies 0 \not= 0$, which is obviously false. Also, as we just saw, $(0,0)$ is a solution for $f_{min}$!
The algebra here is confusing me, since we cannot have $y = \dfrac{x}{2}$, but $(0,0)$ is apparently the minima. Is this not a contradiction? Did I do something wrong? Or am I supposed to discard the solution $(0,0)$ and just have $f_{min} = 0$ at $(1, -1)$? What am I misunderstanding?
I would greatly appreciate it if people could please take the time to clarify this.

$(x+y)^2\geq0$ for all real variables. 
The equality occurs for $x=y=0$ for example.
Thus, $0$ is a minimal value.
For $x=y=2$ we'll get a value $16$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$(x+y)^2\leq16$$ or
$$16-(x+y)^2\geq0$$ or
$$16-(x+y)^2+4(2x^2+y^2-xy-4x)\geq0$$ or
$$3(x-y)^2+4(x-2)^2\geq0,$$
which is obvious.
Done!
