complex numbers - Find all $zinBbb C$ such that $|z+1|+ |z-1|=4$

I'd like to find all points of the complex plane which satisfy

$$|z+1| + |z-1| = 4. $$

I know this is an ellipsis with foci $1$ and $-1$, and I know that the answer is

$$3 x^2+4 y^2 \leq 12,$$

but I can't find a correct way of getting there.

First, I write $z$ as $x + i y$ and square both sides of the equation, then divide by 2 and get

$$x^2+y^2+1+\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =8.$$

Pass $x^2+y^2+1$ to the RHS (right hand side), then

$$\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =7-x^2-y^2.\tag{1} $$

Now, I would have to square both sides of the equation like this,

$$((x-1)^2+y^2)((x+1)^2+y^2) = (7-x^2-y^2)^2, \tag{2}$$

but the problem is that I cannot assure that the RHS is not negative, so there could be a value for $z$ such that (2) is satisfied but not (1), i.e it could exist $z=x + i y$ which satisfies

$$\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =-(7-x^2-y^2)\tag{3} $$

in which case also satisfy (2) but not (1)!

So I would get an incorrect solution.