$|z − 2| + |z + 2| ≤ 2$
I've tried solving it like this:
$|x+iy - 2| + |x+iy + 2| \le 2\\ |x - 2 + iy| + |x + 2 + iy| \le 2\\ \sqrt{(x-2)^2+(y)^2} + \sqrt{(x+2)^2+(y)^2} \le 2\\ \sqrt{(x-2)^2+(y)^2} \le 2 - \sqrt{(x+2)^2+(y)^2}\\ \sqrt{x^2 - 4x + 4 + y^2} \le 2 - \sqrt{x^2 + 4x + 4 + y^2}\\ x^2 - 4x +4 + y^2 \le 4 - 4\sqrt{x^2 + 4x + 4 + y^2} + x^2 + 4x + 4 + y^2\\ -8x \le 4 - 4\sqrt{x^2 + 4x + 4 + y^2}\\ -2x \le 1 - \sqrt{x^2 + 4x + 4 + y^2}\\ -2x - 1\le\sqrt{x^2 + 4x + 4 + y^2}\\ 4x^2+4x + 1 \le x^2 + 4x + 4 + y^2\\ 3x^2-y^2 \le 3$
Which seems to be incorrect, what am I doing incorrectly in my procedure? Is my logic correct?
Solve the inequality geometrically.
Since $|z-w|$ is the distance between the points representing $z$ and $w$, the inequality $$|z − 2| + |z + 2| ≤ 2$$ can be interpreted as:
The sum of distances from $z$ to $2$ and from $z$ to $-2$ is smaller than $2.$
Put on $x$-axis $-2$ and $2$ and various points $z$ in the plane .... wherever is $z\in \mathbb{C},$ the sum of distances to $2$ and $-2$ is at least $4,$ obtained when $z\in [-2,2].$