simple complex number proofs Here is one question on my text book:

$P$ is a point on an argand diagram corresponding to a complex number $z$ which satisfies equation $|4+z |-|4-z  |=6$, prove that$$| 4+z |^2-| 4-z |^2 \ge 48$$ 

Here is my proof but it did not ends up well:   
By $|4+z |-|4-z  |=6$ , we can rewrite inequality as 
$$| 4+z |^2-| 4-z |^2\ge8(|4+z |-|4-z  |)$$ 
 and hence $ | 4+z |+| 4-z | \ge 8$ and then $$| 4+z |-| 4-z |+2| 4-z |\ge    8 $$ and all we left to prove is $| 4-z |\ge 1$, we then follow a common triangle inequality, deduce$| 4-z |\ge|4 |-|z |$ then we need to prove $4-|z |\ge 1$, and it requires, $|z |\le3$, and I cannot prove this , in fact ,proof following $|4+z |-|4-z  |=6$ actually give $|z |\ge 3$. 
So I kind of failed into this loophole . Can anyone help me with the proof ?
 A: Hint
$$|4-z|+|4+z|\ge|4+z+4-z|$$
and use $a^2-b^2=(a+b)(a-b)$
See this. 
A: Clearly $z$ is on a right branch of a hyperbola with a focuses at $-4$ and $4$. Since it vertex is at $3$ the real part of $z$ is $\ge 3$

Let $w$ be a projection of $z$ on real line, thus $w\geq 3$. Then we have (by Pythagoras theorem) \begin{eqnarray}| 4+z |^2-| 4-z |^2 &=&| 4+w |^2-| 4-w |^2\\ &=&(16+8w+w^2)-(16-8w+w^2)\\ &=&16w\geq 48\end{eqnarray}
A: Geometrical interpretation in complex numbers' problem are sometimes quite helpful and apart this it unfolds some beauty.
Here, given equation $$ |z + 4| - |z-4| = 6$$ Which can be interpreted as Hyperbola whose foci lies at $( \pm 4 ,0)$
Now, you've to prove that $$|z + 4|^2 - |z-4|^2 \ge 48 $$ $$\Rightarrow (|z+4| + |z-4|)(|z+4| - |z-4|) \ge 48 $$ $$|z+4| + |z-4| \ge 8$$
Which looks like an equation of ellipse with foci at $( \pm 4,0)$
.So the equation is asking to prove that the major axis of ellipse is greater or equal than $8$, which is obviously true.If I define $2c$ to be the distance between the foci, then here $2c = 8$, then,
Let $a$ be the length of semi-major axis, then $a$ and $a$ are related by $c = ae$, where $e$ is the eccentricity of the ellipse. Here, $c = 4 \Rightarrow a = 4/e$,for ellipse $e \le 1 \Rightarrow a \ge 4\Rightarrow 2a \ge 8$.
Which was to be proved.
