Proving result of complex numbers Let $w,z$ be complex such that $wz\neq0$. Let $$t = 2- \left|\frac{w}{|w|} + \frac{z}{|z|}\right|.$$ Show that $0 \leq t \leq 2$, and that 
$$|w| + |z| - t\text{max}(|w|,|z|) \leq |w+z| \leq |w| + |z| - t\text{min}(|w|,|z|).$$
The first part I've shown, but the latter part I've struggled quite a bit with.
 A: Note that scaling both $w$ and $z$ by a same factor does not affect the inequality. Thus without loss of generality we may assume that $w=1$. Then we have
$$t=2-\left|1+z/|z|\right|.$$
Thus it suffices to prove that
$$1+|z|-(2-\left|1+z/|z|\right|)\max(1,|z|)\leq |1+z|\leq 1+|z|-(2-\left|1+z/|z|\right|)\min(1,|z|).$$
Let's look at the first inequality. Again without loss of generality assume that $1\geq |z|$, thus what we want to prove becomes
$$1+|z|-(2-\left|1+z/|z|\right|)\leq|1+z|\quad (*).$$
The left hand side of the inequality is
$$1+|z|-2+|1+z/|z||=|z|-1+|1+z/|z||.$$
Put $\gamma:=\text{Arg}(z)$ and $r=|z|$, we have
$$\text{LHS of }(*)=r-1+|1+e^{i\gamma}|.$$
Moreover the RHS is
$$\text{RHS of }(*)=|1+re^{i\gamma}|.$$
Thus
$$\begin{aligned}
\text{RHS of }(*)-\text{LHS of }(*)&=|1+re^{i\gamma}|-|1+e^{i\gamma}|+1-r\\
&=|1+re^{i\gamma}|-|1+e^{i\gamma}|+|(1-r)e^{i\gamma}|\\
&=|1+re^{i\gamma}|-|1+e^{i\gamma}|+|(1+e^{i\gamma})-(1+re^{i\gamma})|\\
&\geq |1+re^{i\gamma}|-|1+e^{i\gamma}|+(|1+e^{i\gamma}|-|1+re^{i\gamma}|)\\
&=0.
\end{aligned}$$
This proves the first inequality. The second inequality can by proved in a similar way.
