complex inequality with 4 Show that if $z\in C$ and $|z|=1$ show that
$$\sum_{k=1}^{n}(n-k+1)|1+z^k|\ge \lfloor\dfrac{n}{2}\rfloor\left(n-\lfloor\dfrac{n}{2}\rfloor\right)|1-z|$$
(2008 Romanian mathematical competitions)
for $n=1$ which holds true
for $n=2$ 
$$\Longleftrightarrow 2|1+z|+|1+z^2|\ge |1-z|$$
which holds true
because $$|1+z^2|+|1+z|=|z\overline{z}+z^2|+|z\overline{z}+z|=|z+\overline{z}|+|\overline{z}+1|\ge |1-z|$$
for $n=3$,
$$\Longleftrightarrow |1+z^3|+2|1+z^2|+3|1+z|\ge 2|1-z|$$
which is true
because the same as $n=2$ we have
$$2|1+z^2|+2|1+z|\ge 2|1-z|$$
How to prove for $n$?
 A: We'll try induction:
Write down LHS for $n$:
$$S_n=\sum_{k=1}^n (n-k+1)|1+z^k|$$
For $n+1$, you just add one instance of $|\ldots|$ for all $k$ up to the new one:
$$S_{n+1}=S_n+\sum_{k=1}^{n+1}|1+z^k|$$
The RHS prefactor has a clear meaning. $\lfloor n/2\rfloor$ and $n-\lfloor n/2\rfloor$ are the most symmetric division of $n$ into two integer parts. Their product determines the number of squares in a tiling with half-perimeter equal to $n$ (that is, the biggest tiling with that perimeter). The sequence goes: 1×1,2×1,2×2,3×2,3×3,4×3,... and the difference between the terms at n and n+1 is $\lfloor (n+1)/2\rfloor$. Write the recursion for RHS:
$$P_{n}=\left\lfloor\frac{n}{2}\right\rfloor\left(n-\left\lfloor\frac{n}{2}\right\rfloor\right)|1-z|$$
$$P_{n+1}=P_n+\left\lfloor\frac{n+1}{2}\right\rfloor|1-z|$$
Induction step:
$$\sum_{k=1}^{n+1}|1+z^k|\geq \left\lfloor\frac{n+1}{2}\right\rfloor|1-z|$$
You can prove this inequality using induction too (it holds on its own for $n>0$) and using this one, go back to prove the original one.
For this step, we might as well jump in $n$ by twos - it works, but the initialization must be done for two consecutive $n$, for instance for $n=1$ and $n=2$. In this case, the induction step is simply to prove:
$$|1+z^{n+1}|+|1+z^{n+2}|\geq |1-z|$$
Write $z^{n+1}=w$.
Now that just reads
$$|1+wz|+|1+w|\geq |1-z|$$
In geometric terms, because $|z|=1$, multiplication by $z$ is rotation. $|1+w|$ is a chord between $-1$ and $w$. $|1+wz|$ is a chord between $-1$ and rotated $w$. $|1-z|$ is the chord between $1$ and $z$, but you can rotate that chord to see its length is the same as the chord between $w$ and $wz$. Therefore, this is triangle inequality for the complex points on the unit circle $-1$, $w$, $wz$ and is thus always true.
This basically proves the entire chain... you've already done the initialization step for the original inequality, and you can do $n=1$ and $n=2$ for the reduced inequality by yourself.
