Does anyone know how to prove any of these three identities? Complex numbers/analysis, $z_i \in \mathbb C$ $$1.)\ \ (n-2)\sum_{k=1}^{n}|z_k|^2+|\sum_{k=1}^{n}z_k|^2=\sum_{1\leq k < s \leq n }|z_k+z_s|^2$$
$$2.)\ \ n\sum_{k=1}^{n}|z_k|^2-|\sum_{k=1}^{n}z_k|^2=\sum_{1 \leq k < s \leq n}|z_k-z_s|^2$$
These first two seem closely related.
$$3.) \ \ \ 2(|z_1|^n+|z_2|^n)\leq|z_1+z_2|^n+|z_1-z_2|^n\leq 2^{n-1}(|z_1|^n+|z_2|^n),n \geq 2$$
I've tried many things from trig, exponential substitutions\representations of complex numbers, all to no avail, because of a lack of an idea on solving these. Help is very much appreciated. 
 A: 1).
\begin{align}
\sum_{1\leq k<s \leq n}|z_k+z_s|^2&=\sum_{1\leq k<s \leq n}\overline{(z_k+z_s)}(z_k+z_s)
\\
&=\sum_{1\leq k < s \leq n}(|z_k|^2+\bar{z}_kz_s+\bar{z}_sz_k+|z_s|^2)
\\
&=\sum_{1\leq k < s \leq n}(|z_k|^2+|z_s|^2)+\sum_{1\leq k < s \leq n}(\bar{z}_kz_s+\bar{z}_sz_k)
\\
&=\sum_{1\leq k < s \leq n}|z_k|^2+\sum_{1\leq s < k \leq n}|z_k|^2+\sum_{1\leq k < s \leq n}(\bar{z}_kz_s+\bar{z}_sz_k)
\\
&=(n-1)\sum_{1\leq k \leq n}|z_k|^2+\sum_{1\leq k < s \leq n}(\bar{z}_kz_s+\bar{z}_sz_k)
\end{align}
On the other hand
\begin{align}
(n-2)\sum_{k=1}^{n}|z_k|^2+\left|\sum_{k=1}^{n}z_k\right|^2&=(n-2)\sum_{k=1}^{n}|z_k|^2+\left(\sum_{k=1}^{n}\bar{z}_k\right)\left(\sum_{k=1}^{n}z_k\right)
\\
&=(n-2)\sum_{k=1}^{n}|z_k|^2+\sum_{k=1}^{n}\sum_{s=1}^{n}\bar{z}_kz_s
\\
&=(n-2)\sum_{k=1}^{n}|z_k|^2+\sum_{1\leq k < s \leq n}(\bar{z}_kz_s+\bar{z}_sz_k)+\sum_{k=1}^{n}|z_k|^2
\\
&=\sum_{1\leq k<s \leq n}|z_k+z_s|^2
\end{align}
2).
\begin{align}
\sum_{1\leq k<s \leq n}|z_k-z_s|^2&=\sum_{1\leq k<s \leq n}\overline{(z_k-z_s)}(z_k-z_s)
\\
&=\sum_{1\leq k < s \leq n}(|z_k|^2-\bar{z}_kz_s-\bar{z}_sz_k+|z_s|^2)
\\
&=\sum_{1\leq k < s \leq n}(|z_k|^2+|z_s|^2)-\sum_{1\leq k < s \leq n}(\bar{z}_kz_s+\bar{z}_sz_k)
\\
&=\sum_{1\leq k < s \leq n}|z_k|^2+\sum_{1\leq s < k \leq n}|z_k|^2-\sum_{1\leq k < s \leq n}(\bar{z}_kz_s+\bar{z}_sz_k)
\\
&=(n-1)\sum_{1\leq k \leq n}|z_k|^2-\sum_{1\leq k < s \leq n}(\bar{z}_kz_s+\bar{z}_sz_k)
\end{align}
On the other hand
\begin{align}
n\sum_{k=1}^{n}|z_k|^2-\left|\sum_{k=1}^{n}z_k\right|^2&=n\sum_{k=1}^{n}|z_k|^2-\left(\sum_{k=1}^{n}\bar{z}_k\right)\left(\sum_{k=1}^{n}z_k\right)
\\
&=n\sum_{k=1}^{n}|z_k|^2-\sum_{k=1}^{n}\sum_{s=1}^{n}\bar{z}_kz_s
\\
&=n\sum_{k=1}^{n}|z_k|^2-\sum_{1\leq k < s \leq n}(\bar{z}_kz_s+\bar{z}_sz_k)-\sum_{k=1}^{n}|z_k|^2
\\
&=\sum_{1\leq k<s \leq n}|z_k-z_s|^2
\end{align}
A: Recall that $|z|^2=zz^*$. Therefore $|z_k+z_s|^2=|z_k|^2+z_kz_s^*+z_sz_k^*+|z_s|^2$. On the other hand,
$$
\left|\sum_{k=1}^n z_k\right|^2=\sum_{k=1}^n \sum_{s=1}^n z_kz_s^*.
$$
Now compare terms on both sides to conclude (1). Similarly for (2). For (3), you could either use the Binomial theorem or fancy inequalities like Karamata's.
