Inequality $\frac{1}{2}\sum\limits_{i=1}^n|r_i^2-s_i^2|\leq\sqrt{\sum\limits_{i=1}^n|r_i-s_i|^2}$

Question

Let $r_i,s_i$ be complex numbers. Edit: Let $n\geq 2$ and $|r_i|,|s_i|\in[0,1]$ such that $\sum\limits_{i=1}^n|r_i|^2=1$ and $\sum\limits_{i=1}^n|s_i|^2=1$.

How do we prove $$\frac{1}{2}\sum_{i=1}^n|r_i^2-s_i^2|\leq\sqrt{\sum_{i=1}^n|r_i-s_i|^2}\text{ ?}$$

I know that $|r_i^2-s_i^2|=|r_i+s_i|\cdot|r_i-s_i|$
What else can we do?

Answer 1

Just use the fact that $\sum |r_i^{2}-s_i^{2}| =\sum |r_i-s_i| |r_i+s_i| \leq \sqrt {\sum|r_i-s_i|^{2}} {\sqrt {\sum|r_i+s_i|^{2}}}$ and $|r_i+s_i|^{2} \leq 2|r_i|^{2} +2|s_i|^{2}$. [ The second factor is therefore less than or equal to $\sqrt {(2+2)}=2$].