<s>[INST] If complex number $a, b, c, d,$ and $|a|=|b|=|c|=|d|=1$, why $|a(c+d)|+|b(c-d)|\leq 2\sqrt{2}$? If we have 4 complex number $a, b, c, d,$ and $|a|=|b|=|c|=|d|=1$, So, how to prove that $|a(c+d)|+|b(c-d)|\leq 2\sqrt{2}$？
I try to separate $|a(c+d)|+|b(c-d)|$ to $|a||(c+d)|+|b||(c-d)|$ than I get       $|(c+d)|+|(c-d)|$. SO, if $c=d, c+d=2c=2b, c-d=0$
Is my idea good?
 [/INST] Let $c=\operatorname{cis}\alpha$ and $d=\operatorname{cis}\beta$.
Thus, by C-S we obtain: $$|a(c+d)|+|b(c-d)|=|c+d|+|c-d|=\sqrt{2+2\cos(\alpha-\beta)}+\sqrt{2-2\cos(\alpha-\beta)}=$$
$$=2|\cos\frac{\alpha-\beta}{2}|+2|\sin\frac{\alpha-\beta}{2}|\leq2\sqrt{(1+1)\left(\cos^2\frac{\alpha-\beta}{2}+\sin^2\frac{\alpha-\beta}{2}\right)}=2\sqrt2.$$
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