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?
 A: Let $c=\operatorname{cis}\alpha$ and $d=\operatorname{cis}\beta$.
Thus, by C-S we obtain: $$|a(c+d)|+|b(c-d)|=|c+d|+|c-d|=\sqrt{2+2\cos(\alpha-\beta)}+\sqrt{2-2\cos(\alpha-\beta)}=$$
$$=2|\cos\frac{\alpha-\beta}{2}|+2|\sin\frac{\alpha-\beta}{2}|\leq2\sqrt{(1+1)\left(\cos^2\frac{\alpha-\beta}{2}+\sin^2\frac{\alpha-\beta}{2}\right)}=2\sqrt2.$$
A: You correctly started with
$$
|a(c+d)|+|b(c-d)| = |c+d|+  |c-d| \, .
$$
How can we estimate that if $|c|=|d|=1$ is known? Here helps the parallelogram law:
$$
|c+d|^2+|c-d|^2 = 2|c|^2 + 2|d|^2 = 4 \, .
$$
It remains to show that
$$
 |c+d|+  |c-d| \le \sqrt2 \sqrt{|c+d|^2+|c-d|^2}
$$
and that is exactly the  Cauchy-Schwarz inequality.
A: $$|a(c+d)| +|b(c-d)|=|a||c+d| +|b||c-d|=|c+d| +|c-d|=\sqrt{(\cos \alpha +\cos \phi )^2 +(\sin\alpha +\sin\phi )^2 } +\sqrt{(\cos \alpha -\cos \phi )^2 +(\sin\alpha -\sin\phi )^2 }=\sqrt{2+ 2\cos \alpha \cos\phi +2\sin\alpha \sin \phi} +\sqrt{2- 2\cos \alpha \cos\phi -2\sin\alpha \sin \phi} =\sqrt{2+2\cos (\alpha -\phi )}+\sqrt{2-2\cos (\alpha -\phi )}=2\sqrt{\frac{1+\cos (\alpha -\phi )}{2}}+2\sqrt{\frac{1-\cos (\alpha -\phi )}{2}}=2\sqrt{\cos^2 \left(\frac{\alpha -\phi }{2}\right)} +2\sqrt{\sin^2 \left(\frac{\alpha -\phi }{2}\right)}=2\left(\left|\cos \left(\frac{\alpha -\phi }{2}\right)\right|+\left|\sin \left(\frac{\alpha -\phi }{2}\right)\right|\right)\leq 2 \sqrt{1^2 +1^2 }\sqrt{\left|\cos \left(\frac{\alpha -\phi }{2}\right)\right|^2+\left|\sin \left(\frac{\alpha -\phi }{2}\right)\right|^2} =2\sqrt{2}$$
where $c=\cos \alpha +i \sin\alpha , d=\cos\phi +i \sin \phi .$
A: $$|a(c+d)|+|b(c-d)|=\left|1+\dfrac dc\right|+\left|1-\dfrac dc\right|=|1+e^{i\theta}|+|1-e^{i\theta}|$$ where $\theta$ is the angle between $c$ and $d$.
Now
$$1+e^{i\theta}=2\cos^2\frac\theta2+2i\sin\frac\theta2\cos\frac\theta2=2\cos\frac\theta2 e^{i\theta/2}$$
and
$$1-e^{i\theta}=2\sin^2\frac\theta2-2i\sin\frac\theta2\cos\frac\theta2=2\sin\frac\theta2e^{-i\theta/2}.$$
Finally, the maximum of $\left|2\cos\dfrac\theta2\right|+\left|2\sin\dfrac\theta2\right|$ is $2\sqrt2$.
