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?

$$|a(c+d)| +|b(c-d)|=|a||c+d| +|b||c-d|=|c+d| +|c-d|=\sqrt{(\cos \alpha +\cos \phi )^2 +(\sin\alpha +\sin\phi )^2 } +\sqrt{(\cos \alpha -\cos \phi )^2 +(\sin\alpha -\sin\phi )^2 }=\sqrt{2+ 2\cos \alpha \cos\phi +2\sin\alpha \sin \phi} +\sqrt{2- 2\cos \alpha \cos\phi -2\sin\alpha \sin \phi} =\sqrt{2+2\cos (\alpha -\phi )}+\sqrt{2-2\cos (\alpha -\phi )}=2\sqrt{\frac{1+\cos (\alpha -\phi )}{2}}+2\sqrt{\frac{1-\cos (\alpha -\phi )}{2}}=2\sqrt{\cos^2 \left(\frac{\alpha -\phi }{2}\right)} +2\sqrt{\sin^2 \left(\frac{\alpha -\phi }{2}\right)}=2\left(\left|\cos \left(\frac{\alpha -\phi }{2}\right)\right|+\left|\sin \left(\frac{\alpha -\phi }{2}\right)\right|\right)\leq 2 \sqrt{1^2 +1^2 }\sqrt{\left|\cos \left(\frac{\alpha -\phi }{2}\right)\right|^2+\left|\sin \left(\frac{\alpha -\phi }{2}\right)\right|^2} =2\sqrt{2}$$
where $c=\cos \alpha +i \sin\alpha , d=\cos\phi +i \sin \phi .$
