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?

$$|a(c+d)|+|b(c-d)|=\left|1+\dfrac dc\right|+\left|1-\dfrac dc\right|=|1+e^{i\theta}|+|1-e^{i\theta}|$$ 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 $2\sqrt2$.
