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I'd like to prove that, for $\theta,\lambda\in[0,\pi/2)$ and $\phi\in[0,2\pi)$, one always has

$$\left| \cos\left(\frac{\theta}{2}\right)\cos\left(\frac{\lambda}{2}\right)+\sin\left(\frac{\theta}{2}\right)\sin\left(\frac{\lambda}{2}\right)e^{-i\phi} \right|> \left| \cos\left(\frac{\theta}{2}\right)\sin\left(\frac{\lambda}{2}\right)+\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\lambda}{2}\right)e^{-i\phi} \right|,$$ where $|\cdot|$ denotes the complex modulus. By plotting the two functions as functions of two parameters ($\lambda$,$\phi$), with the third ($\phi$) chosen randomly, it is clear that the relation is true. However, I found it quite difficult to demonstrate it.

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By factorizing out the cosines (they are always positive), you have to show that $$ |1 + \tan(\theta/2) \tan(\lambda/2) e^{-i\phi}| > |\tan(\lambda/2)+\tan(\theta/2)e^{-i\phi}|,$$ or, equivalently, that for any $x,y \in [0,1)$, $\phi \in [0, 2\pi]$, $$ |1 + x\,y\, e^{i\phi}| > |x+y\,e^{i\phi}|.$$ By squaring each side and developing, this is equivalent to $$1 + x^2 y^2 + 2 xy\cos(\phi) > x^2 + y^2 + 2xy \cos(\phi), $$ so it remains to show that $1 + x^2 y^2 > x^2 + y^2$ for $x,y\in [0,1)$. But this follows simply from $$ x^2 + y^2 - x^2y^2 = x^2(1-y^2) + y^2 < 1-y^2 + y^2 = 1.$$

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    Seems right to me. The only minor correction I have is that in the right hand side of your first equation $tan(\theta/2)$ and $tan(\lambda/2)$ should be switched (hence $x$ and $y$ should be switched in the right hand side of the equations that follow), but the result still holds. Thank you! – 2017-01-20
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    Or that $1+x^2y^2>x^2+y^2\iff (1-x^2)(1-y^2)>0.$................+1. – 2017-01-20
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    That's even better, thanks! – 2017-03-08