$(1-x)^{3}|1-xe^{iy}|^{4}|1-xe^{2iy}|^{2}<1$ for real $x,y$ with $x\in(0,1)$

Question

For real $x,y$ with $x\in(0,1)$ we have $$(1-x)^{3}|1-xe^{iy}|^{4}|1-xe^{2iy}|^{2}<1$$

My attempt :

Obviously, we have the identity

$(1-x)^{3}|1-xe^{iy}|^{4}|1-xe^{2iy}|^{2}=(1-x^{3})|1-x\cos y-ix\sin y|^{4}|1-x\cos 2y-ix\sin 2y|^{2}$.

But this seems no help for our goal. Is there any hint I can follow $?$

Any suggestions I will be grateful.

Answer 1

By AM-GM $$(1-x)^3|1-xe^{iy}|^4|1-xe^{2iy}|^2=(1-x)^3(1-2x\cos{y}+x^2)^2(1-2x\cos2y+x^2)\leq$$ $$=(1-x)^3\left(\frac{2(1-2x\cos{y}+x^2)+1-2x\cos2y+x^2}{3}\right)^3=$$ $$=(1-x)^3\left(1+x+x^2-\frac{x(2\cos{y}+1)^2}{3}\right)^3\leq(1-x)^3(1+x+x^2)^3=$$ $$=(1-x^3)^3<1.$$

Answer 2

We have $$|1-x e^{2iy}|^2 |1-x e^{iy}|^4 = (1+x^2 -2x\cos(2y)) (1+x^2 -2x \cos y)^2 = ((1+x)^2 -4x (\cos y)^2) (1+x^2 -2x \cos y)^2 = 16x^3 (\frac{(1+x)^2}{4x} -(\cos y)^2)(\frac{1+x^2}{2x} -\cos y)^2.$$ Put $a = \frac{(1+x)^2}{4x}$, $b = \frac{1+x^2}{2x}$, $t = \cos y \in [-1,1]$ and $f(t) = (a-t^2)(b-t)^2$. Notice that $b > a >1$ since $x \in (0,1)$, and $f'(t) = -2(b-t)(t(b-t) + (a-t^2)) < 0$, hence $f(t) < f(-1) = \frac{(1-x)^2(1+x)^4}{16 x^3}$. Therefore, we get $$(1-x)^3|1-x e^{2iy}|^2 |1-x e^{iy}|^4 < (1-x)^5(1+x)^4 = (1-x)(1-x^2)^4 < 1.$$