Suppose I have non-zero complex numbers $a,b,c,d,f \in \mathbb{C} \notin \{ 0 \}$ where the following holds true $$ |a| \gg |b| \ \ \ \ \ \ \mathrm{and} \ \ \ \ \ \ |c| \gg |d| $$
Is the above enough to ensure that the identity $$ \big| f a + f^{\ast} c \big| \gg \big| f b + f^{\ast} d \big| $$ is also true?
My Attempt: Using the triangle identity I have been able to prove so far that $$ \big| f b + f^{\ast} d \big| \ \leq \ |f| \left( |b| + |d| \right) \ \ll \ |f| \left( |a| + |c| \right) $$ but I can't get further than this.
Is this true? And if so, how do you prove it?
Hint Pick $a,c,f$ such that $$f^*c=-fa$$
This is done easily, you just pick $a$ such that $|a|$ is "large enough", $f\neq 0$ and set $$c=\frac{-f}{f^*}a$$
Note that $|c|=|a|$ is also "large enough".
Added: If you are trying to ask which $f'$ work, note that
$$\big| f a + f^{\ast} c \big| \gg \big| f b + f^{\ast} d \big| \Leftrightarrow \\ \big| f a + f^{\ast} c \big|^2 \gg \big| f b + f^{\ast} d \big|^2\Leftrightarrow \\ (f a + f^{\ast} c)(f^\ast a^\ast +f c^\ast) \gg (f b + f^{\ast} d)(f^\ast a^\ast +f b^\ast) \Leftrightarrow \\ f a f^\ast a^\ast +f a f c^\ast + f^{\ast} cf^\ast a^\ast +f^ast c f c^\ast \gg f b f^\ast b^\ast +f b f d^\ast + f^{\ast} df^\ast b^\ast +f^ast df d^\ast\Leftrightarrow \\ |f|^2 \left(|a|^2+|c|^2 -|b^2|-|d|^2\right)+ f^2\left(ac^\ast +bd^\ast \right)+ (f^ast)^2 \left( ca^\ast-db^\ast \right) $$
Now, setting $k:=|a|^2+|c|^2 -|b^2|-|d|^2$, which is a large positive number, and $\alpha =ac^\ast +bd^\ast$ you are really asking when is $$k|f|^2 +\alpha f^2 +(\alpha f^2)^2 > >0$$
Note now that if $f=x+iy$ and $\alpha = u+iv$ your inequality becomes $$(k+u)x^2+(k-u)y^2-2vuxy >>0$$
So your question simply becomes: Given $\alpha =ac^\ast +bd^\ast=u+iv$ and $k=|a|^2+|c|^2-|b|^2-|d|^2$, and knowing that $k >>0$, for which $(x,y) \in \mathbb R^2$ do we have $$(k+u)x^2+(k-u)y^2-2vuxy >>0 \,?$$
Also note that, for fixed $k,u,v$ the equation $(k+u)x^2+(k-u)y^2-2vuxy=M$ is an ellipse or a hyperbola/parabola. In the first case, "generic" $f$ will be outside the curve and your inequality will hold "generically". In the second case, generic $f$ will not be ourtside the curve.
To make sure that you have an ellipse, you need to ensure that $$4(k-u)(k+u) > 4uv \Leftrightarrow k^2>u^2+uv \\ \Leftrightarrow (|a|^2+|c|^2 -|b^2|-|d|^2) > (\mbox{Re}(ac^\ast +bd^\ast))^2+ (\mbox{Re}(ac^\ast +bd^\ast))(\mbox{Im}(ac^\ast +bd^\ast))$$ which can only be enforced with extra conditions on $a,b,c,d$.
So again, the answer is NO in general, but depending on your willigness to add extra conditions, and the meaning of generic (for example generic here would mean outside a compact set) it could be enforced.