TITLE: Minimise $|z_az_b + z_cz_d|$ where $\{z\}$ are the roots of $x^4 + 14x^3 + 52x^2 + 56x + 16$
QUESTION [15 upvotes]: Let $f(x) = x^4 + 14x^3 + 52x^2 + 56x + 16.$ Let  $z_1, z_2, z_3, z_4$ be the four roots of $f$. Find the smallest possible value of $|z_az_b + z_cz_d|$ where ${a, b, c, d} = {1, 2, 3, 4}.$
So I have tried Vieta, but it is try too complex and plugging in a simple value to try to find a root failed, so I think this is going to be much harder than finding the four roots-- I wonder if there are more involved techniques to solve this problem.

REPLY [3 votes]: Alternative approach (just for fun!):
[Subsequent edit: It was not especially fun...]
Let us broach this with the hope that it will actually factor.
The leading term is $x^4$ and the constant term is $16$, so something like $(x+2)^4$ would be great. Clearly this will not work, though; so, instead, let us attempt the next most likely suspect in these setups: a difference of squares factorization. 
We will have leading term $x^2$ so that we get $x^4$; we will have constant term $4$ so that we get $16$; and we will do our best to get around $52x^2$: in particular, the best we can do may be squaring $7x$ to get $49x^2$, as a bit more will be added on to this in the process. Anyway, we delve deeper based on an unreasonable hope derived from knowing this is a pre-cooked contest problem:
$$\big((x^2 + 7x + 4) + a\big)\big((x^2 + 7x + 4) - a\big) = (x^2 + 7x + 4)^2 - a^2$$ 
$$= x^4 + 14x^3 + 57x^2 + 56x + 16 - a^2$$
Setting this equal to the original expression, we find:
$$a^2 = 5x^2$$
... this has worked out absurdly well: set $a = x\sqrt{5}$ (or $-x\sqrt{5}$: it won't matter since our product has an $a$ and $-a$ symmetrically placed in it) and we have the factorization into quadratic terms, from which one can find all four of the roots.