Minimise $|z_az_b + z_cz_d|$ where $\{z\}$ are the roots of $x^4 + 14x^3 + 52x^2 + 56x + 16$

Hint:  the ratios of symmetric coefficients are in increasing powers of $4$ since $\,16/1 = (56/14)^2\,$. (This indicates that the polynomial reduces to a palindromic one with the substitution $x=2y\,$, which can then be explicitly solved though the calculations are tedious.) More useful, however, it indicates that if $z$ is a root then so is $4/z\,$.

Let then the $4$ roots be $z_1=u, z_2=v, z_3=4/u,z_4= 4/v\,$. Using that the roots are real, and $|z+1/z| \ge 2$ for all real $z$ it follows that:

• $\,|z_1z_3+z_2z_4| = 4+4 = 8$

• $\,|z_1z_2+z_3z_4| = |uv + 16/uv| \;\ge\; 4 \cdot 2 = 8$

• $\,|z_1z_4+z_2z_3| = |4(u/v+v/u)| \;\ge\; 4 \cdot 2 = 8$

I claim the minimum is $8$. I will use dxiv suggestion. Let $g(y) = y^4+7y^3+13y^2+7y+1$ Obviously all roots are real and negative since we have

\begin{array}{cccccc} y & -5 & -2 & -1 & -0.5 & 0 \\ g(y) & + & - & + & - & + \\ \end{array}

Let $a,b,{1\over a},{1\over b}$ be all roots for $g$ where $a,b,$ are negative.

So we have next possible expresions \begin{eqnarray} E_1 &=&ab+{1\over ab} >2\\ E_2 &=&{a\over b}+{b\over a} >2\\ E_3 &=&a{1\over a}+b{1\over b} =2\\ \end{eqnarray} and does are all. So $E_{\min}=2$. So the minimum value for the given expression is $4\cdot 2 =8$

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.