Re: HSC 2015 4U Marathon - Advanced Level
Finishing off the cubic question (finally!):
It remains to show that for a polynomial with three real roots, the inequalities:
1. |bd-c| < 1-d^2
and
2. |b+d| < |1+c|
imply that all three roots live in the open unit interval.
As discussed in my previous posts, inequality 2 implies that p(1) > 0 and p(-1) < 0, which means there is at least one real root in the unit interval.
If all three roots DON'T live in the unit interval then either two of them are at least 1 or two of them are at most -1. (By considering the limiting behaviour of p at +inf and -inf.)
Let us assume we are in one of these cases for the sake of contradiction.
For inequality 1 to hold, we must of course have d^2 < 1. (*)
We also have that the polynomial q(1)= 1-d^2 - (c-bd) >= 1-d^2 - |c-bd| > 0. (q was defined earlier as the polynomial with roots the pairwise products of the roots of p).
This means that either zero or two of the quantities
exceed 1, with the others being less than 1. (**)
As at least two of these roots have the same sign, this observation implies that all three roots must have the same sign.
So we are left with the possibilities:
In the first of these, we have
from (**).
This means we can multiply
which contradicts (**).
Similarly, in the latter of the two possibilities, we have:
So
again in contradiction with (**).
This completes the proof.