Every even degree polynomial is eventually symmetric Let  we have  a  polynomial function $F:\mathbb{R} \to \mathbb{R}$ with $F(x)=ax^{2n}+bx^{2n-1}+\ldots+px+q$.
We assume  that $a>0$. For    sufficiently large $y$,  let  $A(y), B(y)$ be two distinct right inverses of  $F$, that  is  $F(A(y))=F(B(y))=y$ please see     the picture of this  linked page
Prove that $$\lim_{y\to{\infty}} A(y)+B(y)=-b/na$$

I  had  and I  have a  proof  for  this exercise but  I am searching for  some  other proofs or some other elementary proofs. Moreover   I wish to check and examine  whether it is really a very trivial elementary exercise or it is a bit nontrivial. Note that for  higher degrees, according to Galois, we have  no  a  precise  formula for $A(y)$ and  $B(y)$.
Pleease see page 4, item III line  $-3$ of    my paper  below. The journal who accepted my paper (year 2002), did not  asked me to provide any  proof. Regarding this  limit, inside the paper below I wrote that  "it is a simple  exercise". I did not write any proof of this limit in my thesis. No one in my defense committee asked me any proof of this limit. After all I think that it is  quite easy to proof. Thought it is  a  very simple  limit but playt a  crucial role to determine the  stability of  the  homoclinic loop based at equattor of Poincare sphere:

https://arxiv.org/pdf/math/0409594.pdf
RemarK: This  actualy gives us some information on the  sum of  complex  preimages $F^{-1}(y) \subset \mathbb{C}$ as $y$ goes to $\infty$. On the other hand, inspired by this post one may think to upper and  lower bound on  the norm of  subsets of $F^{-1}(y)$
 A: If $A(y)>0$, such that 
$$
a\big(A(y)\big)^{2n}+b\big(A(y)\big)^{2n-1}+\cdots+q=y, \tag{1}
$$
then $A(y)\to\infty$, as $yas\to\infty$, and hence $\big(A(y)\big)^{2n}\gg \big(A(y)\big)^{2n-1}$. So from $(1)$ we obtain $A(y)\approx\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}$. In particular, there exists a function $e_+(y)$, such that
$$
A(y)=\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}+e_+(y), \quad |e_+(y)|\ll y^\frac{1}{2n}.
$$
To get a better estimate for $e_+$, we have
$$
a\bigg(\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}+e_+(y)\bigg)^{2n}+b\bigg(\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}+e_+(y)\bigg)^{2n-1}+\cdots+q=y,
$$
or
$$
\bigg(y+2na\Big(\frac{y}{a}\Big)^{\frac{2n-1}{2n}}e_+(y)+\frac{2n(2n-1)}{2}a\Big(\frac{y}{a}\Big)^{\frac{2n-2}{2n}}e_+^2(y)+\cdots\bigg)+\bigg(b\Big(\frac{y}{a}\Big)^{\frac{2n-1}{2n}}+b\Big(\frac{y}{a}\Big)^{\frac{2n-2}{2n}}e_+(y)+\cdots\bigg)\\+\cdots+q=y.
$$
The assumption $|e_+(y)|\ll y^\frac{1}{2n}$ implies now that
$$
e_+(y)=-\frac{b}{2an}+d_+(y)
$$
where $\lim_{y\to\infty} d_+(y)=0$.
Repeating this argument for $B(y)<0$, with
$$
a\big(B(y)\big)^{2n}+b\big(B(y)\big)^{2n-1}+\cdots+q=y,
$$
we obtain that 
$$
B(y)=-\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}+e_-(y), \quad |e_+(y)|\ll y^\frac{1}{2n}.
$$
and similarly we obtain that
$$
e_-(y)=-\frac{b}{2an}+d_-(y)
$$
where $\lim_{y\to\infty} d_-(y)=0$.
Hence
$$
A(y)+B(y)=-\frac{b}{na}+d_-(y)+d_+(y)\to -\frac{b}{na}.
$$
A: It is clear that $A(y), B(y) \sim \pm (y/a)^{1/2n}$, and the problem is to determine the lower order term in their expansion. Wlog suppose $B(y) < 0 < A(y)$. By the mean value theorem,
$$(A(y) - (y/a)^{1/2n}) \cdot F'(\xi) \sim y - F((y/a)^{1/2n}) \,,$$
for some $\xi \sim (y/a)^{1/2n}$.
The LHS is $\sim 2an \cdot (y/a)^{(2n-1)/2n}$ times what we are looking for, and the RHS is $b(y/a)^{(2n-1)/2n} (1 + o(1))$. We conclude that
$$A(y) - (y/a)^{1/2n} \sim \frac{b}{2an} \,.$$
Replacing $F(x)$ by $F(-x)$, we obtain
$$-B(y) - (y/a)^{1/2n} \sim \frac{-b}{2an} \,.$$
The conclusion follows.
A: Let's replace $A(y), B(y) $ by $A, B$ to simplify typing and let $A>0>B$ and we write $C=-B$ so that $C>0$. Then we have $$y=aA^{2n}+bA^{2n-1}+\dots$$ and $$y=aC^{2n}-bC^{2n-1}+\dots$$ From these equations we get $$y/A^{2n}\to a, y/C^{2n}\to a$$ so that $A/C\to 1$.
Subtracting these equations we get $$a(A^{2n}-C^{2n})+b(A^{2n-1}+C^{2n-1})+\dots=0$$ or $$a(A-C) (A^{2n-1}+A^{2n-2}C+\dots+C^{2n-1})+b(A^{2n-1}+C^{2n-1})+\dots =0$$ Dividing the above equation by $C^{2n-1}$ we get $$a(A-C)\{1+(A/C)+(A/C)^2+\dots+(A/C)^{2n-1}\} +b\{1+(A/C)^{2n-1}\} + \text{ (terms tending to zero)} =0$$ Letting $y\to \infty $ in above equation we get $$2na\lim_{y\to\infty} (A-C) +2b=0$$ or $$A-C\to-\frac{b} {na} $$ as $y\to\infty $.
One has to observe that when we divide by $C^{2n-1}$ the terms like $A^{r} / C^{2n-1}$ tend to $0$ for $r<2n-1$ because we can write it as $(A/C) ^r(C^r/C^{2n-1})$.

The argument above is entirely elementary and simple. We just have to understand that $A, C$ are functions of $y$ which are strictly increasing as $y\to\infty $ and $$A\to\infty, C\to\infty, A/C\to 1$$ as $y\to\infty $. Further they satisfy the relation $F(A) =y=F(-C) $.
In general most of the algebraic limits do not involve anything more algebraic manipulation. 
A: Proof: It is  sufficient to prove the result when $b=0$. Otherwise we set the  change  of  variable $x:=x-b/2na$ to obtain a polynomial with $b=0$.
Assume that $b=0$. Then for  every given   $\epsilon>0$ if  $x>0$ is  sufficiently large we have $F(x-\epsilon)<F(-x)<F(x+\epsilon)$. Now  apply intermediate value theorem. Putting $y=F(-x)$ we have $B(y)=-x$ now intermediate value theorem implies $x-\epsilon< A(y)<x+\epsilon$  thus  $-\epsilon<A(y)+B(y)<\epsilon$. Q.E.D

As we said in the  question, despite of  its  simplicity, this  limit played a  crucial role  to prove the main result of the paper( stability of the  homoclinic  loop under consideration of the paper). But this limit was not questioned any where, neither by the journal nor by defense committee, etc.

A: COMMENT.-Let $F_n(x)$ defined as in your problem
$$F_n(x)=ax^{2n}+bx^{2n-1}+cx^{2n-2}+dx^{2n-3}\ldots+px+q$$ Assuming $c\ne0$ one has 
$$F_n(x)=ax^{2n}+bx^{2n-1}\pm F_{n-1}(x)\hspace{10mm}(*)$$ (the sign $+$ when $c$ is positive and the sign $-$ when $c$ is negative).
Besides the property is easily verified for $n=1$. In fact $$F_1(x)=ax^2+bx+c\Rightarrow a(x_1^2-x_2^2)+b(x_1-x_2)=0\Rightarrow x_1+x_2=\frac{-b}{a}$$
Can you now apply induction in any way using the equation $(*)$?
