$\displaystyle |g(x)| \leq |f(x)|$ for polynomials I have got a polynomial:
$$f(x) = x^4 - 5x^2 + 5$$
And the condition for a polynomial $g(x)$:
$$\forall x \in \mathbb{R}, |g(x)|\leq |f(x)|$$
Prove that $f(x) = a \cdot g(x)$
It's quite easy to see that $f(x)$ has four real roots, with the general form:
$$x_{1,2,3,4} = \pm \sqrt{\frac{5 \pm \sqrt{5}}{2}}$$
where the two $\pm$ signs act independently.
And also, replacing $x = x_1, x_2, x_3, x_4$ in the inequality, we can spot that $x_1, x_2, x_3, x_4$ are also roots of $g(x)$. I know that it's enough to prove that $\deg g(x) = \deg f(x)$, because if this is true, both polynomials could be written as:
$$\alpha(x-x_1)(x-x_2)(x-x_3)(x-x_4)$$
And also, form the condition it follows from $|g(x)|\leq |f(x)|$ that:
$$-1 \leq \frac{\alpha_f}{\alpha_g} \leq 1$$
because the graph of $|g(x)|$ should be under the graph of $|f(x)|$.
 A: Claim:
If $q(x)$ is a non-constant polynomial, then there is a real number $x_0$ such that $|q(x_0)|>1.$
Proof:
If $$q(x)=a_nx^n+a_{n-1}x^{n-1}+\dots +a_0$$ with $a_n\neq 0$ and $n>0.$
Then $$\frac{q(x)}{a_nx^n}=1+\frac{a_{n-1}}{a_nx^1}+\frac{a_{n-2}}{a_nx^2}+\cdots+\frac{a_0}{a_nx^n}\tag{1}$$
Then for any $$x>x_1=\max\{1\}\cup\left\{\left|2n\frac{a_{n-i}}{a_n}\right|\,\Big\vert\, i=1\dots,n\right\}$$
for $i=1,\dots,n.$
Then the $n$ non-constant terms of (1) are bounded below by $\frac{-1}{2n},$ and we get $$\left|\frac{a_{n-i}}{a_nx^i}\right|\leq\frac{1}{2n}$$
So $$\frac{q(x)}{a_nx^n}\geq 1-\frac{n}{2n}=\frac{1}{2}$$
Then if $$x_0>\max\left(x_1,\sqrt[n]{\frac{2}{|a_n|}}\right)$$
Then $$|q(x_0)|\geq \frac{|a_n|}{2}x_0^n>1$$

Claim: More generally, if $q(x)$ is a non-constant polynomial, then for any constant $C,$ there is some $x_0$ such that $|q(x)|>C$ for all $x>x_0.$
Proof:  Induction on the degree $n$ of $q.$
The case $n=1$ is easy.
If $q(x)$ is of degree $n+1,$ then $q(x)=xq_1(x)+a$ for some polynomial $q_1$ of degree $n$ and some real number $a.$
Then $\frac{q(x)}{x}=q_1(x)+\frac{a}{x},$ so: $$\left|{q(x)}\right|\geq |x||q_1(x)|-|a|.$$
Letting $C'=C+a,$ there is an $x_0'$ such that $|q_1(x)|>C+a$ for $x>x_0'.$ Let $x_0>\max(x_0',1).$ Then for $x>x_0,:$ $$|q(x)|> C+a-a=C.$$
